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Stat 110

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Ch 5+ch6

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Ch 5 Stat 110

0507017098/0580535304مـحـمد عـمـران ريـاضـيات واحـصاء للـمرحله الجـامعيه 1

Example For these exercises, state whether the variable is discrete or continuous.

1-the speed of a jet airplane. (continous)

2- The number of cheeseburgers a fast – food restaurant serves each day.

(discreter)

3- The number of people who play the state lottery each day.(discrete)

4- The weight of a Siberian tiger.(continuous)

5- The time it takes to complete a marathon. (continuous)

6- The number of mathematics major in your school.(discrete)

7-the blood pressures of all patients admitted to hospital on a specific day.

(continuous)

Probability distribution

-A random variable is a variable whose value are determined

by a chance

-Variable that can assume all values in the interval between

any two given value are called continuous variable. For

example if the temperature go from 𝟔𝟎° to𝟕𝟎°

-if a variable can assume only a specific number of value,

such as the outcomes for the roll of a die or the outcomes for

the toss of a coin, then the variable is called a discrete

variable.

Ch 5 Stat 110


Example

1)Construct a probability distribution for rolling a single die

Sol

S={𝟏, 𝟐, 𝟑, 𝟒, 𝟓, 𝟔}

𝒙 1 2 3 4 5 6

𝒑(𝒙) 1/6 1/6 1/6 1/6 1/6 1/6

2)Construct aprobability distribution for tossed acoin ,x is the

number of head

Sol

S={𝑯, 𝑻}

𝒙 (𝒕𝒉𝒆 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒉𝒆𝒂𝒅) 0 1

𝒑(𝒙) 1/2 1/2

3) construct a probability distribution for tossed two coin , x is

the number of head

Sol

S={𝑯𝑯, 𝑯𝑻, 𝑻𝑯, 𝑻𝑻}

(𝑿 𝒏𝒖𝒎𝒃𝒆𝒓 𝒐𝒇 𝒉𝒆𝒂𝒅) 0 1 2

𝒑(𝒙) 1/4 2/4 1/4

Ch 5 Stat 110


4)Represent graphically the probability for the sample space for

tossing three coin

Number of head X 0 1 2 3

Probability :p(x) 1/8 3/8 3/8 1/8

Sol

S={𝑯𝑯𝑯, 𝑯𝑯𝑻, 𝑯𝑻𝑯, 𝑯𝑻𝑯, 𝑻𝑯𝑯, 𝑻𝑯𝑻, 𝑻𝑻𝑯, 𝑻𝑻𝑻}

note

Two requirements for a probability distribution

1- The sum of the probabilities of all the events in the sample space must be equal 1 ∑ 𝑷(𝒙) = 𝟏

2-The probability of each event in the sample space must between or equal to 0 and 1

𝟎 ≤ 𝑷(𝒙) ≤ 𝟏

Ch 5 Stat 110


Example Determine whether each distribution is a probability distribution

1-

∑ 𝑷(𝑿) =𝟏

𝟓+

𝟏

𝟓+

𝟏

𝟓+

𝟏

𝟓+

𝟏

𝟓= صحيحه 𝟏

2-

خاطئه بها عدد سالب

3-

∑ 𝒑(𝒙) =𝟏

𝟒+

𝟏

𝟖+

𝟏

𝟏𝟔+

𝟗

𝟏𝟔= 𝟏

صحيحه

4-

∑ 𝒑(𝒙) = 𝟎. 𝟓 + 𝟎. 𝟑 + 𝟎. 𝟒 = 𝟏. 𝟐

خاطئه

X 0 5 10 15 20

P(x) 1/5 1/5 1/5 1/5 1/5

X 0 2 4 6

P(x) -1 1.5 0.3 0.2

X 1 2 3 4

P(x) 1/4 1/8 1/16 9/16

X 2 3 7

P(x) 0.5 0.3 0.4

Ch 5 Stat 110


Formula for the mean of a probability distribution

The mean of random variable with a discrete probability distribution

𝜇 = 𝑥1𝑝(𝑥1) + 𝑥2𝑝(𝑥2) + 𝑥3𝑝(𝑥3) + ⋯ … … … … … … … … . . +𝑥𝑛𝑝(𝑥𝑛)

𝜇 = ∑ 𝑥 . 𝑝(𝑥)

Mean variance, standard deviation and expectation

Formula for the variance of a probability distribution

𝜎2 = ∑[𝑥2. 𝑝(𝑥)] − 𝜇2

The standard deviation of probability distribution is

𝜎 = √𝜎2 Or √∑[𝑥2. 𝑝(𝑥) − 𝜇2]

The expected value:

𝜇 = 𝐸(𝑥) = ∑ 𝑥. 𝑝(𝑥)

Remember that the variance and standard deviation cannot be negative

Ch 5 Stat 110


Example 1) Find the mean variance and standard deviation for the probability

distribution for the number of spot when the die is tossed?

Sol

Probability distrbution

𝒙 1 2 3 4 5 6

𝒑(𝒙) 1/6 1/6 1/6 1/6 1/6 1/6

Sol

𝒎𝒆𝒂𝒏 𝝁 = ∑ 𝒙𝒑(𝒙)

= 𝟏 (𝟏

𝟔) + 𝟐 (

𝟏

𝟔) + 𝟑 (

𝟏

𝟔) + 𝟒 (

𝟏

𝟔) + 𝟓 (

𝟏

𝟔) + 𝟔 (

𝟏

𝟔) = 𝟑. 𝟓

∑ 𝒙𝟐𝒑(𝒙) = 𝟏𝟐 (𝟏

𝟔) + 𝟐𝟐 (

𝟏

𝟔) + 𝟑𝟐 (

𝟏

𝟔) + 𝟒𝟐 (

𝟏

𝟔) + 𝟓𝟐 (

𝟏

𝟔) + 𝟔𝟐 (

𝟏

𝟔)

= 𝟏𝟓. 𝟏𝟔𝟕

𝒗𝒂𝒓𝒊𝒂𝒏𝒄𝒆 = ∑ 𝒙𝟐𝒑(𝒙) − 𝝁𝟐 = 𝟏𝟓. 𝟏𝟔𝟕 − (𝟑. 𝟓)𝟐 = 𝟐. 𝟗𝟏𝟕

2)If a family with two children, find the mean of the number of children

who will be girl

Sol

𝑺 = {𝑩𝑩, 𝑩𝑮, 𝑮𝑩, 𝑮𝑮}

Probability distrbution

X 0 1 2

P(x) 1/4 2/4 1/4

𝒎𝒆𝒂𝒏 𝝁 = ∑ 𝒙𝒑(𝒙) = 𝟎 (𝟏

𝟒) + 𝟏 (

𝟐

𝟒) + 𝟐 (

𝟏

𝟒) = 𝟏

Ch 5 Stat 110


3)A pizza shop owner determine the number of pizza that are delivered

each day. Find the mean variance, and standard deviation for the

distribution shown. if the manger stated that 45 pizzas were delivered on

one day . Do you think that this is a believable claim?

𝒔𝒐𝒍

𝝁 = ∑ 𝒙. 𝒑(𝒙)

= 𝟑𝟓(𝟎. 𝟏) + 𝟑𝟔(𝟎. 𝟐) + 𝟑𝟕(𝟎. 𝟑) + 𝟑𝟖(𝟎. 𝟑)

+ 𝟑𝟗(𝟎. 𝟏) = 𝟑𝟕. 𝟏

∑ 𝒙𝟐. 𝒑(𝒙)

= (𝟑𝟓)𝟐(𝟎. 𝟏) + (𝟑𝟔)𝟐(𝟎. 𝟐) + (𝟑𝟕)𝟐(𝟎. 𝟑)

+ (𝟑𝟖)𝟐(𝟎. 𝟑) + (𝟑𝟗)𝟐(𝟎. 𝟏) = 𝟏𝟑𝟑𝟕𝟕. 𝟕

𝒗𝒂𝒓𝒊𝒂𝒎𝒄𝒆 𝝈𝟐 = 𝟏𝟑𝟑𝟕𝟕. 𝟕 − (𝟑𝟕. 𝟏)𝟐 = 𝟏. 𝟐𝟗

𝒔𝒕𝒂𝒏𝒅𝒓𝒂𝒅 𝒅𝒆𝒗𝒂𝒕𝒊𝒐𝒏 𝝈 = √𝟏. 𝟐𝟗 = 𝟏. 𝟏𝟑

𝒏𝒐 𝒃𝒆𝒄𝒂𝒖𝒔𝒆 𝑬(𝒙) = 𝝁 = 𝟑𝟕. 𝟏 < 𝟒𝟓

Number of

deliveries X

35 36 37 38 39

Probability : p(x) 0.1 0.2 0.3 0.3 0.1

Ch 5 Stat 110


4)The number of suits sold per day at a retail store is shown in the table,

with the corresponding probabilities. Find the mean, variance and

standard deviation of the distribution.

Number of suits

sold X

19 20 21 22 23

Probability: p(x) 0.2 0.2 0.2 0.3 0.1

If a manager of the retail store wants to be sure that he has enough

suits for the next 5 days, how many should the manager purchase?

𝒔𝒐𝒍

𝝁 = ∑ 𝒙. 𝒑(𝒙)

= 𝟏𝟗(𝟎. 𝟐) + 𝟐𝟎(𝟎. 𝟐) + 𝟐𝟏(𝟎. 𝟐) + 𝟐𝟐(𝟎. 𝟑) + 𝟐𝟑(𝟎. 𝟏)

= 𝟐𝟎. 𝟗

∑ 𝒙𝟐. 𝒑(𝒙)

= (𝟏𝟗)𝟐(𝟎. 𝟐) + (𝟐𝟎)𝟐(𝟎. 𝟐) + (𝟐𝟏)𝟐(𝟎. 𝟐) + (𝟐𝟐)𝟐(𝟎. 𝟑)

+ (𝟐𝟑)𝟐(𝟎. 𝟏) = 𝟒𝟑𝟖. 𝟓

𝒗𝒂𝒓𝒊𝒂𝒏𝒄𝒆 𝝈𝟐 = 𝟒𝟑𝟖. 𝟓 − (𝟐𝟎. 𝟗)𝟐 = 𝟏. 𝟔𝟗

𝒔𝒕𝒂𝒏𝒅𝒓𝒂𝒅 𝒅𝒆𝒗𝒂𝒕𝒊𝒐𝒏 𝝈 = √𝟏. 𝟔𝟗 = 𝟏. 𝟑

𝒔𝒖𝒊𝒕𝒔 = 𝟐𝟎. 𝟗(𝟓) = 𝟏𝟎𝟒. 𝟓 = 𝟏𝟎𝟒 𝒔𝒖𝒊𝒕𝒆

Ch 5 Stat 110


5)If three coins are tossed, find the mean variance of the number of the

heads that occur?

Sol

𝒔 = {𝑯𝑯𝑯, 𝑯𝑯𝑻, 𝑯𝑻𝑯, 𝑯𝑻𝑻, 𝑻𝑯𝑯, 𝑻𝑯𝑻, 𝑻𝑻𝑯, 𝑻𝑻𝑻}

X 0 1 2 3

P(x) 1/8 3/8 2/8 1/8

𝝁 = 𝟎 (𝟏

𝟖) + 𝟏 (

𝟑

𝟖) + 𝟐 (

𝟐

𝟖) + 𝟑 (

𝟏

𝟖) = 𝟏. 𝟏𝟐𝟓

∑ 𝒙𝟐. 𝒑(𝒙) = (𝟎)𝟐 (𝟏

𝟖) + (𝟏)𝟐 (

𝟑

𝟖) + (𝟐)𝟐 (

𝟐

𝟖) + (𝟑)𝟐 (

𝟏

𝟖) = 𝟐. 𝟑𝟕𝟓

𝒗𝒂𝒓𝒊𝒂𝒏𝒄𝒆 𝝈𝟐 = 𝟐. 𝟑𝟕𝟓 − (𝟏. 𝟏𝟐𝟓)𝟐 = 𝟏. 𝟏𝟏

𝒔𝒕𝒂𝒏𝒅𝒓𝒂𝒅 𝒅𝒆𝒗𝒂𝒕𝒊𝒐𝒏 𝝈 = √𝟏. 𝟏𝟏 = 𝟏. 𝟎𝟓

6)From the past experience, a company has found that in cartons of

transistor,

92% contain no defective transistor, 3%contain one defective transistor,

3%contain two defective transistor, and 2%contain three defective

transistor.

Find the mean, variance, and standard deviation.

