6.2 Using the Normal Distribution - Introductory Statistics

The shaded area in the following graph indicates the area to the left of x. This area is represented by the probability P(X < x). Normal tables, computers, and calculators provide or calculate the probability P(X < x).

This diagram shows a bell-shaped curve with uppercase X at the extreme right end of the X axis. The X axis also contains a lowercase x about one-quarter of the way across the X axis from the right. The area under the bell curve to the right of the lowercase x is shaded. The label states: shaded area represents 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 ( X ≤ x ) and P ( X > x ) is the same as P ( X ≥ x ) for continuous distributions.

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 Appendix H Tables without the use of technology. The tables include instructions for how to use them.

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.

Problem

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

Solution

a. Let X = a score on the final exam. X ~ N(63, 5), where μ = 63 and σ = 5.

Draw a graph.

Then, find P(x > 65).

P(x > 65) = 0.3446

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

Figure 6.5

The probability that any student selected at random scores more than 65 is 0.3446.

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

Go into 2nd DISTR.
After pressing 2nd DISTR, press 2:normalcdf.

The syntax for the instructions are as follows:

normalcdf(lower value, upper value, mean, standard deviation) For this problem: normalcdf(65,1E99,63,5) = 0.3446. You get 1E99 (= 10⁹⁹) by pressing 1, the EE key (a 2nd key) and then 99. Or, you can enter 10^99 instead. The number 10⁹⁹ is way out in the right tail of the normal curve. We are calculating the area between 65 and 10⁹⁹. In some instances, the lower number of the area might be –1E99 (= –10⁹⁹). The number –10⁹⁹ is way out in the left tail of the normal curve.

Historical Note

The TI probability program calculates a z-score and then the probability from the z-score. Before technology, the z-score was looked up in a standard normal probability table (because the math involved is too cumbersome) to find the probability. In this example, a standard normal table with area to the left of the z-score was used. You calculate the z-score and look up the area to the left. The probability is the area to the right.

z = $\frac{65 – 63}{5}$ = 0.4

Area to the left is 0.6554.

P(x > 65) = P(z > 0.4) = 1 – 0.6554 = 0.3446

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.

Problem

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

Solution

b. Draw a graph.

Then find P(x < 85), and shade the graph.

Using a computer or calculator, find P(x < 85) = 1.

normalcdf(0,85,63,5) = 1 (rounds to one)

The probability that one student scores less than 85 is approximately one (or 100%).

Problem

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

Solution

c. Find the 90^th percentile. For each problem or part of a problem, draw a new graph. Draw the x-axis. Shade the area that corresponds to the 90^th percentile.

Let k = the 90^th percentile. The variable k is located on the x-axis. P(x < k) is the area to the left of k. The 90^th percentile k separates the exam scores into those that are the same or lower than k and those that are the same or higher. Ninety percent of the test scores are the same or lower than k, and ten percent are the same or higher. The variable k is often called a critical value.

k = 69.4

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

Figure 6.6

The 90^th percentile is 69.4. This means that 90% of the test scores fall at or below 69.4 and 10% fall at or above. To get this answer on the calculator, follow this step:

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

Problem

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

Solution

d. Find the 70^th percentile.

Draw a new graph and label it appropriately. k = 65.6

The 70^th percentile is 65.6. This means that 70% of the test scores fall at or below 65.5 and 30% fall at or above.

invNorm(0.70,63,5) = 65.6

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.

Problem

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

Solution

a. Let X = the amount of time (in hours) a household personal computer is used for entertainment. X ~ N(2, 0.5) where μ = 2 and σ = 0.5.

Find P(1.8 < x < 2.75).

The probability for which you are looking is the area between x = 1.8 and x = 2.75. P(1.8 < x < 2.75) = 0.5886

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.

Figure 6.7

normalcdf(1.8,2.75,2,0.5) = 0.5886

The probability that a household personal computer is used between 1.8 and 2.75 hours per day for entertainment is 0.5886.

Problem

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

Solution

b. To find the maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment, find the 25^th percentile, k, where P(x < k) = 0.25.

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.

Figure 6.8

invNorm(0.25,2,0.5) = 1.66

The maximum number of hours per day that the bottom quartile of households uses a personal computer for entertainment is 1.66 hours.

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.

Problem

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 80^th percentile of this distribution, and interpret it in a complete sentence.

Solution

a. normalcdf(23,64.7,36.9,13.9) = 0.8186

b. normalcdf(–10⁹⁹,50.8,36.9,13.9) = 0.8413

invNorm(0.80,36.9,13.9) = 48.6
The 80^th percentile is 48.6 years.
80% of the smartphone users in the age range 13 – 55+ are 48.6 years old or less.

Try It 6.10

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

Find the 30^th 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).

Problem

a. Calculate the interquartile range (IQR).

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

Solution

IQR = Q₃ – Q₁
Calculate Q₃ = 75^th percentile and Q₁ = 25^th percentile.
invNorm(0.75,36.9,13.9) = Q₃ = 46.2754
invNorm(0.25,36.9,13.9) = Q₁ = 27.5246
IQR = Q₃ – Q₁ = 18.8

Find k where P(x ≥ k) = 0.40 ("At least" translates to "greater than or equal to.")
0.40 = the area to the right.
Area to the left = 1 – 0.40 = 0.60.
The area to the left of k = 0.60.
invNorm(0.60,36.9,13.9) = 40.4215.
k = 40.4.
Forty percent of the smartphone users from 13 to 55+ are at least 40.4 years.

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.

Problem

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

Solution

a. normalcdf(6,10^99,5.85,0.24) = 0.2660

Figure 6.9

1 – 0.20 = 0.80
The tails of the graph of the normal distribution each have an area of 0.40.
Find k1, the 40^th percentile, and k2, the 60^th percentile (0.40 + 0.20 = 0.60).
k1 = invNorm(0.40,5.85,0.24) = 5.79 cm
k2 = invNorm(0.60,5.85,0.24) = 5.91 cm

c. 6.16: Ninety percent of the diameter of the mandarin oranges is at most 6.16 cm.

Try It 6.12

Using the information from Example 6.12, answer the following:

The middle 40% of mandarin oranges from this farm are between ______ and ______.
Find the 16^th percentile and interpret it in a complete sentence.