Content check — Business Statistics, Chapter 5: Continuous Random Variables

5-1 §5.1 source ↗

Which type of distribution does the graph illustrate?

The horizontal axis ranges from 0 to 10. The distribution is modeled by a rectangle extending from x = 3 to x =8. — Figure 5.28 in the book ↗

Answer

Uniform Distribution

5-3 §5.1 source ↗

Which type of distribution does the graph illustrate?

This graph shows a bell-shaped graph. The symmetric graph reaches maximum height at x = 0 and slopes downward gradually to the x-axis on each side of the peak. — Figure 5.30 in the book ↗

Answer

Normal Distribution

5-5 §5.1 source ↗

What does the shaded area represent? P(___< x < ___)

This graph shows an exponential distribution. The graph slopes downward. It begins at a point on the y-axis and approaches the x-axis at the right edge of the graph. The region under the graph from x = 6 to x = 7 is shaded. — Figure 5.32 in the book ↗

Answer

P(6 < x < 7)

5-7 §5.1 source ↗

What is the area under f(x) if the function is a continuous probability density function?

Answer

one

5-9 §5.1 source ↗

A continuous probability function is restricted to the portion between x = 0 and 7. What is P(x = 10)?

Answer

zero

5-11 §5.1 source ↗

f(x), a continuous probability function, is equal to

\frac{1}{12}

, and the function is restricted to 0 ≤ x ≤ 12. What is P (0 < x < 12)?

Answer

one

5-13 §5.1 source ↗

Find the probability that x falls in the shaded area.

A graph illustrating a uniform probability distribution where f(x) = 1/8 for x between 0 and 8. The area under the curve from x=0 to x=5 is highlighted in blue. — Figure 5.34 in the book ↗

Answer

0.625

5-15 §5.1 source ↗

f(x), a continuous probability function, is equal to

\frac{1}{3}

and the function is restricted to 1 ≤ x ≤ 4. Describe

P (x > \frac{3}{2}) .

Answer

The probability is equal to the area from x =

\frac{3}{2}

to x = 4 above the x-axis and up to f(x) =

\frac{1}{3}

.

5-17 §5.2 source ↗

Use the following information to answer the next ten questions. The data that follow are the square footage (in 1,000 feet squared) of 28 homes.

The sample mean = 2.50 and the sample standard deviation = 0.8302.

The distribution can be written as X ~ U(1.5, 4.5).

In this distribution, outcomes are equally likely. What does this mean?

Table 5.1 in the book ↗

1.5	2.4	3.6	2.6	1.6	2.4	2.0
3.5	2.5	1.8	2.4	2.5	3.5	4.0
2.6	1.6	2.2	1.8	3.8	2.5	1.5
2.8	1.8	4.5	1.9	1.9	3.1	1.6

Answer

It means that the value of x is just as likely to be any number between 1.5 and 4.5.

5-19 §5.2 source ↗

Use the following information to answer the next ten questions. The data that follow are the square footage (in 1,000 feet squared) of 28 homes.

The sample mean = 2.50 and the sample standard deviation = 0.8302.

The distribution can be written as X ~ U(1.5, 4.5).

What are the constraints for the values of x?

Table 5.1 in the book ↗

1.5	2.4	3.6	2.6	1.6	2.4	2.0
3.5	2.5	1.8	2.4	2.5	3.5	4.0
2.6	1.6	2.2	1.8	3.8	2.5	1.5
2.8	1.8	4.5	1.9	1.9	3.1	1.6

Answer

1.5 ≤ x ≤ 4.5

5-21 §5.2 source ↗

Use the following information to answer the next ten questions. The data that follow are the square footage (in 1,000 feet squared) of 28 homes.

The sample mean = 2.50 and the sample standard deviation = 0.8302.

The distribution can be written as X ~ U(1.5, 4.5).

What is P(2 < x < 3)?

Table 5.1 in the book ↗

1.5	2.4	3.6	2.6	1.6	2.4	2.0
3.5	2.5	1.8	2.4	2.5	3.5	4.0
2.6	1.6	2.2	1.8	3.8	2.5	1.5
2.8	1.8	4.5	1.9	1.9	3.1	1.6

Answer

0.3333

5-23 §5.2 source ↗

Use the following information to answer the next ten questions. The data that follow are the square footage (in 1,000 feet squared) of 28 homes.