Sol

X 0 1 2 3

P(x) 0.92 0.03 0.03 0.02

𝝁 = ∑ 𝒙. 𝒑(𝒙) = (𝟎)(𝟎. 𝟗𝟐) + 𝟏(𝟎. 𝟎𝟑) + 𝟐(𝟎. 𝟎𝟑) + 𝟑(𝟎. 𝟎𝟐) = 𝟎. 𝟏𝟓

∑ 𝒙𝟐𝒑(𝒙)

= (𝟎)𝟐(𝟎. 𝟗𝟐) + (𝟏)𝟐(𝟎. 𝟎𝟑) + (𝟐)𝟐(𝟎. 𝟎𝟑) + (𝟑)𝟐(𝟎. 𝟎𝟐) = 𝟎. 𝟑𝟑

𝒗𝒂𝒓𝒊𝒂𝒏𝒄𝒄𝒆 𝝈𝟐 = 𝟎. 𝟑𝟑 − (𝟎. 𝟏𝟓)𝟐 = 𝟎. 𝟑𝟎𝟕𝟓

𝒔𝒕𝒂𝒏𝒅𝒓𝒂𝒅 𝒅𝒆𝒗𝒂𝒕𝒊𝒐𝒏 𝝈 = √𝟎. 𝟑𝟎𝟕𝟓 = 𝟎. 𝟓𝟓𝟒

Ch 5 Stat 110


7)Five balls numbered 0,2,4,6, and8 are placed in a bag .after the balls are

mixed one is selected, its number is noted and then it is replaced many

times, find the variance and standard deviation of the numbers on the

balls?

Sol

X 0 2 4 6 8

P(x) 1/5 1/5 1/5 1/5 1/5

𝝁 = 𝟎 (𝟏

𝟓) + 𝟐 (

𝟏

𝟓) + 𝟒 (

𝟏

𝟓) + 𝟔 (

𝟏

𝟓) + 𝟖 (

𝟏

𝟓) = 𝟒

∑ 𝒙𝟐. 𝒑(𝒙)

= (𝟎)𝟐 (𝟏

𝟓) + (𝟐)𝟐 (

𝟏

𝟓) + (𝟒)𝟐 (

𝟏

𝟓) + (𝟔)𝟐 (

𝟏

𝟓)

+ (𝟖)𝟐 (𝟏

𝟓) = 𝟐𝟒

𝒗𝒂𝒓𝒊𝒂𝒏𝒄𝒆𝝈𝟐 = 𝟐𝟒 − 𝟒𝟐 = 𝟖

𝒔𝒕𝒂𝒏𝒅𝒓𝒂𝒅𝒅𝒆𝒗𝒂𝒕𝒊𝒐𝒏 𝝈 = √𝟖 = 𝟐. 𝟐𝟖𝟐𝟖

Ch 5 Stat 110


Example

1)One thousand tickets are sold at $1 each for a color television valued at

$350. What is the expected value of the gain if you purchase one ticket?

Sol

Win Lose

Gain x 𝟑𝟒𝟗$ −𝟏$

P(x) 1/1000 999/1000

𝐸(𝑥) = 349 (1

1000) + (−1) (

999

1000) = −0.65$

2)One thousand tickets are sold at $1 each for four prizes of $100, $50,

$25, and $10. After each prize drawing, the winning ticket is then

returned to the pool of tickets. What is the expected value if you

purchase two tickets?

Sol

Gain x 𝟗𝟖$ 𝟒𝟖$ 𝟐𝟑$ 𝟖$ −𝟐$

P(x) 2/1000 2/1000 2/1000 2/1000 992/1000

𝑬(𝒙) = 𝟗𝟖 (𝟐

𝟏𝟎𝟎𝟎) + 𝟒𝟖 (

𝟐

𝟏𝟎𝟎𝟎) + 𝟐𝟑 (

𝟐

𝟏𝟎𝟎𝟎) + 𝟖 (

𝟐

𝟏𝟎𝟎𝟎)

+ (−𝟐) (𝟗𝟗𝟐

𝟏𝟎𝟎𝟎) = −𝟏. 𝟔𝟑$

Ch 5 Stat 110


3)If a player rolls one die and when gets a number greater than

4, he wins 12$, the cost to play the game is 5$. What is the

expectation of the gain?

A) 2$ 𝑩) − 𝟏$ 𝑪) − 𝟐$ 𝑫)𝟏$

Sol

𝒈𝒓𝒆𝒂𝒕𝒆𝒓 𝒕𝒉𝒂𝒏 𝟒 = {𝟓, 𝟔} → 𝒑 =𝟐

𝟔=

𝟏

𝟑

𝑬(𝒙) = (𝟏

𝟑) (𝟏𝟐) − 𝟓 = −𝟏$

4)If X is a discrete random variable with ∑[𝒙𝟐. 𝒑(𝒙)] = 𝟔 𝒂𝒏𝒅 𝑬(𝒙) = 𝟐.

The variance for the probability distribution of X is

A)1.732 B)2 C )4 D)1.141

Sol

𝒗𝒂𝒓𝒊𝒂𝒏𝒄𝒆 𝝈𝟐 = ∑ 𝒙𝟐. 𝒑(𝒙) − 𝝁𝟐 = 𝟔 − 𝟐𝟐 = 𝟔 − 𝟒 = 𝟐

5)If X is a discrete random variable with ∑[𝒙𝟐 . 𝒑(𝒙) = 𝟒 𝒂𝒏𝒅 𝑬(𝒙) = −𝟐

the standard deviation for the probability distribution of X is

A)8 B)1.41 C)2.828 D)0

Sol

𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒅𝒆𝒗𝒂𝒕𝒊𝒐𝒏 𝝈 = √∑ 𝒙𝟐. 𝒑(𝒙) − 𝝁𝟐 = √𝟒 − (−𝟐)𝟐

= 𝟎

Ch 5 Stat 110


6)If X a discrete random variable with ∑[𝒙𝟐. 𝒑(𝒙)] = 𝟕 𝒂𝒏𝒅 𝝈𝟐 = 𝟐,

then the mean for the probability distribution of X is

A)2.24 B)5 C)1.141 D)2

Sol

𝝈𝟐 = ∑ 𝒙𝟐. 𝒑(𝒙) − 𝝁𝟐

𝝁𝟐 = ∑ 𝒙𝟐𝒑(𝒙) − 𝝈𝟐

𝝁𝟐 = 𝟕 − 𝟐 = 𝟓

𝝁 = √𝟓 = 𝟐. 𝟐𝟒

7)Find the mean of the distribution shown

X 1 2

P(x) 0.40 0.60

A)1.60 B)0.87 C)1.09 D)1.33

Sol

𝝁 = 𝟏(𝟎. 𝟒𝟎) + 𝟐(𝟎. 𝟔𝟎) = 𝟏. 𝟔𝟎

8)What the sample size for the following probability distribution?

X 1 3 5 7 9

P(x) 1/7 1/7 2/7 2/7 1/7

A)it cannot be determined B)25 C)5 D)1

Ch 5 Stat 110


The Binomial distribution

A binomial experiment is a probability experiment

that satisfies the following four requirements

1-there must be a fixed number of trial

2-each trial has only two outcomes: success or fail

3-the outcomes of each trial must be independent of

each other.

4-the probability of a success must remain the same

for each trial.

Note Two out comes yes or no → (binomial )

More than two outcomes→ (not binomial)

Ch 5 Stat 110


Which of the following are binomial experiments or can

reduced to binomial experiments?

1- Surveying 100people to determine if they like study

soap(binomail)

2- Tossing a coin 100 times to see how many heads

occur(binomail)

3- Asking 1000 people which brand of cigarettes they smoke.

(not bonomail)

4- Testing one brand of aspirin by 10 people to determine

whether it is effective (not binomail)

5- Asking 100 people if they smoke. (binomial)

6- Checking 1000 applicants to see how many different

crimes they were convicted of .(not binomail)

7- Surveying 300 prisoners to see whether this is their first

offense. (binomail)

Ch 5 Stat 110


Notation for the binomial distribution

The symbol for the probability of success p(s).

The symbol for the probability of failure p(f).

The numerical probability of success is P.

The numerical probability of failure is q.

P(s)=1-p(f) P= 1-q

N the number of trials

X number of success

The Binomial distribution In a binomial experiment the probability X success in n trial is

𝒑(𝒙) =𝒏!

(𝒏 − 𝒙)! 𝒙! 𝒑𝒙 𝒒𝒏−𝒙

𝒑(𝒙) = 𝒏𝑪𝒙 𝒑𝒙 𝒒𝒏−𝒙

Ch 5 Stat 110


Example 1)A coin is tossed 3 times .find the probability of getting exactly 2 head

Sol

𝒏 = 𝟑 𝒑 =𝟏

𝟐 𝒒 = 𝟏 −

𝟏

𝟐=

𝟏

𝟐 𝒙 = 𝟐

𝒑(𝒙 = 𝟐) = 𝟑𝑪𝟐(𝟏

𝟐)𝟐(

𝟏

𝟐)𝟑−𝟐 =

𝟑

𝟖= 𝟎. 𝟑𝟕𝟓

2)A survey found that one out of five American says he or she visited a

doctor in any given month. If 10 people are selected at random. Find the

probability that exactly 3 would have visited a doctor last month ?

Sol

𝒏 = 𝟏𝟎 𝒑 =𝟏

𝟓 𝒒 = 𝟏 −

𝟏

𝟓=

𝟒

𝟓 𝒙 = 𝟑

𝒑(𝒙 = 𝟑) = 𝟏𝟎𝑪𝟑(𝟏

𝟓)𝟑(

𝟒

𝟓)𝟏𝟎−𝟑 = 𝟎. 𝟐𝟎𝟏𝟑

Ch 5 Stat 110


3)A survey from teenage researcher unlimited found that 30%of teenage

consumer receive their spending money from part-time jobs. If 5 teenager

are selected at random. Find the probability that at least 3 of them will

have part time-jobs.

Sol

𝒏 = 𝟓 𝒑 = 𝟎. 𝟑𝟎 𝒒 = 𝟏 − 𝟎. 𝟑𝟎 = 𝟎. 𝟕𝟎

𝒑(𝒂𝒍 𝒍𝒆𝒂𝒔𝒕 𝟑) = 𝒑(𝒙 = 𝟒) + 𝒑(𝒙 = 𝟓)

= 𝟓𝒄𝟒(𝟎. 𝟑𝟎)𝟒(𝟎. 𝟕𝟎)𝟓−𝟒 + 𝟓𝒄𝟓(𝟎. 𝟑𝟎)𝟓(𝟎. 𝟕𝟎)𝟓−𝟓

= 𝟎. 𝟎𝟑𝟎𝟕𝟖

4)A student takes a 20- question , true/false exam and guesses on each

question .find the probability of passing if the lowest passing grade is 15

correct out 20. Would you consider this event likely to occur? Explain

why?

Sol

𝒏 = 𝟐𝟎 𝒑 =𝟏

𝟐 𝒒 = 𝟏 −

𝟏

𝟐=

𝟏

𝟐

𝒑(𝒙 ≥ 𝟏𝟓)

= 𝒑(𝒙 = 𝟏𝟓) + 𝒑(𝒙 = 𝟏𝟔) + 𝒑(𝒙 = 𝟏𝟕)

+ 𝒑(𝒙 = 𝟏𝟖) + 𝒑(𝒙 = 𝟏𝟗) + 𝒑(𝒙 = 𝟐𝟎)

= 𝟐𝟎𝑪𝟏𝟓(𝟏

𝟐)𝟏𝟓(

𝟏

𝟐)𝟐𝟎−𝟏𝟓 + 𝟐𝟎𝑪𝟏𝟔(

𝟏

𝟐)𝟏𝟔(

𝟏

𝟐)𝟐𝟎−𝟏𝟔

+ 𝟐𝟎𝑪𝟏𝟕(𝟏

𝟐)𝟏𝟕(

𝟏

𝟐)𝟐𝟎−𝟏𝟕 + 𝟐𝟎𝑪𝟏𝟖(

𝟏

𝟐)𝟏𝟖(

𝟏

𝟖)𝟐𝟎−𝟏𝟖

+ 𝟐𝟎𝑪𝟏𝟗(𝟏

𝟐)𝟏𝟗(

𝟏

𝟐)𝟐𝟎−𝟏𝟗 + 𝟐𝟎𝑪𝟐𝟎(

𝟏

𝟐)𝟐𝟎(

𝟏

𝟐)𝟐𝟎−𝟐𝟎

= 𝟎. 𝟎𝟐𝟏 < 𝟎. 𝟓 𝑵𝑶𝑻 𝑳𝑰𝑲𝑬𝑳𝒀

Ch 5 Stat 110


5)A survey found that 86%of Americans have never been a victim of

violent crime. If a sample of 12 Americans is selected at random, find the

probability that 10 or more have never been victims of violent crime. Does

it seem reasonable that 10 or more have never been victims of violent

crime?

SOL

𝒏 = 𝟏𝟐 𝒑 = 𝟎. 𝟖𝟔 𝒒 = 𝟏 − 𝟎. 𝟖𝟔 = 𝟎. 𝟏𝟒

𝒑(𝒙 ≥ 𝟏𝟎) = 𝒑(𝒙 = 𝟏𝟎) + 𝒑(𝒙 = 𝟏𝟏) + 𝒑(𝒙 = 𝟏𝟐)

= 𝟏𝟐𝑪𝟏𝟎(𝟎. 𝟖𝟔)𝟏𝟎(𝟎. 𝟏𝟒)𝟏𝟐−𝟏𝟎

+ 𝟏𝟐𝑪𝟏𝟏(𝟎. 𝟖𝟔)𝟏𝟏(𝟎. 𝟏𝟒)𝟏𝟐−𝟏𝟎

+ 𝟏𝟐𝑪𝟏𝟐(𝟎. 𝟖𝟔)𝟏𝟐(𝟎. 𝟏𝟒)𝟏𝟐−𝟏𝟐 = 𝟎. 𝟕𝟕

> 𝟎. 𝟓 𝒊𝒕 𝒔𝒆𝒆𝒎 𝒓𝒆𝒂𝒔𝒐𝒏𝒂𝒃𝒍𝒆

Ch 5 Stat 110


7)If 80%of the people in a community have internet access from

their home find these probabilities for a sample of 10 people

1- At most 6 have internet access.

