Section 5.2 ~ Properties of the Normal Distribution

Introduction to Probability and StatisticsMs. Young

Objective

Sec. 5.2

In this section you will learn how to interpret the normal distribution in terms of the 68-95-99.7 rule, standard scores, and percentiles.

Properties of the Normal Distribution

Sec. 5.2

Recall the following notations: The population mean is represented using the Greek letter mu

The population standard deviation is represented using the Greek letter sigma

The following diagram is an example of a distribution that represents the replacement time (in years) for a TV

Properties of the Normal Distribution

Sec. 5.2

Knowing the mean and standard deviation of a distribution allows us to say a lot about where the data values lie Example:

The mean TV replacement time is 8.2 years with a standard deviation of 1.1 years

About two-thirds of all values fall within one standard deviation, so the replacement time for about two-thirds of the TV’s is between 7.1 years and 9.3 years (8.2 - 1.1 = 7.1 and 8.2 + 1.1 = 9.3)

About 95% of the data fall within two standard deviations of the mean, so the replacement time for 95% of the TV’s in this sample is between 6.0 years and 10.4 years (8.2 – 2.2 = 6.0 and 8.2 + 2.2 10.4)

The 68-95-99.7 Rule

Sec. 5.2

The 68-95-99.7 Rule gives precise guidelines for the percentage of data values that lie within 1, 2, and 3 standard deviations of the mean for any normal distribution About 68% (more precisely, 68.3%), or just over two-

thirds, of the data points fall within 1 standard deviation of the mean

About 95% (more precisely, 95.4%) of the data points fall within 2 standard deviations of the mean

About 99.7% of the data points fall within 3 standard deviations of the mean

Example 1

Sec. 5.2

If there are 600 total values. Approximately, how many values occur within 1 standard of deviation from the mean? 2 standards of deviation? 3 standards of deviation?

Since 68% of the data values lie within 1 standard deviation, 408 values fall within 1 standard deviation (.68 * 600 = 408)

Since 95% of the data values lie within 2 standard deviations, 570 values fall within 2 standard deviations (.95 * 600 = 570)

Since 99.7% of the data values lie within 3 standard deviations, 598.2 values fall within 3 standard deviations (.997 * 600 = 598.2)

Example 2

Sec. 5.2

Vending machines can be adjusted to reject coins above and below certain weights. The weights of legal U.S. quarters have a normal distribution with a mean of 5.67 grams and a standard deviation of 0.0700 gram. If a vending machine is adjusted to reject quarters that weigh more than 5.81 grams and less than 5.53 grams, what percentage of legal quarters will be rejected by the machine?

A weight of 5.81 is 0.14 gram (5.81 – 5.67 = 0.14), or 2 standard deviations above the mean and a weight of 5.53 is 0.14 gram (5.53 - 5.67 = -0.14), or 2 standard deviations below the mean

Since 95% of the data values fall within 2 standard deviation of the mean, the machine will accept 95% of the quarters inserted and will reject 5%

Unusual Values

Sec. 5.2

Unusual values are values that are more than 2 standard deviations away from the mean

Example 3 ~ Traveling and Pregnancy Suppose your friend is pregnant and is trying to determine if she

should schedule an important business meeting two weeks before her due date. Data suggests that the number of days between the birth date and due date is normally distributed with a mean of μ = 0 days and a standard deviation of σ = 15 days. How would you help your friend make the decision? Would a birth two weeks before the due date be considered “unusual”?

A birth 2 weeks before the due date is approximately 1 standard deviation below the mean

Since 68% of births will occur within 15 days of the due date (in either direction), that means that a total of 32% of births will occur more than 15 days from the due date (in either direction)

Since the distribution is symmetric, that means that 16% of births will occur more than 15 days early

To help your friend decide if it would be safe to schedule a meeting 14 days before her due date, you could tell her that 16% of births will occur more than 15 days early, which equates to about 1 in 6

Since a birth date that is two weeks early falls well within 2 standard deviations, this would not be considered unusual

Example 4

Sec. 5.2

Suppose you measure your heart rate at noon every day for a year and record the data. You discover that the data has a normal distribution with a mean of 66 and a standard of deviation of 4. On how many days was your heart rate below 58 beats per minute?

