module 18 probability distributions

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PROBABILITYDISTRIBUTIONS

ADDITIONAL MATHEMATICSFORM 5

MODULE 13

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MODULE 13 : PROBABILITY DISTRIBUTIONS

CONTENT PAGE

13.1. CONCEPT MAP 1

13.2. PROBABILITY IN BINOMIAL DISTRIBUTION 2

13.3. ACTIVITY 1 5

1 13.4 BINOMIAL DISTRIBUTION GRAPH 6

13.5. MEAN, VARIANCE, STANDARD DEVIATIONOF BINOMIAL DISTRIBUTION

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13.6. ACTIVITY 2 8

13.7. ACTIVITY 3 : SPM FOCUS PRACTICE 10

13.8. ANSWERS 12



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13.1 CHAPTER 8

CONCEPT MAP

PROBABILITY DISTRIBUTIONS

BINOMIAL DISTRIBUTIONX ~B( n, p)

Binomial DistributionGraph

Probabilityin Binomial Distribution

Mean, Variance, StandardDeviation of Binomial

Distribution



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13.2 Probability in Binomial Distribution

Example 1 :

In a town, the probability that rain will fall on any day is 0.3. Calculate the probabilitythat rain will fall on exactly 2 days in a certain week.

Solution :

Step 1 : Identify the parameters:

p = 0.3q = 1 – 0.3 = 0.7r = 2n = 7

Step 2 : Substitute into the formula

P(X = r) = n C r p r q rn

P(X = 2) = 7 C 2 (0.3) 2 (0.7) 27

= (21)x(0.09)x(0.1681)

= 0.3177

Example 2 :

The probability that a durian chosen at random from a basket is rotten is20

1. Calculate

the probability that exactly 3 durians are rotten if a sample of 10 durians is chosen.


WhereP = probabilityX = binomial discrete random variabler = number of successes (r = 0, 1, 2, …., n)n = number of trialsp = probability of success (0 < p < 1)q = probability of failure (q = 1 - p)



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p =20

1

q = 1 –20

1=

20

19

r = 3n = 10



P(X = 3) = 10 C 3 (20

1) 3 (

20

19) 310

= (120)x(0.000125)x(0.6983)

= 0.0105

Example 3 :

65% of Form Five students of a school pass the SPM Additional Mathematics paper. If asample of 5 student is chosen at random, calculate the probability that all of them pass theSPM Additional Mathematics paper.

Solution :


p = 65% = 0.65q = 1 – 0.65 = 0.35r = 5n = 5



P(X = 5) = 5 C 5 (0.65) 5 (0.35) 55

= (1)x(0.116)x(1)

= 0.116



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Example 4 :

The probability that Danial will win a tennis competition is 0.8. If a total of 5 games are played,find the probability that Danial will win

(a) exactly 3 games,(b) at least three games,(c) not more than 3 games.

Solution :

(a) P(X = r) = n C r p r q rn

P(X = 3) = 5 C 3 (0.8) 3 (0.2) 2

= (10)x(0.512)x(0.04)

= 0.2048

(b) P(X 3) = P(X = 3) + P(X = 4) + P(X = 5)

= [ 5 C 3 (0.8) 3 (0.2) 2 ] + [ 5 C 4 (0.8) 4 (0.2) 1 ] + [ 5 C 5 (0.8) 5 (0.2) 0 ]

= 0.2048 + 0.4096 + 0.3277

= 0.9421

(c) P(X 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

= [ 5 C 0 (0.8) 0 (0.2) 5 ] + [ 5 C 1 (0.8)1 (0.2) 4 ] + [ 5 C 2 (0.8) 2 (0.2) 3 ] + [ 5 C 3 (0.8) 3 (0.2) 2 ]

= 0.00032 + 0.0064 + 0.0512 + 0.2048

= 0.2627

ALTERNATIVE METHOD

(d) P(X 3) = 1 - P(X = 4) - P(X = 5)

= 1 - 0.4096 - 0.3277

= 0.2627



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13.3 ACTIVITY 1

1. During a shooting competition, the probability that Rasyidi will strike the targetis 0.8. If Rasyidi fires 8 shoots, calculate the probability that exactly 7 shotsstrike the target.

2. In a certain school, 4 out 10 students have a computer at home. Calculate theprobability that from a sample of 5 students, none of them have a computer athome.

3. The probability that Rafieq will win a badminton competition is 60%. If a totalof 7 games are played, find the probability that Rafieq will win

(a) exactly 4 games,(b) at least 5 games,(c) not more than 4 games.