2- Exactly 6 have internet access.

3- At least 6 have internet access.

4- Which event 1, 2.and 3 are most likely to occur?explain why?

Sol

𝒏 = 𝟏𝟎 𝒑 = 𝟎. 𝟖𝟎 𝒒 = 𝟏 − 𝟎. 𝟖𝟎 = 𝟎. 𝟐𝟎

𝟏) 𝒑(𝒂𝒕 𝒎𝒐𝒔𝒕 𝟔) = 𝒑(𝒙 = 𝟎) + 𝒑(𝒙 = 𝟏) + 𝒑(𝒙 = 𝟐) + 𝒑(𝒙

= 𝟑) + 𝒑(𝒙 = 𝟒) + 𝒑(𝒙 = 𝟓) + 𝒑(𝒙 = 𝟔)

= 𝟏𝟎𝒄𝟎(𝟎. 𝟖𝟎)𝟎(𝟎. 𝟐𝟎)𝟏𝟎 + 𝟏𝟎𝒄𝟏(𝟎. 𝟖𝟎)𝟏(𝟎. 𝟐𝟎)𝟗

+ 𝟏𝟎𝒄𝟐(𝟎. 𝟖𝟎)𝟐(𝟎. 𝟐𝟎)𝟖 + 𝟏𝟎𝒄𝟑(𝟎. 𝟖𝟎)𝟑(𝟎. 𝟐𝟎)𝟕

+ 𝟏𝟎𝒄𝟒(𝟎. 𝟖𝟎)𝟒(𝟎. 𝟐𝟎)𝟔 + 𝟏𝟎𝒄𝟓(𝟎. 𝟖𝟎)𝟓(𝟎. 𝟐𝟎)𝟓

+ 𝟏𝟎𝒄𝟔(𝟎. 𝟖𝟎)𝟔(𝟎. 𝟐𝟎)𝟒 = 𝟎. 𝟏𝟐𝟏

𝟐)𝒑(𝒙 = 𝟔) = 𝟏𝟎𝒄𝟔(𝟎. 𝟖𝟎)𝟔(𝟎. 𝟐𝟎)𝟒 = 𝟎. 𝟎𝟖𝟖 )

𝟑)𝒑(𝒂𝒕 𝒍𝒆𝒂𝒔𝒕 𝟔)

= 𝒑(𝒙 = 𝟔) + 𝒑(𝒙 = 𝟕) + 𝒑(𝒙 = 𝟖) + 𝒑(𝒙 = 𝟗)

+ 𝒑(𝒙 = 𝟏𝟎)

= 𝟏𝟎𝒄𝟔(𝟎. 𝟖𝟎)𝟔(𝟎. 𝟐𝟎)𝟒 + 𝟏𝟎𝒄𝟕(𝟎. 𝟖𝟎)𝟕(𝟎. 𝟐𝟎)𝟑

+ 𝟏𝟎𝒄𝟖(𝟎. 𝟖𝟎)𝟖(𝟎. 𝟐𝟎)𝟐 + 𝟏𝟎𝒄𝟗(𝟎. 𝟖𝟎)𝟗(𝟎. 𝟐𝟎)

+ 𝟏𝟎𝒄𝟏𝟎(𝟎. 𝟖𝟎)𝟏𝟎(𝟎. 𝟐𝟎)𝟎 = 𝟎. 𝟗𝟔𝟕

𝟒)𝟑 𝒊𝒔 𝒎𝒐𝒔𝒕 𝒍𝒊𝒌𝒆𝒍𝒚

Ch 5 Stat 110


Example 1)A dice is rolled 480 times. Find the mean, variance, and standard

deviation of the number of 2 that will be rolled?

Sol

𝒏 = 𝟒𝟖𝟎 𝒑 =𝟏

𝟔 𝒒 = 𝟏 −

𝟏

𝟔=

𝟓

𝟔

𝝁 = 𝒏. 𝒑 = 𝟒𝟖𝟎 (𝟏

𝟔) = 𝟖𝟎

𝝈𝟐 = 𝒏. 𝒑. 𝒒 = 𝟒𝟖𝟎 (𝟏

𝟔) (

𝟓

𝟔) = 𝟔𝟔. 𝟕

𝝈 = √𝟔𝟔. 𝟕 = 𝟖. 𝟐

Mean , variance , and standard deviation for the binomial distribution

𝒎𝒆𝒂𝒏 ∶ 𝝁 = 𝒏. 𝒑

𝒗𝒂𝒓𝒊𝒂𝒏𝒄𝒆 ∶ 𝝈𝟐 = 𝒏. 𝒑 . 𝒒

𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏 ∶

𝝈 = √𝒏. 𝒑. 𝒒

Ch 5 Stat 110


2)A coin is tossed 4 times. Find the mean , variance ,and standard deviation

of the number of head that will be obtained?

Sol

𝒏 = 𝟒 𝒑 =𝟏

𝟐 𝒒 =

𝟏

𝟐

𝝁 = 𝒏. 𝒑 = 𝟒 (𝟏

𝟐) = 𝟐

𝝈𝟐 = 𝒏. 𝒑. 𝒒 = 𝟒 (𝟏

𝟐) (

𝟏

𝟐) = 𝟏

𝝈 = √𝟏 = 𝟏

3)If 3% of calculator are defective. Find the mean, variance, and

standard deviation of a lot of 300 calculator.

Sol

𝒏 = 𝟑𝟎𝟎 𝒑 = 𝟎. 𝟎𝟑 𝒒 = 𝟏 − 𝟎. 𝟎𝟑 = 𝟎. 𝟗𝟕

𝝁 = 𝒏. 𝒑 = 𝟑𝟎𝟎(𝟎. 𝟎𝟑) = 𝟗

𝝈𝟐 = 𝒏. 𝒑. 𝒒 = 𝟑𝟎𝟎(𝟎. 𝟎𝟑)(𝟎. 𝟕𝟗) = 𝟖. 𝟕

𝝈 = √𝟖. 𝟕 = 𝟐. 𝟗

Ch 5 Stat 110


4)In a restaurant, a study found that 42% of all patrons smoked. If the

seating capacity of the restaurant is 80 people, find the mean, variance,

and standard deviation of the number smokers. About how many seats

should be available. For smoking customers?

Sol

𝒏 = 𝟖𝟎 𝒑 = 𝟎. 𝟒𝟐 𝒒 = 𝟏 − 𝟎. 𝟒𝟐 = 𝟎. 𝟓𝟖

𝝁 = 𝒏. 𝒑 = 𝟖𝟎(𝟎. 𝟒𝟐) = 𝟑𝟑. 𝟔

𝝈𝟐 = 𝒏. 𝒑. 𝒒 = 𝟖𝟎(𝟎. 𝟒𝟐)(𝟎. 𝟓𝟖) = 𝟏𝟗. 𝟒𝟖𝟖

𝝈 = √𝟏𝟗. 𝟒𝟖𝟖 = 𝟒. 𝟒𝟏𝟓

Ch 5 Stat 110


1)A die is rolled 5 times .the probability of getting 4 one time only is …

A) 0.402 B)0.167 C)0.015 D)0.386

Sol

𝒏 = 𝟓 𝒑 =𝟏

𝟔 𝒒 = 𝟏 −

𝟏

𝟔=

𝟓

𝟔

𝒑(𝒙 = 𝟏) = 𝟓𝑪𝟏(𝟏

𝟔)𝟏(

𝟓

𝟔)𝟒 = 𝟎. 𝟒𝟎𝟐

2)A die rolled 5 times. The probability of getting a number 5 exactly

two times only is

A) 0.598 B)0.161 C)0.839 D)0.402

𝑺𝑶𝑳

𝒏 = 𝟓 𝒑 =𝟏

𝟔 𝒒 = 𝟏 −

𝟏

𝟔=

𝟓

𝟔

𝒑(𝒙 = 𝟐) = 𝟓𝒄𝟐(𝟏

𝟔)𝟐(

𝟓

𝟔)𝟑 = 𝟎. 𝟏𝟔𝟏

Ch 5 Stat 110


3)A student takes 7-question multiple-choice quiz with 4 choices for

each question .if the student guesses at random on each question. What

is the probability that the student gets exactly 3 questions correct?

A) 0.130 B)0.345 C)0.173 D)0.043

Sol

𝒏 = 𝟕 𝒑 =𝟏

𝟒 𝒒 = 𝟏 −

𝟏

𝟒=

𝟑

𝟒

𝒑(𝒙 = 𝟑) = 𝟕𝑪𝟑(𝟏

𝟒)𝟑(

𝟑

𝟒)𝟒 = 𝟎. 𝟏𝟕𝟑

4)A study shows that 70% of drivers consider themselves above average

in driving ability . if 10 drive at random are chosen , what is the mean and

variance of the number drivers who consider themselves above average?

A) Mean=7 and variance =7

B) Mean =10 and variance=1.45

C) Mean =7 and variance =2.1

D) Mean =10 and variance =10

Sol

𝒏 = 𝟏𝟎 𝒑 = 𝟎. 𝟕𝟎 𝒒 = 𝟏 − 𝟎. 𝟕𝟎 = 𝟎. 𝟑𝟎

𝝁 = 𝒏. 𝒑 = 𝟏𝟎(𝟎. 𝟕𝟎) = 𝟕

𝝈𝟐 = 𝒏. 𝒑. 𝒒 = 𝟏𝟎(𝟎. 𝟕𝟎)(𝟎. 𝟑𝟎) = 𝟐. 𝟏

Ch 5 Stat 110


5)The outcomes of each trial in a binomial experiment

A)are unlimited B) are independent C)are dependent D)must be fixed

6)The number of trial in binomial experiment ………..

A)are independent B)are dependent C)must be fixed D) are unlimited

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Stat 110 Ch 6

0580535304-0507017098 ران ــد عـمــمـحـمـ 1

The normal distribution

* When the data values are evenly distributed about the mean. A

distribution is said to be asymmetric distribution. (a normal distribution's

symmetric)

*when the majority of the data values fall to the right of the mean , the

distribution is said to be a negatively or left skewed distribution

*when the majority of the data values fall to the left of the mean, a

distribution is said to be a positively or right skewed distribution.

Stat 110 Ch 6


Properties of a normal distribution A normal distribution a continuous, symmetric, bell-shaped

distribution of a variable

Properties

1- A normal distribution curve is bell-shaped

2- The mean ,median and mode are equal and are located at the center

of the distribution

3- A normal distribution curve is unimodal (it has only one mode)

4- The curve is symmetric about the mean

5- The curve is continuous, that is , there are no gaps or holes.

6- The curve never touches the X axis.

7- The total area under a normal distribution curve is equal to 1 or 100%

8- The area under the part of a normal curve that lies within 1 standard

deviation of the mean is approximately 0.68 or 68%. Within 2

standard deviation, about 0.95 or 95% and within 3 standard

devotions about 0.977 or 99.7%

Stat 110 Ch 6


ሺ𝑧 𝑡𝑎𝑏𝑙𝑒 ሻمالحظات هامه لقراءه جدول

( اذا كان فالقراءه مباشره من الجدول االفقى هو 1

الجزء من مائه والراسى هوالعدد الصحيح والجزء العشر

( اذا كان فتتحول الى 2

قراءه الجدول -1وتصبح

الناتج ( اذا كانت تتحول الى ف3

قراءه الجدول للعدد الصغير (–هو )قراءه الجدول للعدد الكبير

𝒑ሺ𝒛 < 𝒂ሻ

𝒑ሺ𝒛 > 𝒂ሻ 𝟏 − 𝒑ሺ𝒛 < 𝒂ሻ

𝒑ሺ𝒂 < 𝒛 < 𝒃ሻ 𝒑ሺ𝒛 < 𝒃ሻ − 𝒑ሺ𝒛 < 𝒂ሻ

Stat 110 Ch 6


Example 1)Find the area under the standard normal distribution curve between

Z=0 and Z=2.34

Sol

𝒑ሺ𝟎 < 𝒛 < 𝟐. 𝟑𝟒ሻ = 𝒑ሺ𝒛 < 𝟐. 𝟑𝟒ሻ − 𝒑ሺ𝒛 < 𝟎ሻ = 𝟎. 𝟗𝟗𝟎𝟒 − 𝟎. 𝟓

= 𝟎. 𝟒𝟗𝟎𝟒

2)Find the area to the lift of Z =-1.93

Sol

𝒑ሺ𝒛 < −𝟏. 𝟗𝟑ሻ = 𝟎. من الجدول مباشره 𝟎𝟐𝟔𝟖

Stat 110 Ch 6


3)Find the area between

a)Z=0 and Z =-1.69

Sol

𝒑ሺ−𝟏. 𝟗𝟔 < 𝒛 < 𝟎ሻ = 𝒑ሺ𝒛 < 𝟎ሻ − 𝒑ሺ𝒛 < −𝟏. 𝟗𝟔ሻ = 𝟎. 𝟓 − 𝟎. 𝟐𝟓𝟎

= 𝟎. 𝟒𝟕𝟓𝟎

b)Z=1.69 and Z =2.24

Sol

𝒑ሺ𝟏. 𝟔𝟗 < 𝒛 < 𝟐. 𝟐𝟒ሻ = 𝒑ሺ𝒛 < 𝟐. 𝟐𝟒ሻ − 𝒑ሺ𝒛 < 𝟏. 𝟔𝟗ሻ

= 𝟎. 𝟗𝟖𝟕𝟓 − 𝟎. 𝟗𝟓𝟒𝟓 = 𝟎. 𝟎𝟑𝟑𝟎

Stat 110 Ch 6


4)Find the area to the right of Z =1.69

Sol

𝒑ሺ𝒛 > 𝟏. 𝟔𝟗ሻ = 𝟏 − 𝒑ሺ𝒛 < 𝟏. 𝟔𝟗ሻ = 𝟏 − 𝟎. 𝟗𝟓𝟒𝟓 = 𝟎. 𝟎𝟒𝟓𝟓

Stat 110 Ch 6


Sol

a)

b)

c)

𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒑𝒓𝒐𝒃𝒂𝒃𝒊𝒍𝒊𝒕𝒚 𝒇𝒐𝒓𝒆𝒂𝒄𝒉

𝒂ሻ 𝒑ሺ𝟎 < 𝒛 < 𝟐. 𝟑𝟐ሻ

𝒃ሻ 𝒑ሺ𝒛 < 𝟏. 𝟔𝟓ሻ

𝒄ሻ 𝒑 ሺ𝒛 > 𝟏. 𝟗𝟏ሻ

𝒑ሺ𝟎 < 𝒛 < 𝟐. 𝟑𝟐ሻ

= 𝒑ሺ𝒛 < 𝟐. 𝟑𝟐ሻ − 𝒑ሺ𝒛 < 𝟎ሻ

= 𝟎. 𝟗𝟗𝟎𝟏 − 𝟎. 𝟓 = 𝟎. 𝟒𝟗𝟎𝟏

𝒑ሺ𝒛 < 𝟏. 𝟔𝟓ሻ = 𝟎. 𝟗𝟓𝟎𝟓

𝒑ሺ𝒛 > 𝟏. 𝟗𝟏ሻ = 𝟏 − 𝒑ሺ𝒛 < 𝟏. 𝟗𝟏ሻ = 𝟏 − 𝟎. 𝟗𝟕𝟏𝟗 = 𝟎. 𝟎𝟐𝟖𝟏

Stat 110 Ch 6


Sol

Z is postive عدد موجب

𝒑ሺ𝟎 < 𝒙 < 𝒛ሻ = 𝟎. 𝟐𝟏𝟐𝟑

𝒑ሺ𝒙 < 𝒛ሻ − 𝒑ሺ𝒙 < 𝟎ሻ = 𝟎. 𝟐𝟏𝟐𝟑

𝒑ሺ𝒙 < 𝒛ሻ − 𝟎. 𝟓 = 𝟎. 𝟐𝟏𝟐𝟑

𝒑ሺ𝒙 < 𝒛ሻ = 𝟎. 𝟐𝟏𝟐𝟑 + 𝟎. 𝟓 = 𝟎. 𝟕𝟏𝟐𝟑

ومن ثم نذهب للجدول للبحث العدد المقابل للقراءه

𝒛 = 𝟎. 𝟓𝟔

𝒛 𝒊𝒔 𝒏𝒆𝒈𝒂𝒕𝒊𝒗𝒆 عدد سالب

𝒑ሺ𝒛 < 𝒙 < 𝟎ሻ = 𝒑ሺ𝒙 < 𝟎ሻ − 𝒑ሺ𝒙 < 𝒛ሻ = 𝟎. 𝟐𝟏𝟐𝟑

− 𝒑ሺ𝒙 < 𝒛ሻ = 𝟎. 𝟐𝟏𝟑 − 𝒑ሺ𝒙 < 𝟎ሻ

−𝒑ሺ𝒙 < 𝒛ሻ = 𝟎. 𝟐𝟏𝟐𝟑 − 𝟎. 𝟓

−𝒑ሺ𝒙 < 𝒛ሻ = −𝟎. 𝟐𝟖𝟕𝟕 𝒑ሺ𝒙 < 𝒛ሻ = 𝟎. 𝟐𝟖𝟖𝟕

𝟎. ومن ثم نذهب للبحث العدد المقابل للقراءه 𝟐𝟖𝟖𝟕

𝒛 = −𝟎. 𝟓𝟔

𝒛 = ±𝟎. 𝟓𝟔

𝒇𝒊𝒏𝒅 𝒕𝒉𝒆 𝒛 𝒗𝒂𝒍𝒖𝒆 𝒔𝒖𝒄𝒉 𝒕𝒉𝒂𝒕 𝒕𝒉𝒆 𝒂𝒓𝒆𝒂 𝒖𝒏𝒅𝒆𝒓 𝒕𝒉𝒆 𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅

𝒏𝒐𝒓𝒎𝒂𝒍 𝒅𝒊𝒔𝒕𝒓𝒃𝒖𝒕𝒊𝒐𝒏 𝒄𝒖𝒓𝒗𝒆 𝒃𝒆𝒕𝒘𝒆𝒆𝒏 𝟎 𝒂𝒏𝒅 𝒛 𝒗𝒂𝒍𝒖𝒆

𝒊𝒔 𝟎. 𝟐𝟏𝟐𝟑

𝟎. 𝟕𝟏𝟐𝟑

Stat 110 Ch 6


Distribution of sample means

A sampling distribution of sample means is

A distribution obtained by using the means computed from random

samples of a specific taken from a population

Sampling error

Is the difference between the sample measure and corresponding

population measure because the sample is not a perfect representation

of the population

Applications of the normal distribution The standard normal distribution is normal distribution

with 𝝁 = 𝟎𝒂𝒏𝒅 𝝈 = 𝟏

To solve problems by using the standard normal

distribution, transform the original variable to a standard

normal distribution variable by using the formula

𝒛 =𝒗𝒂𝒍𝒖𝒆 − 𝒎𝒆𝒂𝒏

𝒔𝒕𝒂𝒏𝒅𝒂𝒓𝒅 𝒅𝒆𝒗𝒊𝒂𝒕𝒊𝒐𝒏 𝒐𝒓 𝒛 =

𝒙 − 𝝁

𝝈

𝒑ሺ𝒙 > 𝒙𝟎ሻ = 𝒑ሺ𝒛 >𝒙𝟎 − 𝝁

𝝈ሻ

𝒙 = 𝒛𝝈 + 𝝁

Stat 110 Ch 6


Example

1)The mean number of hours an American worker spends on the

computer is 3.1 hours per workday. Assume the standard deviation is

0.5 hour. Find the percentage of workers who spend less than 3.5

hours on the computer. Assume the variable is normally distribution.

𝑠𝑜𝑙

𝝁 = 𝟑. 𝟏 𝝈 = 𝟎. 𝟓

𝒑ሺ𝒙 < 𝟑. 𝟓ሻ = 𝒑 (𝒙 − 𝝁

𝝈<

𝟑. 𝟓 − 𝟑. 𝟏

𝟎. 𝟓)

= 𝒑ሺ𝒛 < 𝟐. 𝟐𝟔ሻ = 𝟎. 𝟗𝟖𝟖𝟏

Stat 110 Ch 6


Sol

𝝁 = 𝟏𝟒𝟔. 𝟐𝟏 𝝈 = 𝟐𝟗. 𝟒𝟒

𝒑ሺ𝒙 < 𝟏𝟔𝟎ሻ =

𝒑 (𝒙 − 𝝁

𝝈<

𝟏𝟔𝟎 − 𝟏𝟒𝟔. 𝟐𝟏

𝟐𝟗. 𝟒𝟒)

= 𝒑ሺ𝒛 < 𝟎. 𝟒𝟕ሻ

= 𝟎. 𝟔𝟖𝟎𝟖

2)A survey founded that woman spend on average $146.21on beauty product

during the summer month. Assume the standard deviation is $29.44. Find the

percentage of woman who spend less than $160.00

Stat 110 Ch 6


3)A survey found that people keep their microwave ovens an average of 3.2

years. The standard deviation is 0.56 year. If a person decides to buy anew

microwave oven.

Find the probability that he or she has owned the old oven for the following

amount of time. Assume the variable is normally distribution

a- Less than 1.5 years

b- Between 2 and 3 years

c- More than 3.2 years

Sol

𝝁 = 𝟑. 𝟐 𝝈 = 𝟎. 𝟓𝟔

𝒂ሻ𝒑ሺ𝒙 < 𝟏. 𝟓ሻ = 𝒑 (𝒙 − 𝝁

𝝈<

𝟏. 𝟓 − 𝟑. 𝟐

𝟎. 𝟓𝟔) = 𝒑ሺ𝒛 < −𝟑. 𝟎𝟒ሻ = 𝟎. 𝟎𝟎𝟏𝟐

𝒃ሻ𝒑ሺ𝟐 < 𝒙 < 𝟑ሻ = 𝒑 (𝟐 − 𝟑. 𝟐

𝟎. 𝟓𝟔<

𝒙 − 𝝁

𝝈<

𝟑 − 𝟑. 𝟐

𝟎. 𝟓𝟔)

= 𝒑ሺ−𝟐. 𝟏𝟒 < 𝒛 < −𝟎. 𝟑𝟔ሻ

= 𝒑ሺ𝒛 < −𝟎. 𝟑𝟔ሻ − 𝒑ሺ𝒛 < −𝟐. 𝟏𝟒ሻ = 𝟎. 𝟑𝟓𝟗𝟒 − 𝟎. 𝟎𝟏𝟔𝟐

= 𝟎. 𝟑𝟒𝟑𝟐

𝒄ሻ𝒑ሺ𝒙 > 𝟑. 𝟐ሻ = 𝒑ሺ𝒙 − 𝝁

𝝈>

𝟑. 𝟐 − 𝟑. 𝟐

𝟎. 𝟓𝟔ሻ = 𝒑ሺ𝒛 > 𝟎ሻ = 𝟏 − 𝒑ሺ𝒛 < 𝟎ሻ

= 𝟏 − 𝟎. 𝟓 = 𝟎. 𝟓

Stat 110 Ch 6


Sol

𝝁 = 𝟐𝟖 𝝈 = 𝟐

𝒂ሻ 𝒑ሺ𝟐𝟕 < 𝒙 < 𝟑𝟏ሻ = 𝒑 (𝟐𝟕 − 𝟐𝟖

𝟐<

𝒙 − 𝝁

𝝈<

𝟑𝟏 − 𝟐𝟖

𝟐)

= 𝒑ሺ−𝟎. 𝟓 < 𝒛 < 𝟏. 𝟓ሻ = 𝒑ሺ𝒛 < 𝟏. 𝟓ሻ − 𝒑ሺ𝒛 < −𝟎. 𝟓ሻ

= 𝟎. 𝟗𝟑𝟑𝟐 − 𝟎. 𝟑𝟎𝟖𝟓 = 𝟎. 𝟔𝟐𝟒𝟕

𝒃ሻ 𝒑ሺ𝒙 > 𝟑𝟎. 𝟐ሻ = 𝒑ሺ𝒙 − 𝝁

𝝈>

𝟑𝟎. 𝟐 − 𝟐𝟖

𝟐ሻ

= 𝒑ሺ𝒛 > 𝟏. 𝟏ሻ = 𝟏 − 𝒑ሺ𝒛 < 𝟏. 𝟏ሻ = 𝟏 − 𝟎. 𝟖𝟔𝟒𝟑 = 𝟎. 𝟏𝟑𝟓𝟕

4)Each month an American house hold(مجلس النواب ) an average of 28 pound

of newspaper for garbage or recycling. Assume the standard divination is 2

pound if a household selected randomly fined probability

a) Between 27 and 31 pound per month

b) More than 30.2 pound per month

Stat 110 Ch 6


5)The average time for a mail carrier to cover his route is 380 minutes,

and the standard deviation is 16 minutes. If one of these trips is

selected at random, find the probability that the carrier will have the

following route time. Assume the variable is normally distribution.

a- At most 350 minutes

b- At least 395 minutes

Sol

𝝁 = 𝟑𝟖𝟎 𝝈 = 𝟏𝟔

𝒂ሻ 𝒑ሺ𝒙 < 𝟑𝟓𝟎ሻ

= 𝒑 (𝒙 − 𝝁

𝝈<

𝟑𝟓𝟎 − 𝟑𝟖𝟎

𝟏𝟔)

= 𝒑ሺ𝒛 < −𝟏. 𝟖𝟖ሻ = 𝟎. 𝟎𝟑𝟎𝟏

𝒃ሻ 𝒑ሺ𝒙 > 𝟑𝟗𝟓ሻ

= 𝒑 (𝒙 − 𝝁

𝝈>

𝟑𝟗𝟓 − 𝟑𝟖𝟎

𝟏𝟔)