The sample mean = 2.50 and the sample standard deviation = 0.8302.

The distribution can be written as X ~ U(1.5, 4.5).

What is P(x = 1.5)?

Table 5.1 in the book ↗

1.5	2.4	3.6	2.6	1.6	2.4	2.0
3.5	2.5	1.8	2.4	2.5	3.5	4.0
2.6	1.6	2.2	1.8	3.8	2.5	1.5
2.8	1.8	4.5	1.9	1.9	3.1	1.6

Answer

zero

5-24 §5.2 source ↗

Use the following information to answer the next ten questions. The data that follow are the square footage (in 1,000 feet squared) of 28 homes.

The sample mean = 2.50 and the sample standard deviation = 0.8302.

The distribution can be written as X ~ U(1.5, 4.5).

Find the probability that a randomly selected home has more than 3,000 square feet given that you already know the house has more than 2,000 square feet.

Table 5.1 in the book ↗

1.5	2.4	3.6	2.6	1.6	2.4	2.0
3.5	2.5	1.8	2.4	2.5	3.5	4.0
2.6	1.6	2.2	1.8	3.8	2.5	1.5
2.8	1.8	4.5	1.9	1.9	3.1	1.6

Answer

0.6

5-26 §5.2 source ↗

Use the following information to answer the next eight exercises. A distribution is given as X ~ U(0, 12).

What is b? What does it represent?

Answer

b is 12, and it represents the highest value of x.

5-28 §5.2 source ↗

Use the following information to answer the next eight exercises. A distribution is given as X ~ U(0, 12).

What is the theoretical mean?

Answer

six

5-30 §5.2 source ↗

Use the following information to answer the next eight exercises. A distribution is given as X ~ U(0, 12).

Draw the graph of the distribution for P(x > 9).

Answer

This graph shows a uniform distribution. The horizontal axis ranges from 0 to 12. The distribution is modeled by a rectangle extending from x = 0 to x = 12. A region from x = 9 to x = 12 is shaded inside the rectangle. — Figure 5.43 in the book ↗

5-33 §5.2 source ↗

Use the following information to answer the next eleven exercises. The age of cars in the staff parking lot of a suburban college is uniformly distributed from six months (0.5 years) to 9.5 years.

In words, define the random variable X.

Answer

X = The age (in years) of cars in the staff parking lot

5-35 §5.2 source ↗

Use the following information to answer the next eleven exercises. The age of cars in the staff parking lot of a suburban college is uniformly distributed from six months (0.5 years) to 9.5 years.

The interval of values for x is ______.

Answer

0.5 to 9.5

5-37 §5.2 source ↗

Use the following information to answer the next eleven exercises. The age of cars in the staff parking lot of a suburban college is uniformly distributed from six months (0.5 years) to 9.5 years.

Write the probability density function.

Answer

f(x) =

\frac{1}{9}

where x is between 0.5 and 9.5, inclusive.

5-39 §5.2 source ↗

Use the following information to answer the next eleven exercises. The age of cars in the staff parking lot of a suburban college is uniformly distributed from six months (0.5 years) to 9.5 years.

Find the average age of the cars in the lot.

Answer

μ = 5

5-41 §5.2 source ↗

Use the following information to answer the next eleven exercises. The age of cars in the staff parking lot of a suburban college is uniformly distributed from six months (0.5 years) to 9.5 years.

Considering only the cars less than 7.5 years old, find the probability that a randomly chosen car in the lot was less than four years old.

a. Sketch the graph, shade the area of interest.

b. Find the probability. P(x < 4 $|$ x < 7.5) = _______

This is a blank graph template. The vertical and horizontal axes are unlabeled. — Figure 5.38 in the book ↗

Answer

a. Answers may vary.

b. $\frac{3.5}{7}$

5-43 §5.2 source ↗

Use the following information to answer the next eleven exercises. The age of cars in the staff parking lot of a suburban college is uniformly distributed from six months (0.5 years) to 9.5 years.

