# Introduction

### Introduction

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

### 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.

Solution 6.8

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

Draw a graph.

Calculate the z-score:

$z=x−μσ=65−635=25=.40z=x−μσ=65−635=25=.40$

The z-table shows that the area to the left of z is 0.6554. Subtracting this area from 1 gives 0.3446.

Then, find P(x > 65).

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 is 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. We chose the exponent of 99 because this produces such a large number that we can reasonably expect all of the values under the curve to fall below it. This is an arbitrary value and one that works well, for our purpose.

### 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, also known as a Z-table—the math involved to find probability is cumbersome. 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.

### 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

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.

Solution 6.8

b. Draw a graph.

Then find P(x

Using a computer or calculator, find P(x

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

The probability that one student scores less than 85 is approximately one, or 100 percent.

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.

Solution 6.8

c. Find the 90th percentile. For each problem or part of a problem, draw a new graph. Draw the x-axis. Shade the area that corresponds to the 90th percentile. This time, we are looking for a score that corresponds to a given area under the curve.

Let k = the 90th percentile. The variable k is located on the x-axis. P(x k) is the area to the left of k. The 90th 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 10 percent are the same or higher. The variable k is often called a critical value.

We know the mean, standard deviation, and area under the normal curve. We need to find the z-score that corresponds to the area of 0.9 and then substitute it with the mean and standard deviation into our z-score formula. The z-table shows a z-score of approximately 1.28, for an area under the normal curve to the left of z (larger portion) of approximately 0.9. Thus, we can write the following:

$1.28=x−6351.28=x−63.5$

Multiplying each side of the equation by 5 gives

$6.4=x−636.4=x−63$

Adding 63 to both sides of the equation gives

$69.4=x.69.4=x.$

Thus, our score, k, is 69.4.

Figure 6.6

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

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.

Solution 6.8

d. Find the 70th percentile.

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

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

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

Solution 6.9

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

First, calculate the z-scores for each x-value.

$z=1.8−20.5=−0.20.5=−0.40z=2.75−20.5=0.750.5=1.5z=1.8−20.5=−0.20.5=−0.40z=2.75−20.5=0.750.5=1.5$

Now, use the Z-table to locate the area under the normal curve to the left of each of these z-scores.

The area to the left of the z-score of −0.40 is 0.3446. The area to the left of the z-score of 1.5 is 0.9332. The area between these scores will be the difference in the two areas, or $0.9332−0.34460.9332−0.3446$, which equals 0.5886.

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.

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

Solution 6.9

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

Figure 6.8

invNorm(0.25,2,0.5) = 1.66

We use invNorm because we are looking for the k-value.

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 smartphone users 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.

Solution 6.10

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

The z-scores are calculated as

$z=23−36.913.9=−13.913.9=−1z=64.7−36.913.9=27.813.9=2z=23−36.913.9=−13.913.9=−1z=64.7−36.913.9=27.813.9=2$

The Z-table shows the area to the left of a z-score with an absolute value of 1 to be 0.1587. It shows the area to the left of a z-score of 2 to be 0.9772. The difference in the two areas is 0.8185.

This is slightly different than the area given by the calculator, due to rounding.

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

Solution 6.10

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

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

Solution 6.10

c.

• invNorm(0.80,36.9,13.9) = 48.6
• The 80th percentile is 48.6 years.
• 80 percent 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:

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 smartphone users 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. Using this information, answer the following questions—round answers to one decimal place:

a. Calculate the interquartile range (IQR).

Solution 6.11

a.

• IQR = Q3Q1
• Calculate Q3 = 75th percentile and Q1 = 25th percentile.
• Recall that we can use invNorm to find the k-value. We can use this to find the quartile values.
• invNorm(0.75,36.9,13.9) = Q3 = 46.2754
• invNorm(0.25,36.9,13.9) = Q1 = 27.5246
• IQR = Q3Q1 = 18.8

b. Forty percent of the ages that range from 13 to 55+ are at least what age?

Solution 6.11

b.

• Find k where P(xk) = 0.40. At least translates to greater than or equal to.
• 0.40 = the area to the right
• The 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 ages that range 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.

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.

Solution 6.12

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

Figure 6.9

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

Solution 6.12

b.

• 1 – 0.20 = 0.80. Outside of the middle 20 percent will be 80 percent of the values.
• The tails of the graph of the normal distribution each have an area of 0.40.
• Find k1, the 40th percentile, and k2, the 60th percentile (0.40 + 0.20 = 0.60). This leaves the middle 20 percent, in the middle of the distribution.
• k1 = invNorm(0.40,5.85,0.24) = 5.79 cm
• k2 = invNorm(0.60,5.85,0.24) = 5.91 cm

So, the middle 20 percent of mandarin oranges have diameters between 5.79 cm and 5.91 cm.

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

Solution 6.12

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:

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.