= 𝒑ሺ𝒛 > 𝟎. 𝟗𝟒ሻ = 𝟏 − 𝒑ሺ𝒛 < 𝟎. 𝟗𝟒ሻ

= 𝟏 − 𝟎. 𝟖𝟐𝟔𝟒 = 𝟎. 𝟏𝟕𝟑𝟔

Stat 110 Ch 6


sol

𝝁 = 𝟏. 𝟔𝟒 𝝈 = 𝟐. 𝟒

𝒑ሺ𝒙 < 𝟏ሻ = 𝒑 (𝒙 − 𝝁

𝝈<

𝟏 − 𝟏. 𝟔𝟒

𝟐. 𝟒)

= 𝒑ሺ𝒛 < −𝟎. 𝟐𝟔𝟕ሻ = 𝟎. 𝟑𝟗𝟕𝟒

العدد الكلى ×المطلوب هو العدد =االحتمال

𝟎. 𝟗𝟑𝟕𝟒 × 𝟓𝟎𝟎 = 𝟏𝟗𝟖. 𝟕 ≈ 𝟏𝟗𝟗 𝒊𝒏𝒅𝒊𝒗𝒊𝒖𝒂𝒍

6)American consume on average 1.64cup of coffee per day. Assume the

variable is approximately normally distribution with standard deviation

2.4 cup. If 500 individual are selected. How many will drink less than 1

cup of coffee per day

Stat 110 Ch 6


Sol

𝝁 = 𝟐𝟎𝟎 𝝈 = 𝟐𝟎

𝒑ሺ𝒛 < 𝒛𝟏ሻ = 𝟎. 𝟗𝟎ሻ عكسه قراءه عكسيه فى الجدول

𝒛 = 𝟏. 𝟐𝟖

𝒕𝒉𝒆𝒏 𝒙 = 𝒛. 𝝈 + 𝝁

𝒙 = 𝟏. 𝟐𝟖ሺ𝟐𝟎ሻ + 𝟐𝟎𝟎 = 𝟐𝟐𝟓. 𝟔 ≈ 𝟐𝟐𝟔

7)To qualify for a police academy, candidates must score in the top 10%

on a general abilities test. The test has a mean of 200 and standard

deviation of 20 find the lowest possible score to qualify. Assume the test

score is normally distribution

Stat 110 Ch 6


Sol

𝝁 = 𝟏𝟐𝟎 𝝈 = 𝟖

𝒑ሺ𝒛 < 𝒛𝟏ሻ = 𝟎. قراءه عكسيه من الجدول 𝟖

𝒛𝟏 = 𝟎. 𝟖𝟒 𝒕𝒉𝒆𝒏𝒛𝟐 = −𝟎. 𝟖𝟒

𝒛𝟏 = 𝟎. 𝟖𝟒 𝒙 = 𝒛𝟏. 𝝈 + 𝝁 = ሺ𝟎. 𝟖𝟒ሻሺ𝟖ሻ + 𝟏𝟐𝟎 = 𝟏𝟐𝟔. 𝟕𝟐

𝒛𝟐 = −𝟎. 𝟖𝟒 𝒙 = 𝒛𝟐. 𝝈 + 𝝁 = ሺ−𝟎. 𝟖𝟒ሻሺ𝟖ሻ + 𝟏𝟐𝟎 = 𝟏𝟏𝟑. 𝟐𝟖

Upper =126.75

Lower=113.2

8)For a medical study , a researcher wishes to select people on the

middle 60% of the population based on blood pressure .if the mean of

systolic blood pressure is 120 and standard deviation is 8, find the upper

and lower reading that would qualify people to participate in the study .

Stat 110 Ch 6


9)If a one – person house hold spends an average of 40$ per week on

groceries, find the maximum and minimum dollar amounts spent per week

for the middle 50% of one-person households. Assume that the standard

deviation is 5$and the variable is normally distribution.

𝒔𝒐𝒍

𝝁 = 𝟒𝟎 𝝈 = 𝟓

𝒑ሺ𝒛𝟏 < 𝒛 < 𝒛𝟐ሻ = 𝟎. 𝟓𝟎

𝒑ሺ𝒛 < 𝒛𝟏ሻ = 𝟎. قراءه عكسيه من الجدول 𝟕𝟓

𝒛𝟏 = 𝟎. 𝟔𝟖 𝒕𝒉𝒆𝒏 𝒛𝟐 = −𝟎. 𝟔𝟖

𝒘𝒉𝒆𝒏

𝒛𝟏 = 𝟎. 𝟔𝟖 𝒙 = 𝒛𝟏. 𝝈 + 𝝁 = 𝟎. 𝟔𝟖 × 𝟓 + 𝟒𝟎 = 𝟒𝟑. 𝟒

𝒛𝟐 = −𝟎. 𝟔𝟖 𝒙 = ሺ−𝟎. 𝟔𝟖ሻ × 𝟓 + 𝟒𝟎 = 𝟑𝟔. 𝟔

𝒎𝒂𝒙 = 𝟒𝟑. 𝟒

𝒎𝒊𝒏 = 𝟑𝟔. 𝟔

Stat 110 Ch 6


10)The mean lifetime of a wristwatch is 25 months, with a standard

deviation of 5 months. If the distribution is normal.

For how many months should a guarantee be made if the manufacturer

does not want to exchange more than 10% of the watches? Assume the

variable is normally distribution.

Sol

𝝁 = 𝟐𝟓 𝝈 = 𝟓

𝒑ሺ𝒛 > 𝒛𝟏ሻ = 𝟎. 𝟏𝟎

𝟏 − 𝒑ሺ𝒛 < 𝒛𝟏ሻ = 𝟎. 𝟏

𝒑ሺ𝒛 < 𝒛𝟏ሻ = 𝟏 − 𝟎. 𝟏

𝒑ሺ𝒛 < 𝒛𝟏ሻ = 𝟎. قراءه عكسيه من الجدول 𝟗

𝒛𝟏 = 𝟏. 𝟐𝟖

𝒙 = 𝒛𝟏. 𝝈 + 𝝁

𝒙 = 𝟏. 𝟐𝟖 × 𝟓 + 𝟐𝟓 = 𝟑𝟏. 𝟒 ≈ 𝟑𝟏𝒎𝒐𝒏𝒂𝒕𝒉

Stat 110 Ch 6


The central limit theorem

N: sample size taken from population

A sampling distribution of sample means is a distribution using the

means computed from all possible random samples of a specific size

taken from a population property of the distribution of sample means

1- The mean of the sample means will be the same as the population

mean.

2- The standard deviation of the sample means will be smaller than the

standard deviation of the population, and it will be equal to the

population standard deviation divided by the square root of the

sample size.

Notes 1- 𝒛 =

𝒙−𝝁

𝝈 Used to gain information about an individual data

value when the variable is normally distributed.

2- 𝒛 =�̅�−𝝁

𝝈/√𝒏 used to gain information when applying the

central limit theorem about a sample mean when the variable is

normally distributed

3- The standrad error of the mean

𝝈𝒙 =𝝈

√𝒏

Stat 110 Ch 6


Example 1)The mean weight of 15-year-old males is 142 pounds, and the

standard deviation is 12.3 pounds. If a sample of thirty –six 15 –year-

old males is selected, find the probability that the mean of the sample

will be greater than 144.5 pounds. Assume the variable is normally

distributed.

𝒔𝒐𝒍

𝝁 = 𝟏𝟒𝟐 𝝈 = 𝟏𝟐. 𝟑 𝒏 = 𝟑𝟔

𝒑ሺ𝒙 > 𝟏𝟒𝟒. 𝟓ሻ =

𝒑 (𝒙 − 𝝁

𝝈

√𝒏

>𝟏𝟒𝟒. 𝟓 − 𝟏𝟒𝟐

𝟏𝟐. 𝟑

√𝟑𝟔

)

= 𝒑ሺ𝒛 > 𝟏. 𝟐𝟐ሻ

= 𝟏 − 𝒑ሺ𝒛 < 𝟏. 𝟐𝟐ሻ

= 𝟏 − 𝟎. 𝟖𝟖𝟖𝟖

= 𝟎. 𝟏𝟏𝟏𝟐

Stat 110 Ch 6


2)The average age of chemical engineers is 37 years a standard

deviation of 4 years. If an engineering firm employs 25 chemical

engineers, find the probability that the average age of the group is

greater than 38.2 years old.

𝒔𝒐𝒍

𝝁 = 𝟑𝟕 𝝈 = 𝟒 𝒏 = 𝟐𝟓

𝒑ሺ𝒙 > 𝟑𝟖. 𝟐ሻ =

𝒑 (𝒙 − 𝝁

𝝈

√𝒏

>𝟑𝟖. 𝟐 − 𝟑𝟕

𝟒

√𝟐𝟓

) =

𝒑ሺ𝒛 > 𝟏. 𝟓ሻ =

𝟏 − 𝒑ሺ𝒛 < 𝟏. 𝟓ሻ =

𝟏 − 𝟎. 𝟗𝟑𝟑𝟐 =

𝟎. 𝟎𝟔𝟔𝟖

Stat 110 Ch 6


3)The average annual salary in Pennsylvania was 24.393$ in 1992, assume

that salaries were normally distributed for a certain group of wage

earners, and the standard deviation of this group is 4362$.

a- Find the probability that a randomly selected individual earned less than

26.000$

b- Find the probability that , for a randomly selected sample of 25

individuals , the mean salary was less than 26.0000$

c- Why is the probability for part b higher than the probability for apart a.

𝒔𝒐𝒍

𝝁 = 𝟐𝟒𝟑𝟗𝟑 𝝈 = 𝟒𝟑𝟔𝟐

𝒂ሻ 𝒑ሺ𝒙 < 𝟐𝟔𝟎𝟎𝟎ሻ

= 𝒑 (𝒙 − 𝝁

𝝈<

𝟐𝟔𝟎𝟎𝟎 − 𝟐𝟒𝟑𝟗𝟑

𝟒𝟑𝟔𝟐)

= 𝒑ሺ𝒛 < 𝟎. 𝟑𝟕ሻ = 𝟎. 𝟔𝟒𝟒𝟑

𝒃ሻ 𝒏 = 𝟐𝟓 𝒑ሺ𝒙 < 𝟐𝟔𝟎𝟎𝟎ሻ

= 𝒑 (𝒙 − 𝝁

𝝈

√𝒏

<𝟐𝟔𝟎𝟎𝟎 − 𝟐𝟒𝟑𝟗𝟑

𝟒𝟑𝟔𝟐

√𝟐𝟓

)

= 𝒑ሺ𝒛 < 𝟏. 𝟖𝟒ሻ = 𝟎. 𝟗𝟔𝟕𝟏

𝒄ሻ 𝒔𝒂𝒎𝒑𝒍𝒆 𝒎𝒆𝒂𝒏 𝒂𝒓𝒆 𝒍𝒆𝒔𝒔 𝒗𝒂𝒓𝒊𝒂𝒃𝒍𝒆 𝒕𝒉𝒂𝒏 𝒊𝒏𝒅𝒊𝒗𝒊𝒅𝒖𝒂𝒍 𝒅𝒂𝒕𝒂

Stat 110 Ch 6


4)Assume that the mean systolic blood pressure of normal adults is 120

millimeters of mercury (mm hg) and the standard deviation is 5.6

assume the variable is normally distributed.

a- If an individual is selected, find the probability that the individuals

pressure will be between 120 and 121.8 mm hg.

b- If a sample of 30 adults is randomly selected , find the probability that

the sample mean will be between 120 and 121.8 mm hg

𝒔𝒐𝒍

𝝁 = 𝟏𝟐𝟎 𝝈 = 𝟓. 𝟔

𝒂ሻ 𝒑ሺ𝟏𝟐𝟎 < 𝒙 < 𝟏𝟐𝟏. 𝟖ሻ

= 𝒑 (𝟏𝟐𝟎 − 𝟏𝟐𝟎

𝟓. 𝟔<

𝒙 − 𝝁

𝝈<

𝟏𝟐𝟏. 𝟖 − 𝟏𝟐𝟎

𝟓. 𝟔)

= 𝒑ሺ𝟎 < 𝒛 < 𝟎. 𝟑𝟐ሻ

= 𝒑ሺ𝒛 < 𝟎. 𝟑𝟐ሻ − 𝒑ሺ𝒛 < 𝟎. 𝟓ሻ

= 𝟎. 𝟔𝟐𝟓𝟓 − 𝟎. 𝟓

= 𝟎. 𝟏𝟐𝟓𝟓

𝒃ሻ 𝒏 = 𝟑𝟎

𝒑ሺ𝟏𝟐𝟎 < 𝒙 < 𝟏𝟐𝟏. 𝟖ሻ

= 𝒑 (𝟏𝟐𝟎 − 𝟏𝟐𝟎

𝟓. 𝟔

√𝟑𝟎

<𝒙 − 𝝁

𝝈

√𝒏

<𝟏𝟐𝟏. 𝟖 − 𝟏𝟐𝟎

𝟓. 𝟔

√𝟑𝟎

)

= 𝒑ሺ𝟎 < 𝒛 < 𝟏. . 𝟕𝟔ሻ

= 𝒑ሺ𝒛 < 𝟏. 𝟕𝟔ሻ − 𝒑ሺ𝒛 < 𝟎ሻ

= 𝟎. 𝟗𝟔𝟎𝟖 − 𝟎. 𝟓

= 𝟎. 𝟒𝟔𝟎𝟖

Stat 110 Ch 6


Test bank ch 6

Determine whether each statement is true or false. If the statement is false, explain why?