Find the third quartile of ages of cars in the lot. This means you will have to find the value such that $\frac{3}{4}$ , or 75%, of the cars are at most (less than or equal to) that age.

a. Sketch the graph, and shade the area of interest.

b. Find the value k such that P(x < k) = 0.75.

c. The third quartile is _______

Blank graph with vertical and horizontal axes. — Figure 5.39 in the book ↗

Answer

a. Answers may vary.

b. k = 7.25

c. 7.25

5-45 §5.3 source ↗

Use the following information to answer the next ten exercises. A customer service representative must spend different amounts of time with each customer to resolve various concerns. The amount of time spent with each customer can be modeled by the following distribution: X ~ Exp(0.2)

Are outcomes equally likely in this distribution? Why or why not?

Answer

No, outcomes are not equally likely. In this distribution, more people require a little bit of time, and fewer people require a lot of time, so it is more likely that someone will require less time.

5-47 §5.3 source ↗

Use the following information to answer the next ten exercises. A customer service representative must spend different amounts of time with each customer to resolve various concerns. The amount of time spent with each customer can be modeled by the following distribution: X ~ Exp(0.2)

What is the mean?

Answer

five

5-49 §5.3 source ↗

Use the following information to answer the next ten exercises. A customer service representative must spend different amounts of time with each customer to resolve various concerns. The amount of time spent with each customer can be modeled by the following distribution: X ~ Exp(0.2)

State the probability density function.

Answer

f(x) = 0.2e^-0.2x

5-51 §5.3 source ↗

Use the following information to answer the next ten exercises. A customer service representative must spend different amounts of time with each customer to resolve various concerns. The amount of time spent with each customer can be modeled by the following distribution: X ~ Exp(0.2)

Find P(2 < x < 10).

Answer

0.5350

5-53 §5.3 source ↗

Use the following information to answer the next ten exercises. A customer service representative must spend different amounts of time with each customer to resolve various concerns. The amount of time spent with each customer can be modeled by the following distribution: X ~ Exp(0.2)

Find the 70^th percentile.

Answer

6.02

5-55 §5.3 source ↗

Use the following information to answer the next seven exercises. A distribution is given as X ~ Exp(0.75).

What is the probability density function?

Answer

f(x) = 0.75e^-0.75x

5-57 §5.3 source ↗

Use the following information to answer the next seven exercises. A distribution is given as X ~ Exp(0.75).

Draw the distribution.

Answer

This graph shows an exponential distribution. The graph slopes downward. It begins at the point (0, 0.75) on the y-axis and approaches the x-axis at the right edge of the graph. The decay parameter, m, equals 0.75. — Figure 5.44 in the book ↗

5-59 §5.3 source ↗

Use the following information to answer the next seven exercises. A distribution is given as X ~ Exp(0.75).

Find the 30^th percentile.

Answer

0.4756

5-61 §5.3 source ↗

Use the following information to answer the next seven exercises. A distribution is given as X ~ Exp(0.75).

Which is larger, the mean or the median?

Answer

The mean is larger. The mean is

\frac{1}{m} =

\frac{1}{0.75} \approx

1.33

, which is greater than 0.9242.

5-63 §5.3 source ↗

Use the following information to answer the next 16 exercises. Carbon-14 is a radioactive element with a half-life of about 5,730 years. Carbon-14 is said to decay exponentially. The decay rate is 0.000121. We start with one gram of carbon-14. We are interested in the time (years) it takes to decay carbon-14.

Are the data discrete or continuous?

Answer

continuous

5-65 §5.3 source ↗

Use the following information to answer the next 16 exercises. Carbon-14 is a radioactive element with a half-life of about 5,730 years. Carbon-14 is said to decay exponentially. The decay rate is 0.000121. We start with one gram of carbon-14. We are interested in the time (years) it takes to decay carbon-14.

What is the decay rate (m)?

Answer

m = 0.000121

5-67 §5.3 source ↗

Use the following information to answer the next 16 exercises. Carbon-14 is a radioactive element with a half-life of about 5,730 years. Carbon-14 is said to decay exponentially. The decay rate is 0.000121. We start with one gram of carbon-14. We are interested in the time (years) it takes to decay carbon-14.

Find the amount (percent of one gram) of carbon-14 lasting less than 5,730 years. This means, find P(x < 5,730).

a. Sketch the graph, and shade the area of interest.

b. Find the probability. P(x < 5,730) = __________

Answer

a. Answers may vary.

b. P(x < 5,730) = 0.5001

5-69 §5.3 source ↗

Use the following information to answer the next 16 exercises. Carbon-14 is a radioactive element with a half-life of about 5,730 years. Carbon-14 is said to decay exponentially. The decay rate is 0.000121. We start with one gram of carbon-14. We are interested in the time (years) it takes to decay carbon-14.