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13.4 Binomial Distribution Graph

Example :

A fair coin is tossed 4 times continuously. X represents the number of times a head appears.

(a) List the possible elements of X.(b) Calculate the probability for the occurrence of each element of X.(c) Hence, plot a graph to represent the binomial probability distribution of X.

Solution :

(a) Since the coin is tossed 4 times continuously, X = { 0, 1, 2, 3, 4 }

(b) P(X = r) = n C r p r q rn

n = 4, p =2

1, q = 1 -

2

1=

2

1

P(X = 0) = 4 C 0 (2

1) 0 (

2

1) 4 = 0.0625

P(X = 1) = 4 C 1 (2

1)1 (

2

1) 3 = 0.25

P(X = 2) = 4 C 2 (2

1) 2 (

2

1) 2 = 0.375

P(X = 3) = 4 C 3 (2

1) 3 (

2

1)1 = 0.25

P(X = 4) = 4 C 4 (2

1) 4 (

2

1) 0 = 0.0625

(c)r 0 1 2 3 4

P(X = r) 0.0625 0.25 0.375 0.25 0.0625

The graph that represents the binomial probability distribution of X is as follows.



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13.5 Mean, Variance, Standard Deviation of Binomial Distribution

Example 1 :

40% of the students in a school wear spectacles. From a sample of 10 students, calculate themean, variance and standard deviation of the number of students who wear spectacles.

Solution :

p = 40% = 0.4, q = 1 – 0.4 = 0.6, n = 10

Mean, = np Variance, 2 = npq

= 10 x 0.4 = 10 x 0.4 x 0.6= 4 = 2.4

Standard deviation, = npq

= 4.2= 1.549

0 1 2 3 4r

P(X= r)

0.5

0.4

0.3

0.2

0.1

0.0

= np 2 = npq = npq



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Example 2 :

In a group of teachers, the mean and variance of the number of teachers who own a Proton carare 6 and 2.4 respectively. Find the probability that a teacher chosen at random owns a Protoncar.

Solution :

Mean, = np = 6

np = 6 --------- (1)

Variance, 2 = npq = 2.4npq = 2.4 --------(2)

)1(

)2(,

np

npq=

6

4.2

q = 0.4p = 1 – 0.4 = 0.6

Hence, the probability that a teacher chosen at random owns a Proton car is 0.6.

13.6 ACTIVITY 2

1. The probability that a papaya chosen at random from a basket is rotten is16

1. If there are 20

papayas in the basket, calculate the mean and standard deviation of the rotten papayas in thebasket.



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2. For a binomial distribution, it is given that mean = 10 and variance = 4, p = probability ofsuccess and q = probability of failure. Find

(a) the value of p and q(b) the probability of obtaining 2 successes out of 10 experiments.

3. In a farm, 45% of the chicks hatched from eggs are males.(a) If 7 eggs are chosen at random, calculate the probability that 2 or more male chicks are

hatched.(b) If there are 1000 eggs in the farm, calculate the mean and standard deviation of the number

of male chicks are hatched.



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13.7 ACTIVITY 3 : SPM FOCUS PRACTICE

1. SPM 2003, PAPER 1. QUESTION 25.

In an examination, 70% of the students passed. If a sample of 8 students is randomly selected,find the probability that 6 students from the sample passed the examination.

[3 marks]


(a) Senior citizens make up 20% of the population of a settlement.(i) If 7 people are randomly selected from the settlement, find the probability that at

least two of them are senior citizens.(ii) If the variance of the senior citizens is 128, what is the population of the

settlement?



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(a) A club organizes a practice session for trainees on scoring goals from penalty kicks. Eachtrainee takes 8 penalty kicks. The probability that a trainee scores a goal from a penalty kickis p. After the session, it is found that the mean number of goals for a trainee is 4.8.(i) Find the value of p.(ii) If a trainee is chosen at random, find the probability that he scores at least one goal.


For this question, give your answer correct to three significant figures.(a) The result of a study shows that 20% of the pupils in a city cycle to school. If 8 pupils from

the city are chosen at random, calculate the probability that(i) exactly 2 of them cycle to school,(ii) less than 3 of them cycle to school.