1- The total area under a normal distribution is infinite.(×ሻ

2- The standard normal distribution is a continuous distribution)√ሻ

3- All variables that are approximately normally distributed can be

transformed to standard normal variables.ሺ√)

4- The Z value corresponding to a number below the mean is always

negative.(√ሻ

5- The area under the standard normal distribution to the left of Z=0 is

negative.(×ሻ

6- The central limit theorem applies to means of samples selected from

different populations.(×ሻ

Select the best answer

1-the mean of standard normal distribution is

a)0 b) 1 c)100 d) variable

2-approxmately what percentage of normally distributed data values

will fall with in 1 standard deviation above or below the mean ?

a)68% b)95% c)99.7% d)variable

3-which is not a property of the standard normal distribution

a)its symmetric about the mean b)its uniform

c)its bell-shaped d)its unimodal

Stat 110 Ch 6


4-when a distribution is positively skewed, the relationship of the mean,

median and mode from the left to right will be

a- Mean ,median , mode

b- Mode ,median, mean

c- Median ,mode, mean

d- Mean ,mode, median

5-the standard deviation of all possible sample means equals.

a- The population standard deviation.

b- The population standard deviation divided by the population mean

c- The population standard deviation divided by the square root of the

sample size

d- The square root of the population standard deviation.

Complete the following statements with the best answer. 1- When one is using the standard normal distribution

p(z<0)=………………..(0.5)

2- The difference between a sample mean and a population mean is due

to ……………………..(sampiing error )

3- The mean of the sample means equals ………………(population

mean)

4- The standard deviation of all possible sample means is

called…………………..(stand error of themean )

Stat 110 Ch 6


1. A normal distribution curve ……… A) is continuous b)is positively skewed

c)is negatively skewed

2. A property of the normal distribution is ……… A) mean ≠ 0 B) mean = 0

C) mean = standard deviation

D) mean = median = mode

3. A standard normal distribution curve is ……… A) U-shaped B) J-shaped

C) bell-shaped

D) uniform

4. A normal distribution curve is symmetric about ………

A) 1

B) µ

C) σ

D) 0

5. The standard normal distribution is a normal distribution with ……… A) mean = 1, standard deviation = 0 B) mean = 0, standard deviation = 1

C) mean = -1, standard deviation = 1

D) mean = 1, standard deviation = 1

Stat 110 Ch 6


6. The average number of calories in a chocolate bar is 225. Suppose that

the variable is approximately normally distributed with a standard

deviation 10. Find the probability that a randomly selected chocolate

bar will have less than 200 calories.

b)0.9938 c)0.4938 d)0.0202 0.0062 )A

𝒔𝒐𝒍

𝝁 = 𝟐𝟐. 𝟓 𝝈 = 𝟏𝟎

𝒑ሺ𝒙 < 𝟐𝟎𝟎ሻ

= 𝒑 (𝒙 − 𝝁

𝝈<

𝟐𝟎𝟎 − 𝟐𝟐𝟓

𝟏𝟎)

= 𝒑ሺ𝒛 < −𝟐. 𝟓ሻ = 𝟎. 𝟎𝟎𝟔𝟐

Stat 110 Ch 6


7. Find two z-values so that 90.5% of the middle area is bounded by them.

D)±1.31 )±1.67C B)±0.24 A) 0.12

SOL

𝑷ሺ𝒛 < 𝒙ሻ = 𝟎. قراءه عكسيه من الجدول 𝟗𝟓𝟐𝟓

𝒙 = 𝟏. 𝟔𝟕

𝒕𝒉𝒆𝒏 𝒛 = ±𝟏. 𝟔𝟕

Stat 110 Ch 6


8. Find P( - 0.55 < z < 0.55 ).

B) 0.7088 C)0.2912 D)0.8643 0.4176 A)

𝒔𝒐𝒍

𝒑ሺ−𝟎. 𝟓𝟓 < 𝒛 < 𝟎. 𝟓𝟓ሻ

= 𝒑ሺ𝒛 < 𝟎. 𝟓𝟓ሻ − 𝒑ሺ𝒛 < −𝟎. 𝟓𝟓ሻ

= 𝟎. 𝟕𝟎𝟖𝟖 − 𝟎. 𝟐𝟗𝟏𝟐

= 𝟎. 𝟒𝟏𝟕𝟔

9. Find the z-value if the area shaded below is equal to 0.9875.

A) – 0.03

B) 2.24

C) 0.03

𝒔𝒐𝒍 2.24 –) D

𝒑ሺ𝒙 > 𝒛ሻ = 𝟎. 𝟗𝟖𝟕𝟓

𝟏 − 𝒑ሺ𝒙 < 𝒛ሻ = 𝟎. 𝟗𝟖𝟕𝟓

𝒑ሺ𝒙 < 𝒛ሻ = 𝟏 − 𝟎. 𝟗𝟖𝟕𝟓

𝒑ሺ𝒙 < 𝒛ሻ = 𝟎. قراءه عكسيه من الجدول 𝟎𝟏𝟐𝟓

𝒛 = −𝟐. 𝟐𝟒

Stat 110 Ch 6

ران ــد عـمــمـحـمـ 0507017098-0580535304

10. . In a normal distribution, find σ when µ = 110 and 2.87% of the area lies to right of 112.

0.7 -D) 1.05C) B)0.7 1.05 –A)

Sol

𝒑ሺ𝒙 < 𝒛ሻ = 𝟎. قراءه عكسيه من الجدول 𝟗𝟕𝟏𝟑

𝒛 = 𝟏. 𝟗

𝝁 = 𝟏𝟏𝟎 𝒙 = 𝟏𝟏𝟐

𝒙 − 𝝁

𝝈= 𝒛 → 𝝈 =

𝒙 − 𝝁

𝒛

=𝟏𝟏𝟐 − 𝟏𝟏𝟎

𝟏. 𝟗

𝝈 = 𝟏. 𝟎𝟓.

Stat 110 Ch 6


11. To qualify for a university, candidates must score in the top

20% on a general abilities test. The test has a mean of 150

and a standard deviation of 25. Find the lowest possible score

to qualify. Assume the test scores are normally distributed.

171C) 5 169.7 B)164. A)

D)129

𝒔𝒐𝒍

𝝁 = 𝟏𝟓𝟎 𝝈 = 𝟐𝟓

𝒑ሺ𝒙 < 𝒛ሻ = 𝟎. قراءه عكسيه من الجدول 𝟖𝟎

𝒛 = 𝟎. 𝟖𝟒

𝒙 = 𝒛𝝈 + 𝝁

= ሺ𝟎. 𝟖𝟒ሻሺ𝟐𝟓ሻ + 𝟏𝟓𝟎 = 𝟏𝟕𝟏

Stat 110 Ch 6


1 2 . The average time it takes a group of adults to complete a test

is 46.2 minutes and the standard deviation is 8 minutes. Assume the

variable is normally distributed. If 50 adults are selected at

random, find the probability that the mean time it takes the group

to complete the test will be less than 47 minutes.

C)0.5398 D) 0.5040 B)0.2389 0.7661A)

Sol

𝝁 = 𝟒𝟔. 𝟐 𝝈 = 𝟖 𝒏 = 𝟓𝟎

𝒑ሺ𝒙 < 𝟒𝟕ሻ

= 𝒑 (𝒙 − 𝝁

𝝈

√𝒏

<𝟒𝟕 − 𝟒𝟔. 𝟐

𝟖

√𝟓𝟎

)

= 𝒑ሺ𝒛 < 𝟎. 𝟕𝟏ሻ

= 𝟎. 𝟕𝟔𝟏𝟏

Stat 110 Ch 6


13. If a sample has a size of 9 and a standard deviation

of 2.3, the standard error of the mean is ………

D)0.256 C)0.767B)1.517 2.300 A)

𝒔𝒐𝒍 𝒏 = 𝟗 𝝈 = 𝟐. 𝟑

𝝈𝒙 =𝝈

√𝒏=

𝟐. 𝟑

√𝟗= 𝟎. 𝟕𝟔𝟕

14.The average time it takes students to get to school is 15.1 minutes.

Assume the time has a normal distribution with a variance of 8.1

minutes. If a student is randomly selected, find the probability that

he gets to school in greater than 17 minutes.

C)0.7486 D)0.40 0.2514B) 0.5910 A)

𝑺𝒐𝒍

𝝁 = 𝟏𝟓. 𝟏 𝝈 = √𝟖. 𝟏 = 𝟐. 𝟖𝟓

𝒑ሺ𝒙 > 𝟏𝟕ሻ = 𝒑 (𝒙 − 𝝁

𝝈>

𝟏𝟕 − 𝟏𝟓. 𝟏

𝟐. 𝟖𝟓)

= 𝒑ሺ𝒛 > 𝟎. 𝟔𝟕ሻ = 𝟏 − 𝒑ሺ𝒛 < 𝟎. 𝟔𝟕ሻ

= 𝟏 − 𝟎. 𝟕𝟒𝟖𝟔 = 𝟎. 𝟐𝟓𝟏𝟒



0580535304




Table entry

Table entry for z is the area under the standard normal curveto the left of z.

Standard Normal Probabilities

z

z .00

–3.4–3.3–3.2–3.1–3.0–2.9–2.8–2.7–2.6–2.5–2.4–2.3–2.2–2.1–2.0–1.9–1.8–1.7–1.6–1.5–1.4–1.3–1.2–1.1–1.0–0.9–0.8–0.7–0.6–0.5–0.4–0.3–0.2–0.1–0.0

.0003

.0005

.0007

.0010

.0013

.0019

.0026

.0035

.0047

.0062

.0082

.0107

.0139

.0179

.0228

.0287

.0359

.0446

.0548

.0668

.0808

.0968

.1151

.1357

.1587

.1841

.2119

.2420

.2743

.3085

.3446

.3821

.4207

.4602

.5000

.0003

.0005

.0007

.0009

.0013

.0018

.0025

.0034

.0045

.0060

.0080

.0104

.0136

.0174

.0222

.0281

.0351

.0436

.0537

.0655

.0793

.0951

.1131

.1335

.1562

.1814

.2090

.2389

.2709

.3050

.3409

.3783

.4168

.4562

.4960

.0003

.0005

.0006

.0009

.0013

.0018

.0024

.0033

.0044

.0059

.0078

.0102

.0132

.0170

.0217

.0274

.0344

.0427

.0526

.0643

.0778

.0934

.1112

.1314

.1539

.1788

.2061

.2358

.2676

.3015

.3372

.3745

.4129

.4522

.4920

.0003

.0004

.0006

.0009

.0012

.0017

.0023

.0032

.0043

.0057

.0075

.0099

.0129

.0166

.0212

.0268

.0336

.0418

.0516

.0630

.0764

.0918

.1093

.1292

.1515

.1762

.2033

.2327

.2643

.2981

.3336

.3707

.4090

.4483

.4880

.0003

.0004

.0006

.0008

.0012

.0016

.0023

.0031

.0041

.0055

.0073

.0096

.0125

.0162

.0207

.0262

.0329

.0409

.0505

.0618

.0749

.0901

.1075

.1271

.1492

.1736

.2005

.2296

.2611

.2946

.3300

.3669

.4052

.4443

.4840

.0003

.0004

.0006

.0008

.0011

.0016

.0022

.0030

.0040

.0054

.0071

.0094

.0122

.0158

.0202

.0256

.0322

.0401

.0495

.0606

.0735

.0885

.1056

.1251

.1469

.1711

.1977

.2266

.2578

.2912

.3264

.3632

.4013

.4404

.4801

.0003

.0004

.0006

.0008

.0011

.0015

.0021

.0029

.0039

.0052

.0069

.0091

.0119

.0154

.0197

.0250

.0314

.0392

.0485

.0594

.0721

.0869

.1038

.1230

.1446

.1685

.1949

.2236

.2546

.2877

.3228

.3594

.3974

.4364

.4761

.0003

.0004

.0005

.0008

.0011

.0015

.0021

.0028

.0038

.0051

.0068

.0089

.0116

.0150

.0192

.0244

.0307

.0384

.0475

.0582

.0708

.0853

.1020

.1210

.1423

.1660

.1922

.2206

.2514

.2843

.3192

.3557

.3936

.4325

.4721

.0003

.0004

.0005

.0007

.0010

.0014

.0020

.0027

.0037

.0049

.0066

.0087

.0113

.0146

.0188

.0239

.0301

.0375

.0465

.0571

.0694

.0838

.1003

.1190

.1401

.1635

.1894

.2177

.2483

.2810

.3156

.3520

.3897

.4286

.4681

.0002

.0003

.0005

.0007

.0010

.0014

.0019

.0026

.0036

.0048

.0064

.0084

.0110

.0143

.0183

.0233

.0294

.0367

.0455

.0559

.0681

.0823

.0985

.1170

.1379

.1611

.1867

.2148

.2451

.2776

.3121

.3483

.3859

.4247

.4641

.01 .02 .03 .04 .05 .06 .07 .08 .09

Table entry

Table entry for z is the area under the standard normal curveto the left of z.