Thirty percent (30%) of carbon-14 will decay within how many years?

a. Sketch the graph, and shade the area of interest.

b. Find the value k such that P(x < k) = 0.30.

Answer

a. Answers may vary.

b. k = 2947.73

5-71 §5.1 source ↗

For each probability and percentile problem, draw the picture.

When age is rounded to the nearest year, do the data stay continuous, or do they become discrete? Why?

Answer

Age is a measurement, regardless of the accuracy used.

5-73 §5.2 source ↗

For each probability and percentile problem, draw the picture.

A random number generator picks a number from one to nine in a uniform manner.

a. Graph the probability distribution.

b. f(x) = _________

c. μ = _________

d. σ = _________

e. P(3.5 < x < 7.25) = _________

f. P(x > 5.67)

g. P(x > 5 $|$ x > 3) = _________

Answer

a. Answers may vary.

b. $f (x) = \frac{1}{8}$ where $1 \leq x \leq 9$

c. five

d. 2.3

e. $\frac{15}{32}$

f. $\frac{333}{800}$

g. $\frac{2}{3}$

5-75 §5.2 source ↗

For each probability and percentile problem, draw the picture.

A subway train on the Red Line arrives every eight minutes during rush hour. We are interested in the length of time a commuter must wait for a train to arrive. The time follows a uniform distribution.

a. Define the random variable. X = _______

b. Graph the probability distribution.

c. f(x) = _______

d. μ = _______

e. σ = _______

f. Find the probability that the commuter waits less than one minute.

g. Find the probability that the commuter waits between three and four minutes.

Answer

a. X represents the length of time a commuter must wait for a train to arrive on the Red Line.

b. Graph the probability distribution.

c. $f (x) = \frac{1}{8}$ where 0 ≤ x ≤ 8

d. four

e. 2.31

f. $\frac{1}{8}$

g. $\frac{1}{8}$

5-77 §5.2 source ↗

Use the following information to answer the next three exercises. The Sky Train from the terminal to the rental–car and long–term parking center is supposed to arrive every eight minutes. The waiting times for the train are known to follow a uniform distribution.

What is the average waiting time (in minutes)?

a. zero

b. two

c. three

d. four

Answer

d

5-78 §5.2 source ↗

Use the following information to answer the next three exercises. The Sky Train from the terminal to the rental–car and long–term parking center is supposed to arrive every eight minutes. The waiting times for the train are known to follow a uniform distribution.

The probability of waiting more than seven minutes given a person has waited more than four minutes is?

a. 0.125

b. 0.25

c. 0.5

d. 0.75

Answer

b

5-80 §5.2 source ↗

Use the following information to answer the next three exercises. The Sky Train from the terminal to the rental–car and long–term parking center is supposed to arrive every eight minutes. The waiting times for the train are known to follow a uniform distribution.

Suppose that the value of a stock varies each day from $16 to $25 with a uniform distribution.

a. Find the probability that the value of the stock is more than $19.

b. Find the probability that the value of the stock is between $19 and $22.

c. Given that the stock is greater than $18, find the probability that the stock is more than $21.

Answer

a. The probability density function of X is $\frac{1}{25 - 16} = \frac{1}{9}$ .
P(X > 19) = (25 – 19) $(\frac{1}{9})$ = $\frac{6}{9}$ = $\frac{2}{3}$ .

b. P(19 < X < 22) = (22 – 19) $(\frac{1}{9})$ = $\frac{3}{9}$ = $\frac{1}{3}$ .

c. This is a conditional probability question. P(x > 21 $|$ x > 18). You can do this two ways:

• Draw the graph where a is now 18 and b is still 25. The height is $\frac{1}{(25 - 18)}$ = $\frac{1}{7}$
So, P(x > 21 $|$ x > 18) = (25 – 21) $(\frac{1}{7})$ = 4/7.

• Use the formula: P(x > 21 $|$ x > 18) = $\frac{P (x > 21 \cap x > 18)}{P (x > 18)}$
= $\frac{P (x > 21)}{P (x > 18)}$ = $\frac{(25 - 21)}{(25 - 18)}$ = $\frac{4}{7}$ .