58 falls 2 standard deviations below the mean (58 – 66 = -8), so 95% of time your heart rate will be between 58 bpm and 74 bpm, therefore 5% of the time your heart rate will either be above 74 bpm or below 58 bpm

Since the distribution is symmetric, the area below 2 standard deviations would be 2.5% (5/2 = 2.5), so 2.5% of the time your heart rate will fall below 58 bpm

Since there are 365 days in a year (with the exception of leap year), approximately 9 days (.025 * 365 = 9.125) out of the year your heart rate will be below 58 bpm

Example 5

Sec. 5.2

Suppose that your fourth-grade daughter is told that her height is 1 standard deviation above the mean for her age and sex. What is her percentile for height? Assume that heights of fourth-grade girls are normally distributed

Since 68% of fourth-grade girls lie within 1 standard deviation (both directions), then 34% of fourth-grade girls lie within 1 standard deviation above the mean

Since the data is symmetric, 50% of fourth-grade girls will fall below the mean

If your daughter’s height is 1 standard deviation above the mean, then she is taller than 84% (50% + 34% = 84%) of fourth-grade girls, which means that she is in the 84th percentile

Standard Scores

Sec. 5.2

A standard score (or z-score) is the number of standard deviations a data value lies above or below the mean The standard score is represented using z

The standard score is positive for values above the mean and negative for values below the mean Examples ~

A value that is 2 standard deviations above the mean has a standard score of 2, or z = 2

A value that is 2 standard deviations below the mean has a standard score of -2, or z = -2

The standard score of the mean will be z = 0, since it’s 0 standard deviations from the mean

The standard score is found using the following formula:data value - mean

= standard score = standard deviation

zx

z

Example 6

Sec. 5.2

The Stanford-Binet IQ test is scaled so that scores have a mean of 100 and a standard deviation of 16. Find the standard scores (z-scores) for IQs of 85, 100, and 125.

Standard score for 85:

Standard score for 100:

Standard score for 125:

data value - mean = standard score =

standard deviationz

85 1000.94

16z

100 1000.00

16z

125 1001.56

16z

Standard Scores and Percentiles

Sec. 5.2

Once you know the standard score of a data value, the properties of the normal distribution allow you to find its percentile

This is typically done with a standard score table (refer to Appendix A on p.446) The table gives the percentage of values less than that

value Ex. ~ a value that has a standard score of 0.15, has

55.96% of the total values falling below it, and it therefore lies in the 55th percentile

Example 7

Sec. 5.2

Cholesterol levels in men 18 to 24 years of age are normally distributed with a mean of 178 and a standard deviation of 41. What is the percentile for a 20-year old man with a cholesterol

level of 190? The standard score for a cholesterol level of 190 is:

If you look at the chart, you see that a standard score of .29 has 61.41% of values falling below, which means that it falls in about the 61st percentile

What cholesterol level corresponds to the 90th percentile, the level at which treatment may be necessary?

According to the chart, a standard score of 1.29 has 90.15% of the values falling below it making it fall in the 90th percentile

If the mean cholesterol level is 178 and the standard deviation is 41, then 1.29 standard deviations would equate to a level of 52.89 (41 * 1.29 = 52.89)

A level of about 231 (178 + 52.89 = 230.89) would be in the 90th percentile

190 1780.29

41z

Example 8

Sec. 5.2

The heights of American women aged 18 to 24 are normally distributed with a mean of 65 inches and a standard deviation of 2.5 inches. In order to serve in the U.S. Army, women must be between 58 inches and 80 inches tall. What percentage of women are ineligible to serve based on their height?

If a woman is shorter than 58 inches or taller than 80 inches she would be ineligible to serve in the Army

The standard score for a height of 58 inches is:

After looking at the table, you can see that .26% of women are too short to serve in the Army

The standard score for a height of 80 inches is:

Since a standard score of 6.0 does not appear on the table, it means that it is above the 99.98th percentile in which case fewer than 0.02% (100% - 99.98% = 0.02%) of women are too tall to serve in the Army

All together, fewer than .28% (.26% + .02% = .28%) of women are ineligible to serve in the Army

58 652.8

2.5z

80 656.0

2.5z