z

z .00

0.00.10.20.30.40.50.60.70.80.91.01.11.21.31.41.51.61.71.81.92.02.12.22.32.42.52.62.72.82.93.03.13.23.33.4

.5000

.5398

.5793

.6179

.6554

.6915

.7257

.7580

.7881

.8159

.8413

.8643

.8849

.9032

.9192

.9332

.9452

.9554

.9641

.9713

.9772

.9821

.9861

.9893

.9918

.9938

.9953

.9965

.9974

.9981

.9987

.9990

.9993

.9995

.9997

.5040

.5438

.5832

.6217

.6591

.6950

.7291

.7611

.7910

.8186

.8438

.8665

.8869

.9049

.9207

.9345

.9463

.9564

.9649

.9719

.9778

.9826

.9864

.9896

.9920

.9940

.9955

.9966

.9975

.9982

.9987

.9991

.9993

.9995

.9997

.5080

.5478

.5871

.6255

.6628

.6985

.7324

.7642

.7939

.8212

.8461

.8686

.8888

.9066

.9222

.9357

.9474

.9573

.9656

.9726

.9783

.9830

.9868

.9898

.9922

.9941

.9956

.9967

.9976

.9982

.9987

.9991

.9994

.9995

.9997

.5120

.5517

.5910

.6293

.6664

.7019

.7357

.7673

.7967

.8238

.8485

.8708

.8907

.9082

.9236

.9370

.9484

.9582

.9664

.9732

.9788

.9834

.9871

.9901

.9925

.9943

.9957

.9968

.9977

.9983

.9988

.9991

.9994

.9996

.9997

.5160

.5557

.5948

.6331

.6700

.7054

.7389

.7704

.7995

.8264

.8508

.8729

.8925

.9099

.9251

.9382

.9495

.9591

.9671

.9738

.9793

.9838

.9875

.9904

.9927

.9945

.9959

.9969

.9977

.9984

.9988

.9992

.9994

.9996

.9997

.5199

.5596

.5987

.6368

.6736

.7088

.7422

.7734

.8023

.8289

.8531

.8749

.8944

.9115

.9265

.9394

.9505

.9599

.9678

.9744

.9798

.9842

.9878

.9906

.9929

.9946

.9960

.9970

.9978

.9984

.9989

.9992

.9994

.9996

.9997

.5239

.5636

.6026

.6406

.6772

.7123

.7454

.7764

.8051

.8315

.8554

.8770

.8962

.9131

.9279

.9406

.9515

.9608

.9686

.9750

.9803

.9846

.9881

.9909

.9931

.9948

.9961

.9971

.9979

.9985

.9989

.9992

.9994

.9996

.9997

.5279

.5675

.6064

.6443

.6808

.7157

.7486

.7794

.8078

.8340

.8577

.8790

.8980

.9147

.9292

.9418

.9525

.9616

.9693

.9756

.9808

.9850

.9884

.9911

.9932

.9949

.9962

.9972

.9979

.9985

.9989

.9992

.9995

.9996

.9997

.5319

.5714

.6103

.6480

.6844

.7190

.7517

.7823

.8106

.8365

.8599

.8810

.8997

.9162

.9306

.9429

.9535

.9625

.9699

.9761

.9812

.9854

.9887

.9913

.9934

.9951

.9963

.9973

.9980

.9986

.9990

.9993

.9995

.9996

.9997

.5359

.5753

.6141

.6517

.6879

.7224

.7549

.7852

.8133

.8389

.8621

.8830

.9015

.9177

.9319

.9441

.9545

.9633

.9706

.9767

.9817

.9857

.9890

.9916

.9936

.9952

.9964

.9974

.9981

.9986

.9990

.9993

.9995

.9997

.9998

.01 .02 .03 .04 .05 .06 .07 .08 .09

Standard Normal Probabilities

King Abdul-Aziz UniversityFaculty of Sciences

Statistics Department Final Exam STAT 110 A

First Term 1430-1431 H

_______

40

Name: ID No: Section:

You have 40 questions and 100 minutes to solve the exam. Please mark all your answers on the answer sheet provided to you. You have to submit both question paper and answer sheet but only answer sheets will be graded. Good luck

Choose the best answer for each of the following questions:

1. Statistics is the science of conducting studies toA) hypothesize, experiment, and form conclusions.B) collect, organize, summarize, analyze, and draw conclusions from data.C) monitor, study, and report on a subject.D) solve a system of equations.

2. Calculate the mean for the following numbers: 1.1 2.3 3.4 -1.5A) 2.08B) 5.3C) 2.27D) 1.33

3. Which one of these events is not mutually exclusive?A) Select a student in your university: The student is married, and the student is a

business major.B) Select a ball from bag: It is a football, and it is a basket ball.C) Roll a die: Get an even number, and get an odd number.D) Select any course: It is an Arabic course, and it is a Statistics course.

4. In a binomial experiment, for each trial the probability of success must be...................A) the sameB) differentC) largeD) small

5. If the equation for the regression line is y = –5x + 3, then the sign of the correlation coefficient between the two variables isA) positiveB) –5C) negativeD) can not be determined

Page 1

6. A person owns a collection of 25 movies, five of which are English. If four movies are selected at random, find the probability that three of them are English.

A)1

25

B)3

5

C)1

125

D)4

253

7. If X is a discrete random variable with ( ) 2.4X P X⋅ =∑ and 2 ( ) 8.88X P X⋅ =∑ ,

the mean and variance for the probability distribution of X is

A) 22.4, 6.48µ = σ =

B) 22.74, 5.76µ = σ =

C) 22.4, 8.88µ = σ =

D) 22.4, 3.12µ = σ =

8. The equation of the regression line between a person's height in centimeters (x) and his weight in kilograms (y) is given by : 44 0.2y x′ = + .

Predict Ali's weight if his height is 180 centimeters.A) 180 kgB) 40 kgC) 44.36 kgD) 80 kg

9. If you know that 2 20, 4 and 3X X n= = =∑ ∑ , the sample standard deviation

will be A) 6.67B) 7.33C) 2.71D) 9.33

10. How many 3-digit passwords are possible if digits can be repeated?A) 720B) 729C) 1000D) 120

11. Data such as car types (Toyota, Kia, Honda, Lexus) can be organized into a(n)...................... frequency distribution.A) relativeB) categorical C) groupedD) ungrouped

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12. If the mean of the daily income in your town is $200, the mode is $100 and the median is $150, this means that the distribution shape of the daily income isA) left-skewedB) uniformC) right-skewedD) symmetric

13. If all the points fall on the straight line the value of r will be

A) 0.01±B) 0.5±C) 1.5±D) 1±

14. Which of the following linear regression equations represent the graph below?

A) 0.5y x′ = −

B) 1.5y x′ = − +

C) 2.5y x′ = − +

D) 2.5y x′ = − −

15. If P(a<z<b)=0.95, then the value of the standard deviationσ equals

A) 2B) 3C) 0.05D) 1

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16. For a normal distribution with mean = –16 and a z value = –0.5for the value x = –17, the standard deviation isA) 1B) –1C) 2D) –2

17. Ahmad says:" my relative position in Math exam is higher than my relative position in History exam ". Which of the following z-score values are correct?A) z-score of Math = -1.5 and z-score of History = -1.5B) z-score of Math = -1.5 and z-score of History = -1.2C) z-score of Math = -1.2 and z-score of History = -1.5D) None of the above

18. The following graph represents students marks in two classes A and B

Choose the best statement that compares the variation between mark of them.A) the marks of the students in class A is less spread than the marks of the students

in class B.B) the marks of the students in class A and B is the same spread.C) the marks of the students in class A is more spread than the marks of the

students in class B.D) the relation between spread of the class A and B can not be determined.

19. A correlation coefficient of 0.96 would mean thatA) the values of x decrease as the values of y increase.B) the values of x increase as the values of y decrease.C) there is no relationship between x and y.D) the values of x increase as the values of y increase.

20. A researcher finds that if expectant mothers use Vitamin Pills, the birth weight of the babies will increase. What type of study was this?A) Quasi-experimental study.B) Independent study.C) Observational studyD) Experimental study.

21. A Pareto chart does not have which of the following properties?A) It is used to represent categorical dataB) The bars have the same width C) The frequencies are arranged from highest to lowestD) The frequencies are arranged from lowest to highest

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22. In a certain school, it is known that 81% of instructors are using e-mail to send messages. For a sample of 10 instructors, find the probability that exactly 6 of them are using e-mail.A) 0.0773B) 0.0004C) 0.2824D) 0.0043

23. The average height of Apple trees in a garden is 10.5 feets. If the heights are normally distributed with a standard deviation of 3, find the percentage that a tree is less than 12.5 feets tallA) 74.86%B) 24.86%C) 66.67%D) 25.14%

24. The area under the standard normal distribution curve equals to ………… .A) -1B) 0C) 1D) 0.5

25. The average weight of books on a library shelf is 8.3 grams. The standard deviation

is 0.6 grams. If 20% of the books are oversized , find the minimum weight of the large books on the library shelf. Assume the variable is normally distributed.A) 8.612 gramsB) 0.3 gramC) 0.84 gramD) 8.804 grams

26. How many times was the coin tossed in the following figure?

A) 4B) 2C) 6D) 3

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27. If a normal distribution has a mean 22 and a standard deviation 4, thenA) the median is 26 and the mode is 18B) the median is 22 and the mode is 22C) the median is 22 and the mode is 26D) the median is 18 and the mode is 26

28. Which of the following plots represents the stem and leaf for this data set? 537 512 529 514 521 520 536A)

B)

C)

D)

29. When the subjects are selected by dividing up the population into groups, and subjects within groups are randomly selected, this is sample called aA) stratified sampleB) random sampleC) cluster sampleD) systematic sample

30. Find two z values so that 41.08% of the middle area is bounded by themA) ± 1.35B) ± 0.4108C) ± 0.54D) ± 0.2054

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31. The standard deviation of the sample means will be ……………..the standard deviation of the population.A) equal toB) larger thanC) smaller than D) can not compare the two

32. The average repair cost of a microwave oven is $55, with a standard deviation of $8. The costs are normally distributed. If 12 ovens are repaired, find the probability that the mean of the oven repair will be greater than $60.A) 0.015B) 0.4850C) 0.2643D) 0.985

33. Find the probability P(–0.09<z < 2.37) A) 0.4911B) 0.527C) 0.0359D) 0.4552

34. A TV station interviews five movie viewers after the first showing of a movie. After finding out that all five enjoyed the movie very much, the reporter

states that this movie will definitely be the best movie for the summer. This is an example ofA) detached statisticsB) suspect samplesC) ambiguous averagesD) changing the subject

35. Classify the number of schools in Florida in a given year.A) NominalB) OrdinalC) DiscreteD) Continuous

Page 7

36. For the following graph, the values on the x-axis represent................

A) cumulative frequencies B) class limitsC) class boundariesD) class midpoints

37. The frequency distribution shows the students hair colors in a classroom.

When constructing a pie graph, the angle degree of black hair students is

A) 160o

B) 90o

C) 50o

D) 180o

38. How many times a die is rolled when the mean for the number of 3's that will be rolled=60?A) 20

B)1

6C) 360D) 10

39. What is the set of all possible outcomes of a probability experiment?A) The Venn diagramB) The outcomeC) The sample spaceD) The event

Page 8

Hair Color No.of Students

black 16brown 8blond 8

40. For the following graph, the corresponding probability distribution is

A)

B)

C)

D)

Good LuckSTAT110 TEAM

Thursday 6-2-1431 H

Page 9

1 2P(X) 0.2 0.4 0.4

3P(X)

X 0.2 0.2 0.6P(X) 3 5 6

0.2 0.4 0.2

X 3 5 6P(X) 0.2 0.2 0.6

X 1 2

X 0

Answer Key

1. B2. D3. A4. A5. C6. D7. D8. D9. C

10. C11. B12. C13. D14. D15. D16. C17. C18. A19. D20. C21. D22. A23. A24. C25. D26. B27. B28. C29. A30. C31. C32. A33. B34. B35. C36. D37. D38. C39. C40. D

Page 10

KING ABDULAZIZ UNIVERSITY

Faculty of Sciences

Statistics Department

Final Exam

STAT 110

First Term

1431-1432 40 A

Name: ID #: Section:

You have 40 questions. You have 120 minutes to solve the exam. Please mark all your answers on the

answer sheet provided to you. Make sure that the answer sheet form matches the question form. You have to

submit both question paper and answer sheet but only answer sheets will be graded. Good luck

Choose the best answer for each of the following questions:

1. A statistic that tells the number of standard deviations a data value is above the mean is called a ...

A) percentile. B) z score. C) coefficient of variation. D) quartile.

2. The student's weight is a (an) ... variable.

A) discrete B) ordinal C) nominal D) continuous

3. In order to get a sample of 40 students, a researcher divided the students population into 40 groups

and then selected the fourth student from each group after numbering them randomly. The type of

sampling is ...

A) random. B) stratified. C) systematic. D) cluster.

4. If the numbers 0, 1, 2 represents the number of hourly accidents in a given street during a week.

The frequency table of 30 cars ...