This shows the graph of the function f(x) = 1/9, the pdf for a uniform distribution. A horizontal line ranges from the point (16, 1/9) to the point (25, 1.9). Vertical lines extend from the x-axis to the graph at x = 16 and x = 25 creating a rectangle. A region is shaded inside the rectangle from x = 19 to x = 25. Text notes that the shaded area represents P(x greater than 19) = 2/3. — Figure 5.45 in the book ↗

5-82 §5.2 source ↗

Use the following information to answer the next three exercises. The Sky Train from the terminal to the rental–car and long–term parking center is supposed to arrive every eight minutes. The waiting times for the train are known to follow a uniform distribution.

The number of miles driven by a truck driver falls between 300 and 700, and follows a uniform distribution.

a. Find the probability that the truck driver goes more than 650 miles in a day.

b. Find the probability that the truck drivers goes between 400 and 650 miles in a day.

Answer

a. P(X > 650) = $\frac{700 - 650}{700 - 300} =$ $\frac{50}{400} =$ $\frac{1}{8}$ = 0.125

b. P(400 < X < 650) = $\frac{650 - 400}{700 - 300} =$ $\frac{250}{400}$ = 0.625

5-84 §5.3 source ↗

Suppose that the useful life of a particular car battery, measured in months, decays with parameter 0.025. We are interested in the life of the battery.

a. Define the random variable. X = _________________________________.

b. Is X continuous or discrete?

c. On average, how long would you expect one car battery to last?

d. On average, how long would you expect nine car batteries to last, if they are used one after another?

e. Find the probability that a car battery lasts more than 36 months.

f. Seventy percent of the batteries last at least how long?

Answer

a. X = the useful life of a particular car battery, measured in months.

b. X is continuous.

c. 40 months

d. 360 months

e. 0.4066

f. 14.27

5-86 §5.3 source ↗

The time (in years) after reaching age 60 that it takes an individual to retire is approximately exponentially distributed with a mean of about five years. Suppose we randomly pick one retired individual. We are interested in the time after age 60 to retirement.

a. Define the random variable. X = _________________________________.

b. Is X continuous or discrete?

c. μ = ________

d. σ = ________

e. Draw a graph of the probability distribution. Label the axes.

f. Find the probability that the person retired after age 70.

g. Do more people retire before age 65 or after age 65?

h. In a room of 1,000 people over age 80, how many do you expect will NOT have retired yet?

Answer

a. X = the time (in years) after reaching age 60 that it takes an individual to retire

b. X is continuous.

c. five

d. five

e. Answers may vary.

f. 0.1353

g. before

h. 18.3

5-88 §5.3 source ↗

Use the following information to answer the next three exercises. The average lifetime of a certain new smartphone is three years. The manufacturer will replace any smartphone failing within two years of the date of purchase. The lifetime of these smartphone is known to follow an exponential distribution.

The decay rate is:

a. 0.3333

b. 0.5000

c. 2

d. 3

Answer

a

5-90 §5.3 source ↗

Use the following information to answer the next three exercises. The average lifetime of a certain new smartphone is three years. The manufacturer will replace any smartphone failing within two years of the date of purchase. The lifetime of these smartphone is known to follow an exponential distribution.

What is the median lifetime of these smartphones (in years)?

a. 0.1941

b. 1.3863

c. 2.0794

d. 5.5452

Answer

c

5-92 §5.3 source ↗

Use the following information to answer the next three exercises. The average lifetime of a certain new smartphone is three years. The manufacturer will replace any smartphone failing within two years of the date of purchase. The lifetime of these smartphone is known to follow an exponential distribution.

In major league baseball, a no-hitter is a game in which a pitcher, or pitchers, doesn't give up any hits throughout the game. No-hitters occur at a rate of about three per season. Assume that the duration of time between no-hitters is exponential.

a. What is the probability that an entire season elapses with a single no-hitter?

b. If an entire season elapses without any no-hitters, what is the probability that there are no no-hitters in the following season?

c. What is the probability that there are more than 3 no-hitters in a single season?