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13.8 ANSWERS

ACTIVITY 1:

1. 0.3355 2. 0.07776

3. (a) 0.2903(b) 0.4199(c) 0.5801

ACTIVITY 2:

1. = 1.25

σ = 1.083

2. (a) p = 0.6, q = 0.4(b) 0.9983

3. (a) 0.8976(b) 15.73

ACTIVITY 3:

1. 0.2965 2. (a) (i) 0.4233(ii) 800

3. (a) (i) p = 0.6(ii) 0.993

4. (a) (i) 0.294(ii) 0.797



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PROGRAM DIDIK CEMERLANG AKADEMIK

SPM

PROBABILITYDISTRIBUTIONS

ADDITIONAL MATHEMATICSFORM 5

MODULE 14



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MODULE 14 : PROBABILITY DISTRIBUTIONS

CONTENT PAGE

14.1. CONCEPT MAP2

14.2. PROBABILITY IN NORMAL DISTRIBUTION 3

14.3 Score- z 7

1 14.4 ACTIVITY 1 9

14.5 ACTIVITY 2 10

14.6. ACTIVITY 311

14.7 SPM QUESTIONS 13

14.8. SELF ASSESSMENT15

14.9. ANSWERS18



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14.1 CHAPTER 8

CONCEPT MAP

Standardised NormalDistribution

Z ~N (0,1)

Score- z

Probabilityin Normal Distribution

PROBABILITY DISTRIBUTIONS

NORMAL DISTRIBUTION

X ~N ( , 2 )



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14.2 Probability in Normal Distribution

Standardised Normal Distribution



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Example 1 :

Find the value of each of the following probabilities by reading the standardised normaldistribution table.

(a) P(Z > 0.934)

(b) P(Z 1.25)

Solution

(b) P(Z 1.25) = 1 – P(Z > 1.25)= 1 – 0.1057= 0.8944

1.25 1.25



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(c) P(Z - 0.23)

Solution

(c) P(Z - 0.23) = 1 – P(Z < - 0.23)= 1 – P(Z > 0.23)= 1 – 0.40905= 0.59095

(d) P(Z > - 1.512)

Solution

(d) P(Z < - 1.512) = P(Z > 1.512)= 0.06527

(e) P(0.4 < Z < 1.2)

Solution

(e) P(0.4 < Z < 1.2) = P(Z > 0.4) – P(Z > 1.2)= 0.3446 – 0.1151= 0.2295

-1.512 1.512

-0.23 0.23

0.4 1.2 0.4 1.2



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(f) P(- 0.828 < Z - 0. 555)

Solution

(f) P(- 0.828 < Z - 0. 555) = P(Z > 0.555) – P(Z > 0.828)= 0.28945 – 0.20384= 0.08561

(g) P(- 0.255 Z < 0.13)

Solution

(g) P(- 0.255 Z < 0.13) = 1 – P(Z < - 0.255) – P(Z > 0.13)

= 1 – P(Z > 0.255) – P(Z > 0.13)= 1 – 0.39936 – 0.44828= 0.15236

-0.828 -0.555 0.555 0.828

-0.255 0.13 0.13-0.255



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14.3 Score- z

Example 2 :

Find the value of each of the following :

(a) P(Z z) = 0.2546(b) P(Z < z) = 0.0329(c) P(Z < z) = 0.6623(d) P(z < Z < z 0.548) = 0.4723

Solution

(a) P(Z z) = 0.2546Score-z = 0.66

(b) P(Z < z) = 0.0329Score-z = -1.84

(c) P(Z < z) = 0.66231 - P(Z > z) = 0.6623

P(Z > z) = 1 – 0.6623= 0.3377

Score-z = 0.419

(d) P(z < Z < z 0.548) = 0.47231 – P(Z < z) – P(Z > 0.548) = 0.4723

1 – P(Z < z) – 0.2919 = 0.4723P(Z < z) = 1 – 0.2919 – 0.4723

= 0.2358Score-z = -0.72

z

0.2546



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Example 3 :

The masses of the loaves of bread produced by a bakery are normally distributed with amean of 400 g and a standard deviation of 15 g. Calculate

(a) the standardised score for the mass of 405 g,(b) the probability that a loaf of bread chosen at random will have a mass of more than

405 g,(c) the percentage of loaves of bread that have masses of less than 403 g,(d) the number of the loaves of bread that have masses between 394 g and 409 g if 1000

loaves of bread are produced in a day.

