Chapter 6: Standard Scores and the Normal Curve Normal distributions show up all over the place.

Chapter 6: Standard Scores and the Normal Curve

Normal distributions show up all over the place.

The standard normal distribution

The standard normal distribution is a continuous distribution.It has a mean of 0 and a standard deviation of 1The total area under the curve is equal to 1

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40 45 50 55 60 65 70 75 80 85 90 95 100Preferred outdoor temperature

-4 -3 -2 -1 0 1 2 3 4z-score

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z

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If we assume that our population is distributed normally, then all we need is the population mean (m) and standard deviation (s) to convert to z-scores for the standard normal distribution. We can then find areas under the curve using Table A.

Population (X) Standard normal (z)

All normal distributions can be converted to a ‘standard normal’ by converting to z scores. We can then look up values in a table in the back of the book (or use a computer).

Example: Suppose the population of heights of U.S. females is normally distributed with a mean of 64 inches with a standard deviation of 2.5 inches. What percent of women are between 61.5 and 66.5 inches?

Answer: 61.5 inches is exactly one standard deviation below the mean, and 66.5 inches is exactly one standard deviation greater than the mean.

We learned in Chapter 5 that 68% of scores fall within +/- one standard deviation of the mean, so the answer is 68%

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68%

Using Table A (page 436) to find the area under the standard normal curve

For any given z score, the table gives you two numbers:

Column 2: area between 0 and z Column 3: area above z

Note that the numbers in columns 2 and 3 always add up to 0.5

z score z score

Example: what is the proportion of area under the standard normal above z=1?

Look up z=1 in table A (page 436), Column 3. The area is .1587, so 15.87% of the area of the curve lies beyond one standard deviations above the mean.

-4 -3 -2 -1 0 1 2 3 4z score

Example: Suppose the population of heights of U.S. females is normally distributed with a mean of 64 inches with a standard deviation of 2.5 inches. What percent of women are taller than 66.5 inches?

Answer: 65.5 inches is 2.5 inches greater than the mean, which is exactly one standard deviation. From the last example, we know that the area beyond z=1 is .1587. 15.87% of women are taller than 65.5 inches.

54 56.5 59 61.5 64 66.5 69 71.5 74Height

-4 -3 -2 -1 0 1 2 3 4z score

15.87% 15.87%

-4 -3 -2 -1 0 1 2 3 4z score

-4 -3 -2 -1 0 1 2 3 4z score

54 56.5 59 61.5 64 66.5 69 71.5 74Height

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In general, we can calculate the z-score for 60 inches with our formula:

Example: Suppose the mean height of a U.S. female is 64 inches with a standard deviation of 2.5 inches. What proportion of women are shorter than 5 feet (60 inches)?

Answer: Convert the height to a z score:

The area below z=-1.6 is equal to the area above z=1.6. Table A, Column 3: area = .0548. 5.48% of women are shorter than 5 feet.

5.48% 5.48% 5.48%

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X

xXz

40 55 70 85 100 115 130 145 160IQ

Over 140 - Genius or near genius 120 - 140 - Very superior intelligence 110 - 119 - Superior intelligence 90 - 109 - Normal or average intelligence 80 - 89 - Dullness 70 - 79 - Borderline deficiency Under 70 - Definite feeble-mindedness

Example: Most IQ tests are standardized to have a mean of 100 and a standard deviation of 15. If we assume that IQ scores are normally distributed, what percent of the population has ‘very superior intelligence’, or an IQ between 120 and 140?

Answer: The area between 120 and 140 is equal to the area above 120 minus the area above 140. z scores for IQs of 120 and 140 are:

Looking at table A, column 3:the proportion above z1=1.33 is .0918the proportion above z2=2.67 is .0038The difference is .0918-.0038 = .0880. 8.8% of the population has ‘very superior intelligence’

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Example: Again, assume IQ scores are normally distributed with m = 100, and s = 15. For what IQ does 90% of the population exceed?

Answer: We need to use table A backwards.

First we’ll find the z score for which the area beyond z is 0.1. This is a value of z=1.28. This means that the area below z=-1.28 is 0.1, and the area above z=-1.28 is 0.90.

Next we’ll convert this z score back to IQ scores. This can be done by reversing the formula that converts from X to z. Solving for X:

Using the new formula:

90% of the population has an IQ above 80.8 points

More examples:

The heights of fathers of the 90 students in this class has a mean of 69 inches and a standard deviation of 3.84 inches. If we assume that this sample is normally distributed:

1) Estimate how many fathers are taller than six feet tall (72) inches?

Answer: Find the z value for a height of 72 inches and use column 3

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Table A, column 3: the proportion above z=0.7 is .2148

# fathers: (90)(.2148)= 19.33

Round to the nearest father : 1957.6 61.4 65.2 69 72.8 76.6 80.4

area =0.2148

z

More examples:

The heights of fathers of the 90 students in this class has a mean of 69 inches and a standard deviation of 3.8 inches. If we assume that this sample is normally distributed.

2) Estimate how many fathers are shorter than 73 inches.

Answer: Find the z value for a height of 73 inches and use column 2 and add 50%

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xXz

Table A, Column 2: the proportion between the mean and z=1.05 is .3537.

To find the area below z = 1.05, we need to add another 0.5 to .3531 which gives .8531

#fathers: 90 x .8531= 76.78,Rounds to 77 fathers

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p=0.5

p=0.3537

z

More examples:

The preferred outdoor temperature for the 91 students in this class has a mean of 72.9 degrees and a standard deviation of 9.43 degrees. If we assume that this sample is normally distributed:

1 How many students prefer a temperature between 55 and 65 degrees?

Answer: Find the z values for 55 and 65 and use table A, column 3

44.61 54.04 63.47 72.9 82.33 91.76 101.19

area =0.1718

Temperature (F)

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To find the area below z1 = -1.9, we find the area above z = 1.9 which is .0287

The area below z2 = -.84, is .2005

Rounding to the nearest student: (0.1718)(91) = 16 students

More examples:

The preferred outdoor temperature for the 106 students in this class has a mean of 75 degrees and a standard deviation of 8 degrees. If we assume that this sample is normally distributed.

2) For what temperature do 90% of the preferred temperatures fall below?

Answer: Find the z value that corresponds to the top 90% of the standard normal Then find the corresponding temperature using:

xx zuX

44.61 54.04 63.47 72.9 82.33 91.76 101.19

area =0.9

Temperature (F)

Using table A backwards, the z-score for an area of 0.9 is z = 1.28.

Converting to temperature, 72.9+(1.28)(9.43) = 85 degrees.