Section 7.3

Areas Under Any Normal Curve

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To Work with Any Normal Distributions

• Convert x values to z values using the formula:

Then use Table 3 of the Appendix to find corresponding areas and probabilities.

xz

Rounding

• Round z values to the hundredths positions before using Table 3.

• Leave area results with four digits to the right of the decimal point.

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ExampleLet x be a normal distribution with μ = 10 and σ = 2.

Find the probability that an x value selected at random from this distribution is between 11 and 14.

SolutionSince probabilities correspond to areas under the distribution curve,

we want to find the area under the x curve above the interval x = 11 and x = 14.

To do so, we will convert the x values to standard z values and then use table 3 of the Appendix to find the corresponding area under the standard curve.

For x = 11 for x = 14

The interval Corresponds to interval

11 100.5

2

xz

14 10

22

xz

(11 14)x (0.5 2)z (11 14) (0.50 2.00) ( 2.00) ( 0.50) 0.9772 0.6915 0.2857P x P z P z P z

(11 14)P x

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5

Application of the Normal Curve

The amount of time it takes for a pizza delivery is approximately normally distributed with a mean of 25 minutes and a standard deviation of 2 minutes. If you order a pizza, find the probability that the delivery time will be:

a. between 25 and 27 minutes. a. __________

b. less than 30 minutes. b. __________

c. less than 22.7 minutes. c. __________

.3413

.9938

.1251

Inverse Normal Probability Distribution

• Finding z or x values that correspond to a given area under the normal curve

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7

Look up area A in body of Table 3 and use corresponding z value.

Inverse Normal Left Tail Case

8

Look up the number 1 – A in body of Table 3 and use corresponding z value.

Inverse Normal Right Tail Case:

9

Look up the number (1 – A)/2 in body of Table 3 and use corresponding ± z value.

Inverse Normal Center Case:

Using Table 3 for Inverse Normal Distribution

• Use the nearest area value rather than interpolating.

• When the area is exactly halfway between two area values, use the z value exactly halfway between the z values of the corresponding table areas.

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When the area is exactly halfway between two area values

• When the z value corresponding to an area is smaller than 2, use the z value corresponding to the smaller area.

• When the z value corresponding to an area is larger than 2, use the z value corresponding to the larger area.

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12

Find the indicated z score:

z = _______– 2.57

13

z = _______2.33

Find the indicated z score:

14

Find the indicated z scores:

z = ____–z = _____–1.23 1.23

Look up the number (1 – A)/2 in body of Table 3 and use corresponding ± z value.

(1-.7814)/2 = .1093

Solution:

15

Find the indicated z scores:

± z =__________ ± 2.58

Solution: 99% = .9900 and (1-.9900)/2 = .0100/2 = .0050

16

± z = ________± 1.96

Find the indicated z scores:

Solution: 95% = .9500 and (1-.9500)/2 = .0500/2 = .0250

Application of Determining z Scores

The Verbal SAT test has a mean score of 500 and a standard deviation of 100. Scores are

normally distributed. A major university determines that it will accept only students whose Verbal SAT scores are in the top 4%. What is the minimum score that a student

must earn to be accepted?

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The cut-off score is 500 + 1.75(100) = 675.

Mean = 500

standard deviation = 100

Application of Determining z Scores

The cut-off score is 1.75 standard deviations above the mean.

7.3 / 18Assignment 15, 16 and 17

Example • Graph and examine a situation where the mean score is 46 and the

standard deviation is 8.5 for a normally distributed set of data.• Go to Y= normalpdf(X,46,8.5) • Adjust the window.• [20.5 , 71.5] x [0 , 0.0588] Xscl=8.5 Yscl=0 Xres=1• GRAPH.

Examine: What is the probability of a value falling between the mean and the first standard deviation to the right?

• Answer:  approximately 34%

• Notice how this percentage supports the information found in the chart at the top of this page for the percentage of information falling within one standard deviation above the mean.

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Example• The lifetime of a battery is normally distributed with

a mean life of 40 hours and a standard deviation of 1.2 hours. Find the probability that a randomly selected battery lasts longer than 42 hours.

• The most accurate answer to this problem cannot be obtained by using the chart of the standard normal distribution. One standard deviation above the mean would be located at 41.2 hours, 2 standard deviations would be at 42.4, and one and one-half standard deviations would be at 41.8 standard deviations. None of these locations corresponds to the needed 42 hours. We need more power than we have in the chart. Calculator to the rescue!!

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Example cont.• Let's get a visual look at the situation by examining the graph. The location

of 42 hours indicates that our answer is going to be quite small.• Go to Y= normalpdf(X,40,1.2). • GRAPH.

• NOTE: If you do not graph first, you may not see the normal curve displayed in the answer due to the previously set window on the calculator.

• Now: What is the probability of a value falling to the right of 42 hours (between 42 hours and infinity)? Answer:  approximately 4.8%

•ShadeNorm( go to DISTR andright arrow to DRAW. Choose #1:ShadeNorm(42,1E99,40,1.2).

•ENTER. The percentage isread from the Area =.

• ShadeNorm (lower bound, upperbound, mean, standard deviation) 7.3 / 21