Answer

Let X = the number of no-hitters throughout a season. Since the duration of time between no-hitters is exponential, the number of no-hitters per season is Poisson with mean λ = 3.
Therefore, (X = 0) = $\frac{3^{0} e^{- 3}}{0!}$ = e^–3 ≈ 0.0498

NOTE

You could let T = duration of time between no-hitters. Since the time is exponential and there are 3 no-hitters per season, then the time between no-hitters is $\frac{1}{3}$ season. For the exponential, µ = $\frac{1}{3}$ .
Therefore, m = $\frac{1}{μ}$ = 3 and T ∼ Exp(3).

a. The desired probability is P(T > 1) = 1 – P(T < 1) = 1 – (1 – e^–3) = e^–3 ≈ 0.0498.

b. Let T = duration of time between no-hitters. We find P(T > 2|T > 1), and by the memoryless property this is simply P(T > 1), which we found to be 0.0498 in part a.

c. Let X = the number of no-hitters is a season. Assume that X is Poisson with mean λ = 3. Then P(X > 3) = 1 – P(X ≤ 3) = 0.3528.

5-94 §5.3 source ↗

Use the following information to answer the next three exercises. The average lifetime of a certain new smartphone is three years. The manufacturer will replace any smartphone failing within two years of the date of purchase. The lifetime of these smartphone is known to follow an exponential distribution.

According to the American Red Cross, about one out of nine people in the U.S. have Type B positive blood. Suppose the blood types of people arriving at a blood drive are independent. In this case, the number of Type B positive blood types that arrive roughly follows the Poisson distribution.

a. If 100 people arrive, how many on average would be expected to have Type B positive blood?

b. What is the probability that over 10 people out of these 100 have type B positive blood?

c. What is the probability that more than 20 people arrive before a person with type B positive blood is found?

Answer

a. $\frac{100}{9}$ = 11.11

b. P(X > 10) = 1 – P(X ≤ 10) = 1 – Poissoncdf(11.11, 10) ≈ 0.5532.

c. The number of people with Type B positive blood encountered roughly follows the Poisson distribution, so the number of people X who arrive between successive Type B arrivals is roughly exponential with mean μ = 9 and m = $\frac{1}{9}$ . The cumulative distribution function of X is $P (X < x) =$ $1 -$ $e^{- \frac{x}{9}}$ . Thus, P(X > 20) = 1 - P(X ≤ 20) = $1 -$ $(1 - e^{- \frac{20}{9}}) \approx$ $0.1084.$

NOTE

We could also deduce that each person arriving has a 8/9 chance of not having Type B positive blood. So the probability that none of the first 20 people arrive have Type B positive blood is ${(\frac{8}{9})}^{20} \approx$ $0.0948$ . (The geometric distribution is more appropriate than the exponential because the number of people between Type B positive people is discrete instead of continuous.)

5-96 §5.3 source ↗

Use the following information to answer the next three exercises. The average lifetime of a certain new smartphone is three years. The manufacturer will replace any smartphone failing within two years of the date of purchase. The lifetime of these smartphone is known to follow an exponential distribution.

At an urgent care facility, patients arrive at an average rate of one patient every seven minutes. Assume that the duration between arrivals is exponentially distributed.

a. Find the probability that the time between two successive visits to the urgent care facility is less than 2 minutes.

b. Find the probability that the time between two successive visits to the urgent care facility is more than 15 minutes.

c. If 10 minutes have passed since the last arrival, what is the probability that the next person will arrive within the next five minutes?

d. Find the probability that more than eight patients arrive during a half-hour period.

Answer

Let T = duration (in minutes) between successive visits. Since patients arrive at a rate of one patient every seven minutes, μ = 7 and the decay constant is m = $\frac{1}{7}$ . The cdf is P(T < t) = $1 - e^{\frac{t}{7}}$

a. P(T < 2) = 1 - $1 - e^{- \frac{2}{7}}$ ≈ 0.2485.

b. P(T > 15) = $1 -$ $P (T < 15) =$ $1 -$ $(1 - e^{- \frac{15}{7}}) \approx$ $e^{- \frac{15}{7}} \approx$ $0.1173$ .

c. P(T > 15|T > 10) = P(T > 5) = $1 -$ $(1 - e^{- \frac{5}{7}}) =$ $e^{- \frac{5}{7}} \approx$ $0.4895$ .

d. Let X = # of patients arriving during a half-hour period. Then X has the Poisson distribution with a mean of $\frac{30}{7}$ , X ∼ Poisson $(\frac{30}{7})$ . Find P(X > 8) = 1 – P(X ≤ 8) ≈ 0.0311.

