Statistics

# 6.2Using the Normal Distribution

Statistics6.2 Using the Normal Distribution

The shaded area in the following graph indicates the area to the left of x. This area could represent the percentage of students scoring less than a particular grade on a final exam. This area is represented by the probability P(X < x). Normal tables, computers, and calculators are used to provide or calculate the probability P(X < x).

Figure 6.4

The area to the right is then P(X > x) = 1 – P(X < x). Remember, P(X < x) = Area to the left of the vertical line through x. P(X < x) = 1 – P(X < x) = Area to the right of the vertical line through x. P(X < x) is the same as P(Xx) and P(X > x) is the same as P(Xx) for continuous distributions.

Suppose the graph above were to represent the percentage of students scoring less than 75 on a final exam, with this probability equal to 0.39. This would also indicate that the percentage of students scoring higher than 75 was equal to 1 minus 0.39 or 0.61.

### Calculations of Probabilities

Probabilities are calculated using technology. There are instructions given as necessary for the TI-83+ and TI-84 calculators.

### NOTE

To calculate the probability, use the probability tables provided in Figure G1 without the use of technology. The tables include instructions for how to use them.

The probability is represented by the area under the normal curve. To find the probability, calculate the z-score and look up the z-score in the z-table under the z-column. Most z-tables show the area under the normal curve to the left of z. Others show the mean to z area. The method used will be indicated on the table.

We will discuss the z-table that represents the area under the normal curve to the left of z. Once you have located the z-score, locate the corresponding area. This will be the area under the normal curve, to the left of the z-score. This area can be used to find the area to the right of the z-score, or by subtracting from 1 or the total area under the normal curve. These areas can also be used to determine the area between two z-scores.

### Example 6.7

If the area to the left is 0.0228, then the area to the right is 1 – 0.0228 = 0.9772.

Try It 6.7

If the area to the left of x is 0.012, then what is the area to the right?

### 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.

### Using the TI-83, 83+, 84, 84+ Calculator

Calculate the z-score

*Press 2nd Distr
*Press 3:invNorm(
*Enter the area to the left of z followed by )
*Press ENTER.
For this Example, the steps are
2nd Distr
3:invNorm(.6554) ENTER
The answer is 0.3999, which rounds to 0.4.

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 percent of the scores below k and 10 percent of the scores above k.

### 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 percent of scores are below k and 30 percent 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.

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

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 smartphone users between the ages of 13 and 55+ between the ages of 13 and 55+ 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:

1. Find the 30th percentile, and interpret it in a complete sentence.
2. 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 ages that range 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.

1. Calculate the first- and third-quartile scores for this exam.
2. The middle 50 percent 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 percent 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.

Try It 6.12

Using the information from Example 6.12, answer the following:

1. The middle 45 percent of mandarin oranges from this farm are between ______ and ______.
2. Find the 16th percentile, and interpret it in a complete sentence.