KEY2STATS

EXAMPLE 6.8

The final exam scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of five.

a. Find the probability that a randomly selected student scored more than 65 on the exam.

TThis diagram shows a bell-shaped curve with 63 located at the center of the X axis and 65 located a short distance to the right of 63. The area under the bell curve to the right of 65 is shaded. The label states: shaded area represents probability uppercase P(X > 65) = 0.3446

$\frac{65 – 63}{5}$

USING THE TI-83, 83+, 84, 84+ CALCULATOR

Find the percentile for a student scoring 65:

*Press 2nd Distr
*Press 2:normalcdf(
*Enter lower bound, upper bound, mean, standard deviation followed by )
*Press ENTER.
For this Example, the steps are
2nd Distr
2:normalcdf(65,1,2nd EE,99,63,5) ENTER

The probability that a selected student scored more than 65 is 0.3446.
To find the probability that a selected student scored more than 65, subtract the percentile from 1.

b. Find the probability that a randomly selected student scored less than 85.

c. Find the 90th percentile (that is, find the score k that has 90% of the scores below k and 10% of the scores above k).

This is a normal distribution curve. The peak of the curve coincides with the point 63 on the horizontal axis. A point, k, is labeled to the right of 63. A vertical line extends from k to the curve. The area under the curve to the left of k is shaded. This represents the probability that x is less than k: P(x < k) = 0.90

USING THE TI-83, 83+, 84, 84+ CALCULATOR

invNorm in 2nd DISTR. invNorm(area to the left, mean, standard deviation)
For this problem, invNorm(0.90,63,5) = 69.4

d. Find the 70th percentile (that is, find the score k such that 70% of scores are below k and 30% of the scores are above k).

TRY IT 6.8

The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three.

Find the probability that a randomly selected golfer scored less than 65.

EXAMPLE 6.9

A personal computer is used for office work at home, research, communication, personal finances, education, entertainment, social networking, and a myriad of other things. Suppose that the average number of hours a household personal computer is used for entertainment is two hours per day. Assume the times for entertainment are normally distributed and the standard deviation for the times is half an hour.

a. Find the probability that a household personal computer is used for entertainment between 1.8 and 2.75 hours per day.

This is a normal distribution curve. The peak of the curve coincides with the point 2 on the horizontal axis. The values 1.8 and 2.75 are also labeled on the x-axis. Vertical lines extend from 1.8 and 2.75 to the curve. The area between the lines is shaded.

b. Find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment.

This is a normal distribution curve. The area under the left tail of the curve is shaded. The shaded area shows that the probability that x is less than k is 0.25. It follows that k = 1.67.

TRY IT 6.9

The golf scores for a school team were normally distributed with a mean of 68 and a standard deviation of three. Find the probability that a golfer scored between 66 and 70.

EXAMPLE 6.10

In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years, respectively.

a. Determine the probability that a random smartphone user in the age range 13 to 55+ is between 23 and 64.7 years old.

b. Determine the probability that a randomly selected smartphone user in the age range 13 to 55+ is at most 50.8 years old.

c. Find the 80th percentile of this distribution, and interpret it in a complete sentence.

TRY IT 6.10

Use the information in Example 6.10 to answer the following questions.

Find the 30th percentile, and interpret it in a complete sentence.
What is the probability that the age of a randomly selected smartphone user in the range 13 to 55+ is less than 27 years old.

EXAMPLE 6.11

In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years respectively. Using this information, answer the following questions (round answers to one decimal place).

a. Calculate the interquartile range (IQR).

b. Forty percent of the smartphone users from 13 to 55+ are at least what age?

TRY IT 6.11

Two thousand students took an exam. The scores on the exam have an approximate normal distribution with a mean μ = 81 points and standard deviation σ = 15 points.

Calculate the first- and third-quartile scores for this exam.
The middle 50% of the exam scores are between what two values?

EXAMPLE 6.12

A citrus farmer who grows mandarin oranges finds that the diameters of mandarin oranges harvested on his farm follow a normal distribution with a mean diameter of 5.85 cm and a standard deviation of 0.24 cm.

a. Find the probability that a randomly selected mandarin orange from this farm has a diameter larger than 6.0 cm. Sketch the graph.

b. The middle 20% of mandarin oranges from this farm have diameters between ______ and ______.

c. Find the 90th percentile for the diameters of mandarin oranges, and interpret it in a complete sentence.