Section 6.4

Applications of the Normal Distribution

ObjectivesFind the probabilities for a normally

distributed variable by transforming it into a standard normal variable

Find specific data values given the percentages, using the standard normal distribution

Key ConceptThis section presents methods for working

with normal distributions that are not standard (NON-STANDARD). That is the mean, , is not 0 or the standard deviation, is not 1 or both.

The key concept is that we transform the original variable, x, to a standard normal distribution by using the following formula:

Conversion Formula

placesdecimaltoscoreszRound

x

deviationdards

meanvalueoriginalz

2

tan

Converting to Standard Normal Distribution

x 0 z

x -

z =

(a) (b)

P P

Cautions!!!!Don’t confuse z-

scores and areas Z-scores are

distances on the horizontal scale. Table E lists z-scores in the left column and across the top row

Areas (probabilities or percentages) are regions UNDER the normal curve. Table E lists areas in the body of the table

Choose the correct (left/right) of the graphNegative z-score

implies it is located to the left of the mean

Positive z-score implies it is located to the right of the mean

Area less than 50% is to the left, while area more than 50% is to the right

Areas (or probabilities) are positive or zero values, but they are never negative

Finding Areas given a specified variable, x Draw a normal distribution curve labeling

the mean and x Shade the area desiredConvert x to standard normal distribution

using z-score formula Use procedures on page 287-288

ExampleAccording to the American College Test (ACT),

results from the 2004 ACT testing found that students had a mean reading score of 21.3 with a standard deviation of 6.0. Assuming that the scores are normally distributed:Find the probability that a randomly selected

student has a reading ACT score less than 20Find the probability that a randomly selected

student has a reading ACT score between 18 and 24

Find the probability that a randomly selected student has a reading ACT score greater than 30

ExampleWomen’s heights are normally distributed

with a mean 63.6 inches and standard deviation 2.5 inches. The US Army requires women’s heights to be between 58 inches and 80 inches. Find the percentage of women meeting that height requirement. Are many women being denied the opportunity to join the Army because they are too short or too tall?

Finding variable x given a specific probability Draw a normal distribution labeling the

given probability (percentage) under the curve

Using Table E, find the closest area (probability) to the given and then identify the corresponding z-score (WATCH SIGN!!!!)

Substitute z, mean, and standard deviation in z-score formula and solve for x.

ExampleAccording to the American College Test

(ACT), results from the 2004 ACT testing found that students had a mean reading score of 21.3 with a standard deviation of 6.0. Assuming that the scores are normally distributed:Find the 75th percentile for the ACT reading

scores

ExampleThe lengths of pregnancies are normally distributed

with a mean of 268 days and a standard deviation of 15 days. One classical use of the normal distribution is inspired by

a letter to “Dear Abby” in which a wife claimed to have given birth 308 days after a brief visit from her husband, who was serving in the Navy. Given this information, find the probability of a pregnancy lasting 308 days or longer. What does this result suggest?

If we stipulate that a baby is premature if the length of the pregnancy is in the lowest 4%, find the length that separates premature babies from those who are not premature. Premature babies often require special care, and this result could be helpful to hospital administrators in planning for that care

Example Men’s heights are normally distributed

with a mean of 69.0 inches and standard deviation of 2.8 inches.

The standard casket has an inside length of 78 inchesWhat percentage of men are too tall to fit in

a standard casket? A manufacturer of caskets wants to reduce

production costs by making smaller caskets. What inside length would fit all men except the tallest 1%?

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