Solution

(a) X = 405, = 400, = 15

The standardised score,

Z =

X=

15

400405 =

3

1

(b) P(X > 405)

= P

15

400405Z

= P (Z > 0.3333)= 0.3696

(c) P(X < 403)

= P

15

400403Z

= P (Z < 0.2)= 1 – P(Z > 0.2)= 1 – 0.4207= 0.5793= 0.5793 x 100%= 57.93%

(d) P(394 < X < 409)

= P

15

400409

15

400394Z

= P(-0.4 < Z < 0.6)= 1 - P(Z > 0.4) – P(Z > 0.6)= 1 – 0.3446 – 0.2743= 0.3811



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Hence, if 1000 loaves of bread are produced, the number of loaves that masses between394 g and 409 g is 0.3811 x 1000 = 381.1 = 381

14.4 Activity 8.1

If Z is the variable for standard normal distribution, find the value for each of thefollowing:

1.P( Z> 0.637) 2. P( Z> 0.1)

3. P( Z2.018) 4 P( Z<-0.5).

5 P( Z -0.34) 6. P( Z -3.47)

7. P( -0.225Z< 0.135) 8. P(-0.25< Z< 0)



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14.5 Activity 8.2

Find the z-score for each of the following:

1. P (Z< z) = 0.5987 2. P (Z< z) = 0.1515

3.P (Z< z) = 0.8599 4. P (Z< z) = 0.4247

5.P ( z< Z< 0.683) = 0.4143 6. P (Z z) = 0.99865

7. P (0.5<Z< z) = 0.0342 8. P (Z z) = 0.9898



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14.6 Activity 8.3

1. Determine the probability for each of the following if

Mean =55, standard deviation = 5

(i) P(X < 68)

(ii) P(X > 56)

(iii) P(58<X<63)

2. In a normal distribution of a variable X,the standard scores of2

1and 1 have x

scores .11 and 12 respectively. Find the mean and standard deviation of the normal

distribution.



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3. The additional mathematics mark for a group of students has a normal distribution

and the mean is 48 while the standard deviation is 5. Determine the probability for

a student to have the mark of

(a) greater than 55

(b) between 40 and 52

4. Find the probability if:

Mean = 43; standard deviation = 8

(i) P (X>35)

(ii) P(X<28)

(iii) P( X > 33)



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14.7

1.The mass of the workers in a factory is normally distributed with a mean of 67.86 kgand a variance of 42.25kg².200 of the workers in the factory weigh between 50 kg and 70kg. Find the total number of worker in the factory.

2.A survey on body-mass is done on a group of students has a normal distribution with amean of 50 kg and a standard deviation of 15kg.(i) If a student is chosen at random,calculate the probability that his mass is less than 41kg.(ii) Given that 12% of the students have a mass of more than m kg, find the value of m.



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3.The mass of water- melon chosen randomly from the orchard follows a normaldistribution with a mean of 3.2 kg and a standard deviation of 0.5kg. Find(i) the probability that a water-melon chosen randomly from the orchard has a mass ofnot more than 4.0kg(ii) the value of m if 60% of the water-melons from the orchard have a mass of morethan m kg.



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14.8 SELF ASSESSMENT

1. The life span of a certain machine is normally distributed with mean 1500 days

and standard deviation 30 days.

(a) what is than probability that a machine chosen at random has a life spam of

(i) more than 1532 days

(ii) between 1480 days and 1530 days

(b) Given that 5% of the machines have than spans of more than n days, find the

value of n



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2.The weight of fish reared by a farmer are normally distributed with mean 1.2 kg and

standard deviation 0.1 kg. Calculate

(a) the probability that a fish chosen at random has a weight that is between 0.9kg

and 1.3kg

(b) the percentage of fish with weights less than 1.25 kg

Answer;



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3.The result of a test is normally distributed with mean 55 and standard deviation

10. If the passing mark is at least 40, find the probability of the chosen student

passed the test.

4.The probability for Mazlan whom threw a stone to hit the target is 0.65.Find the

quantity of stones for Mazlan to hit the target at least once so that the probability would

be greater than 0.9



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ANSWERS14.9 Activity 8.1

1.0.2620 2. 0.4602

3. 0.9782 4. 0.3085.

5. 0.6331 6. 0.99974

7. 0.1427 8. 0.0987

Activity 8.2

1. 0.25 2. -1.03

3. -1.08 4. -0.19

5. -0.417 6. -3.00

7. 0.600 8. 2.32

Activity 8.3

1. (i) 0.99534 (ii) 0.4207 (iii) 0.2195

2. 2,10

3. (a) 0.0808(b) 0.7333

4. (i) 0.8413(ii) 0.0303(iii) 0.7888



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SELF ASSESSMENT

1. (a) (i) 0.1430,(ii) 0.589,

(b) n=1549

2. (a) 0.84,(b) 69.15%

3. 9332

4. 3

SPM QUESTIONS

1. 319 workers

2. (i) 0.2743,(ii) m=67.625kg

3. (i) 0.9452,(ii) m=3.0735



module 18 probability distributions

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