section 07 02e (1)

SECTION 7.2APPLICATIONS OF THE NORMAL DISTRIBUTION

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Definition: z-score of x

Let x be a value from a normal distribution with mean µ and standard deviation σ. We can convert x to a z-score by using a method known as standardization.

The z-score of x is(x )z


Definition: z-score of x

Properties of the z-score1. The z-score follows a standard normal distribution.

2. Values below the mean have negative z-scores, and values above the mean have positive z-scores.

3. The z-score tells how many standard deviations the original value is above or below the mean.


Example 7.14Heights in a certain population of women follow a normal distribution with mean µ = 64 inches and standard deviation σ = 3 inches.

a. A randomly selected woman has a height of x = 67 inches. Find and interpret the z-score of this value.Solution: The z-score for x = 67 is z = (67–µ)/σ = (67–64)/3 = 1.00. Because the z-score is positive, we interpret this by saying that a height of 67 inches is 1 standard deviation above the mean height of 64 inches.

b. Another randomly selected woman has a height of x = 63 inches. Find and interpret the z-score of this value.Solution: The z-score for x = 63 is z = (63–µ)/σ = (63–64)/3 = –0.33. Because the z-score is negative, we interpret this by saying that a height of 63 inches is 0.33 standard deviation below the mean height of 64 inches.


Finding an Area Under a Normal Curve Using TablesWhen using tables to compute areas, we first standardize to z-scores, then proceed with the methods from the last section.


Example 7.15 (Tables)

A study reported that the length of pregnancy from conception to birth is approximately normally distributed with mean µ = 272 days and standard deviation σ = 9 days. What proportion of pregnancies last longer than 280 days?



Step 1: Find the z-score for x = 280.The z-score for 280 is

Step 2: Sketch a normal curve, label the mean, x-value, and z-score, and shade in the area to be found.

280 272 0.899

xz



Step 3: Since we are interested in the proportion of pregnancies lasting longer than 280 days, find the area to the right of z = 0.89.

Using Table A.2, we find the area to the left of z = 0.89 to be 0.8133. The are to the right is therefore 1 – 0.8133 = 0.1867. We conclude that the proportion of pregnancies that last longer than 280 days is 0.1867.



The length of a pregnancy from conception to birth approximately normally distributed with mean µ = 272 days and standard deviation σ = 9 days. A pregnancy is considered full-term if it lasts between 252 days and 298 days.

What proportion of pregnancies are full-term?



Step 1: Find the z-score for x = 252 and x = 298.

z-score for x = 252:

z-score for x = 298:

Step 2: Sketch a normal curve, label the mean, the x-values, and the z-scores, and shade in the area to be found.

252 272 2.229

z

298 272 2.899

z



Step 3: Using Table A.2, we find that the area to the left of z = 2.89 is 0.9981 and the area to the left of z = –2.22 is 0.0132. The area between z = − 2.22 and z = 2.89 is therefore 0.9981 – 0.0132 = 0.9849.

The proportion of pregnancies that are full-term, between 252 days and 298 days, is 0.9849.


Find the Value from a Normal Distribution with a Given ProportionSuppose we want to find the value from a normal distribution that has a given proportion of the population above or below it.

The method for doing this is the reverse of the previous method where we found a proportion for a given value. In other words, we want to find the value from the distribution that has a given z-score.

Therefore, we will solve z = (x – µ)/σ for x. This produces x = µ + zσ.The value of x that corresponds to a given z-score is

given by x = µ + zσ



Heights in a group of men are normally distributed with mean µ= 69 inches and standard deviation σ = 3 inches.

a. Find the height whose z-score is 1. Interpret the result.Solution:We use the invNorm command

b. Find the height whose z-score is – 2.0. Interpret the result.

c. Find the height whose z-score is 0.6. Interpret the result.


Example 7.18


a. Find the height whose z-score is 1. Interpret the result.

Solution:We want the height that is equal to the mean plus one standard deviation. Therefore, x = µ + zσ

= 69 + (1)(3) = 72

We interpret this by saying that a man 72 inches tall has a height one standard deviation above the mean.


Example 7.18


b. Find the height whose z-score is – 2.0. Interpret the result.

Solution:We want the height that is equal to the mean minus two standard deviation. Therefore, x = µ + zσ

= 69 + (– 2)(3) = 63

We interpret this by saying that a man 63 inches tall has a height two standard deviation below the mean.


Example 7.18


c. Find the height whose z-score is 0.6. Interpret the result.

Solution: We want the height that is equal to the mean plus 0.6 standard deviation. Therefore, x = µ + zσ

= 69 + (0.6)(3) = 70.8

We interpret this by saying that a man 70.8 inches tall has a height 0.6 standard deviation above the mean.


Finding a Normal Value That has a GivenProportion Above or Below It Using TablesThe following procedure can be use to find the value from a normal distribution that has a given proportion above or below it using Table A.2:

Step 1: Sketch a normal curve, label the mean, label the value x to be found,

and shade in and label the given area.Step 2: If the given area is on the right, subtract it from 1 to get the area on the left.Step 3: Look in the body of Table A.2 to find the area closest to the given area. Find the z-score corresponding to that area.Step 4: Obtain the value from the normal distribution by computing

x = µ + zσ



Mensa is an organization whose membership is limited to people whose IQ is in the top 2% of the population. Assume that scores on an IQ test are normally distributed with mean µ = 100 and standard deviation σ = 15. What is the minimum score needed to qualify for membership in Mensa?



Step 1: The figure below represents a sketch of the normal curve, showing the value x separating the upper 2% from the lower 98%.

Step 2: The area 0.02 is on the right, so we subtract from 1 and work with the area 0.98 on the left.



Step 3: The area closest to 0.98 in Table A.2 is 0.9798, which corresponds to a z-score of 2.05.

Step 4: The IQ score that separates the upper 2% from the lower 98% is

x = µ + zσ = 100 + (2.05)(15) = 130.75

Since IQ scores are generally whole numbers, we will round this to x = 131.


section 07 02e (1)

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