A) has three classes. C) can be represented by bar chart.

B) has a total frequency of 30. D) A, B and C.

5. Which is a part of the five-number summary?

A) The mode. B) The mean. C) The median. D) The midrange.

Use the following to answer questions 6-7:

The coefficient of variation of the height of 20 people selected at random from a given city is found to be

15%. The weight of the selected group has a mean value 72 kg and a standard deviation 8 kg.

6. The coefficient of variation for the weight of the selected group is ...

A) 11.11 B) 8.33% C) 11.11% D) 8.33

7. The obtained results show that ...

A) the weight is more variable than height.

B) the weight is less variable than height.

C) height and weight have the same degree of variation.

D) height and weight are independent.

8. A fair coin is tossed three times. What is the probability of getting 2 heads?

A) 3/8 B) 0 C) 1/8 D) 5/8


For the values 7, 3, 9, 4, 12, 8, 19, 6, 54, answer the following two questions.

9. The outlier value for the given values is ...

A) 53 B) 20.5 C) 54 D) any value greater or less than IQR

10. The inter quartile range (IQR) is ...

A) 12 B) 20.5 C) 19 D) 10.5

11. When the correlation coefficient(r) equals zero, the linear relationship between the variables …

A) is moderate. B) is strong. C) is weak. D) does not exist.

12. Which of the following events is mutually exclusive when rolling a die?

A) get a prime number and an odd number. C) get an even number and an odd number.

B) get a prime number and an even number. D) get an odd number and a number < 2.

13. The number of outcomes in the sample space for the gender of children in a family with 7 children

is ...

A) 128 B) 64 C) 256 D) 32


For a class limit 69.4 - 70.5 answer the following three questions:

14. The class boundaries are ...

A) 69.35 - 70.55 B) 69.9 - 70 C) 69.45 - 70.45 D) 68.9 - 71

15. The class midpoint is ...

A) 69.5 B) 69.45 C) 69.95 D) 69.7

16. The class width ...

A) is 1.2 B) is 1.1 C) is 1.3 D) cannot be calculated

17. Determine the number of all possible outcomes of guessing the last four digits in a telephone

number if repetition among the four digits is allowed.

A) 10000 B) 5040 C) 1000 D) 720


In the study of the relationship between the number of daily studying hours X and the final grade in statistics

Y of 8 students, the data show the following: 2 242, 470, 3143, 354 and 37358X Y XY X Y= = = = =∑ ∑ ∑ ∑ ∑

18. The value of the Pearson correlation coefficient is ...

A) -0.592 B) 0.829 C) 0.592 D) -0.829

19. The value of the Pearson correlation coefficient means that there is a ... linear relationship between

the number of daily studying hours and the final grade.

A) strong negative B) moderate positive C) moderate negative D) strong positive

20. The slope of the regression line is ...

A) 2.17 B) 5.37 C) 8.48 D) 5.06

21. The final grade is called ... variable.

A) response B) explanatory C) predictor D) independent

22. If the variance of a probability distribution is 3.6 grams, what is the standard deviation?

A) 39.69 B) 1.9 C) 12.96 D) 2.51

23. What type of distributions is the binomial distribution?

A) Continuous. C) Neither discrete nor continuous.

B) Discrete. D) Discrete and continuous.

24. Which one of the following is NOT one of the binomial distribution requirements?

A) Only two outcomes.

B) At least 10 observations.

C) Probability of success remains constant from trial to trial.

D) Independent trials.

25. Ten thousand tickets are sold at 100 SAR each for a car valued at 45000 SAR. What is the expected

value of the gain if a person purchases two tickets?

A) -190 B) -150 C) -165 D) -191

26. Determine which one of the following is a probability distribution.

A) X 2 3 4 5 6

P(X) 2/3 2/5 2/7 2/9 2/11

B) X -4 -3 -2 -1 0

P(X) 1/8 1/4 1/8 1/4 1/4

C) X -2 -1 0 1 2

P(X) 1/4 1/4 1/4 1/4 1/4

D) X -2 -1 0 1 2

P(X) 1/5 1/10 1/10 1/5 1/5

27. The Spearman rank correlation coefficient can be calculated for ... variable(s).

A) ordinal B) quantitative C) nominal D) ordinal and quantitative

28. Find the variance of the following probability distribution

X -2 -1 0 1 2

P(X) 1/5 1/5 1/5 1/5 1/5

A) 0 B) 4 C) 1.414 D) 2


A multiple choice quiz consists of 6 questions, each with 5 possible answers. If a student guesses the answer

of each question, then

29. the probability of at least one correct answer is ...

A) 0.738 B) 0.607 C) 0.598 D) 0.709

30. the mean number of correct answers is ...

A) 0.83 B) 1.20 C) 0.875 D) 1.14

31. the probability of guessing exactly two correct questions is ...

A) 0.246 B) 0.161 C) 0.227 D) 0.168

32. The mean of a normal probability distribution is 500 and the standard deviation is 10. About 95

percent of the observations lie between what two values?

A) 490 and 510 B) 480 and 520 C) 470 and 530 D) 450 and 550

33. The standard normal probability distribution is unique because it has ...

A) Mean of 1 and variance of 0. C) Mean of 0 and any variance.

B) Mean of 1 and any variance. D) Mean of 0 and variance of 1.


The scores of a college entrance test is normally distributed with mean 500 and a standard deviation 75.

Answer the following five questions:

34. Find the percentage of students who scored between 485 and 590.

A) 46.42% B) 68.53% C) 58.20% D) 23.94%

35. Find the lowest score for the top 10% of students.

A) 514 B) 451.2 C) 596 D) 464

36. If a sample of 36 students are selected, find the probability that the mean scores of the sample is

below 510.

A) 0.9332 B) 0.7881 C) 0.9641 D) 0.8849

37. Find the percentage of students who scored below 320.

A) 0.47% B) 2.28% C) 0.82% D) 5.48%

38. Find the percentage of students who scored above 455.

A) 46.02% B) 15.87% C) 13.57% D) 72.57%

39. If the standard deviation of a population is 48 and we took a sample of size 32, then the standard

error of the mean (the standard deviation of the sample mean) is ...

A) 8.398 B) 11.685 C) 14.849 D) 8.485

40. Which is NOT a property of the normal distribution?

A) It has a mean of 0 and a standard deviation of 1.

B) It has a single peak.

C) The mean equals the median.

D) It is unimodal.

Good luck

Stat 110 Team

Answer Key

1. B

2. D

3. C

4. D

5. C

6. C

7. B

8. A

9. C

10. D

11. D

12. C

13. A

14. A

15. C

16. A

17. A

18. C

19. B

20. D

21. A

22. B

23. B

24. B

25. D

26. B

27. D

28. D

29. A

30. B

31. A

32. B

33. D

34. A

35. C

36. B

37. C

38. D

39. D

40. A

©2013 STAT 110 Team A Page 1

King Abdulaziz University STAT 110 1433/1434

Faculty of Sciences Final First Term

Statistics Department Exam Time: 120 minutes Date: 21/02/1434H

Please mark all your answers on the answer sheet provided to you. Only the answer sheet will be graded.

Choose the best answer for each of the following questions. Good Luck

If the weight of 8000 men follows a normal distribution with mean 67 kg. and standard deviation 10 kg.

Answer questions (1 - 5)

1- A man was selected randomly. Find the probability that his weight is less than 62 kg.

A) 0.9525 B) 0.3085 C) 0.0475 D) 0.6915

2- If another man was selected at random, find the probability that his weight is between 60 and 80 kg.

A) -0.6612 B) 0.9032 C) 0.2420 D) 0.6612

3- If a sample of 36 men was selected, find the probability that the mean of the sample will be more than 71 kg.

A) 0.0082 B) 0.3446 C) 0.9918 D) 0.6554

4- How many men their weight are less than 60 kg.

A) 774 B) 7226 C) 6064 D) 1936

5- What is the percentage of men that their weight are less than 80 kg.

A) 75.8 % B) 24.2 % C) 90.32 % D) 9.68 %

6- The normal distribution curve is ………… .

A) discrete B) continuous C) left skewed D) right skewed

7- The total area under the normal distribution curve is equal to …………

A) +1 B) +0.5 C) -1 D) -0.5

8- In the binomial experiment, the trials must be ………… from each other.

A) unimodal B) dependent C) fixed D) independent

9- If the scores of three courses for a student, who studied at King Abdulaziz University, are 87, 73, 95, with

credit hours are 4, 2, 3, find the mean of the student’s scores from 100

A) 3.05 B) 85 C) 86.56 D) 5.03

10- From the scatter plot below, what can you say about the type of the relationship between and ?

A) Positive linear B) Negative linear C) No relationship D) Curvilinear

A


If the numbers of sold phones for a store in the past six days are: 3, 6, 4, 9, 3, 8 Answer questions (11 – 22) 11- Find the mean, A) 3 B) 5 C) 6 D) 5.5 12- find the median, A) 5 B) 6 C) 5.5 D) 3 13- Find the mode. A) no mode B) 3 C) 3, 6 D) 3, 4, 6 14- The data set is said to be ………… A) no mode B) a multimodal C) a bimodal D) an unimodal 15- Find the midrange, A) 5.5 B) 6 C) 5 D) 3 16- Find the range, A) 5 B) 12 C) 3 D) 6 17- Find the variance, A) 6.7 B) 3.45 C) 2.59 D) 7.3 18- Find the standard deviation, A) 3.45 B) 2.59 C) 7.3 D) 6.7 19- Find the coefficient of variation, A) 212.48 % B) 52.94 % C) 47.09 % D) 112.48 % 20- Find the first quartile, . A) 5 B) 3 C) 8 D) 6 21- Find the interquartile range, . A) 8 B) -5 C) 3 D) 5 22- Find an outlier if any. A) no outliers B) 3 C) -4.5 D) 9 _________________________________________________________________________________________________________________________ 23- If , then find the z-score, A) 1.67 B) 0.22 C) -0.67 D) -1.22


A survey from King Abdulaziz University found that 23% of patients are smoker. If 7 patients were selected at random, answer questions (24 – 29) 24- Find the probability that 3 of the patients are smoker. A) 0.946 B) 0.203 C) 0.150 D) 0.054 25- Find the probability that less than 2 patients are smoker. A) 0.496 B) 0.301 C) 0.797 D) 0.203 26- Find the probability that at most 2 of the patients are smoker. A) 0.203 B) 0.496 C) 0.301 D) 0.797 27- Find the mean of number of the patients who are smoker. A) 1.24 B) 1.61 C) 1.11 D) 1.98 28- Find the variance of number of the patients who are smoker. A) 1.61 B) 1.11 C) 1.98 D) 1.24 29- Find the standard deviation of number of the patients who are smoker A) 1.11 B) 1.98 C) 1.24 D) 1.61

Use the available information to answer questions (30 – 33):

30- Find the Pearson correlation coefficient, .

A) B) C) D) 31- The value of the correlation coefficient obtained in previous question can be interpreted as follows: There is a ………… linear relationship between the variables and . A) weak negative B) strong positive C) weak positive D) strong negative 32- Find the equation of the regression line. A) B) C) D) 33- Predict the value of when A) B) C) D)

If a single die is rolled, answer questions (34 – 36) 34- Find the probability of getting an odd number or a number can be divided by 3. A) 0.833 B) 0.333 C) 0.500 D) 0.667 35- Find the probability of getting a number less than 2 or can be divided by 3. A) 0.667 B) 0.500 C) 0.833 D) 0.333 36- Find the probability of getting number 5. A) 0.333 B) 0.167 C) 0.667 D) 0.833


If a random variable has the discrete probability distribution as shown below, then answer questions (37 - 40)

1 2 3 4 5 6

0.17 0.03 k 0.39 0.14 0.01

37- Find the missing probability, .

A) 3.33 B) 1.28 C) 1.64 D) 0.26

38- Find the mean of the distribution, .

A) 1.28 B) 0.26 C) 3.33 D) 1.64

39- Find the variance of the distribution, .

A) 1.64 B) 3.33 C) 0.26 D) 1.28

40- Find the standard deviation of the distribution,

A) 0.26 B) 1.64 C) 1.28 D) 3.33


Formulas:

Percentage:

Degree:

Mean:

Midrange:

Weighted mean:

Range:

Variance:

Standard deviation:

Coefficient of variations:

z-score:

Interquartile Range:

Non-outliers interval: [ , ]

Pearson linear correlation coefficient:

Spearman rank correlation coefficient:

Equation of regression line:

intercept of the regression line:

Slope of the regression line:


Classical probability:

Empirical probability:

Rule for complementary events:

Probability of mutually exclusive:

Probability of not mutually exclusive:

Probability of independent:

Mean of a probability distribution:

Variance of a probability distribution:

Standard deviation of a probability distribution:

Binomial distribution:

Probability: Mean:

Variance: Standard deviation:

Z-value:

Z-value:


Answer Key Form A

1-

2-

3-

4-

5-

6-

7-

8-

9-

10-

11-

12-

13-

14-

15-

16-

17-

18-

19-

20-

21-

22-

23-

24-

25-

26-

27-

28-

29-

30-

31-

32-

33-

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35-

36-

37-

38-

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