1.5	2.4	3.6	2.6	1.6	2.4	2.0
3.5	2.5	1.8	2.4	2.5	3.5	4.0
2.6	1.6	2.2	1.8	3.8	2.5	1.5
2.8	1.8	4.5	1.9	1.9	3.1	1.6

1.5	2.4	3.6	2.6	1.6	2.4	2.0
3.5	2.5	1.8	2.4	2.5	3.5	4.0
2.6	1.6	2.2	1.8	3.8	2.5	1.5
2.8	1.8	4.5	1.9	1.9	3.1	1.6

1.5	2.4	3.6	2.6	1.6	2.4	2.0
3.5	2.5	1.8	2.4	2.5	3.5	4.0
2.6	1.6	2.2	1.8	3.8	2.5	1.5
2.8	1.8	4.5	1.9	1.9	3.1	1.6

1.5	2.4	3.6	2.6	1.6	2.4	2.0
3.5	2.5	1.8	2.4	2.5	3.5	4.0
2.6	1.6	2.2	1.8	3.8	2.5	1.5
2.8	1.8	4.5	1.9	1.9	3.1	1.6

1.5	2.4	3.6	2.6	1.6	2.4	2.0
3.5	2.5	1.8	2.4	2.5	3.5	4.0
2.6	1.6	2.2	1.8	3.8	2.5	1.5
2.8	1.8	4.5	1.9	1.9	3.1	1.6

1.5	2.4	3.6	2.6	1.6	2.4	2.0
3.5	2.5	1.8	2.4	2.5	3.5	4.0
2.6	1.6	2.2	1.8	3.8	2.5	1.5
2.8	1.8	4.5	1.9	1.9	3.1	1.6

1.5	2.4	3.6	2.6	1.6	2.4	2.0
3.5	2.5	1.8	2.4	2.5	3.5	4.0
2.6	1.6	2.2	1.8	3.8	2.5	1.5
2.8	1.8	4.5	1.9	1.9	3.1	1.6

1.5	2.4	3.6	2.6	1.6	2.4	2.0
3.5	2.5	1.8	2.4	2.5	3.5	4.0
2.6	1.6	2.2	1.8	3.8	2.5	1.5
2.8	1.8	4.5	1.9	1.9	3.1	1.6

1.5	2.4	3.6	2.6	1.6	2.4	2.0
3.5	2.5	1.8	2.4	2.5	3.5	4.0
2.6	1.6	2.2	1.8	3.8	2.5	1.5
2.8	1.8	4.5	1.9	1.9	3.1	1.6

1.5	2.4	3.6	2.6	1.6	2.4	2.0
3.5	2.5	1.8	2.4	2.5	3.5	4.0
2.6	1.6	2.2	1.8	3.8	2.5	1.5
2.8	1.8	4.5	1.9	1.9	3.1	1.6

1.5	2.4	3.6	2.6	1.6	2.4	2.0
3.5	2.5	1.8	2.4	2.5	3.5	4.0
2.6	1.6	2.2	1.8	3.8	2.5	1.5
2.8	1.8	4.5	1.9	1.9	3.1	1.6

1.5	2.4	3.6	2.6	1.6	2.4	2.0
3.5	2.5	1.8	2.4	2.5	3.5	4.0
2.6	1.6	2.2	1.8	3.8	2.5	1.5
2.8	1.8	4.5	1.9	1.9	3.1	1.6

1.5	2.4	3.6	2.6	1.6	2.4	2.0
3.5	2.5	1.8	2.4	2.5	3.5	4.0
2.6	1.6	2.2	1.8	3.8	2.5	1.5
2.8	1.8	4.5	1.9	1.9	3.1	1.6

1.5	2.4	3.6	2.6	1.6	2.4	2.0
3.5	2.5	1.8	2.4	2.5	3.5	4.0
2.6	1.6	2.2	1.8	3.8	2.5	1.5
2.8	1.8	4.5	1.9	1.9	3.1	1.6

1.5	2.4	3.6	2.6	1.6	2.4	2.0
3.5	2.5	1.8	2.4	2.5	3.5	4.0
2.6	1.6	2.2	1.8	3.8	2.5	1.5
2.8	1.8	4.5	1.9	1.9	3.1	1.6