Question 1

An exponential probability distribution

Accepted Answer

An exponential probability distribution is a continuous distribution. It is used to model the time between events that occur randomly and independently at a constant average rate, and it has a probability density function that decreases exponentially as time increases.

Question 2

Excel's EXPON.DIST function can be used to compute

Accepted Answer

The EXPON.DIST function in Excel can compute both the probability density function (PDF) for exponential distributions and the cumulative distribution function (CDF), which are used for calculating exponential probabilities and cumulative exponential probabilities, respectively.

Question 3

Exhibit 6-1
Consider the continuous random variable x, which has a uniform distribution over the interval from 20 to 28.
-Refer to Exhibit 6-1. The probability that x will take on a value between 21 and 25 is

Accepted Answer

The uniform distribution over the interval from 20 to 28 has a total length of 8 (28-20). The interval from 21 to 25 has a length of 4 (25-21). The probability is the length of the interval of interest divided by the total length, which is 4/8 = 0.500.

Question 4

The exponential probability distribution is used with

Accepted Answer

The exponential probability distribution is used with a continuous random variable. This distribution is commonly used to model the time between events in a Poisson process, such as the time between arrivals at a service station or the time between machine failures. It has a single parameter, lambda, which represents the average rate of occurrence of events.

Question 5

Exhibit 6-4
The starting salaries of individuals with an MBA degree are normally distributed with a mean of $40,000 and a standard deviation of $5,000.
-Refer to Exhibit 6-4. What is the random variable in this experiment?

Accepted Answer

The random variable in this experiment is the starting salaries because it is what is being measured and is the outcome of the experiment. The normal distribution, $40,000, and $5,000 are not the random variables themselves.

Question 6

Exhibit 6-1
Consider the continuous random variable x, which has a uniform distribution over the interval from 20 to 28.
-Refer to Exhibit 6-1. The probability density function has what value in the interval between 20 and 28?

Accepted Answer

For a uniform distribution, the probability density function (pdf) is constant over the interval. The pdf value is calculated as 1/(b-a), where a and b are the interval limits. Here, it is 1/(28-20) = 0.125.

Question 7

Exhibit 6-2
The travel time for a college student traveling between her home and her college is uniformly distributed between 40 and 90 minutes.
-Refer to Exhibit 6-2. The probability that her trip will take longer than 60 minutes is

Accepted Answer

Since the distribution is uniform, we can find the probability by taking the area of the part of the distribution that is greater than 60 minutes, which is a triangle with base 30 minutes (from 60 to 90) and height 1/2 (since it is a uniform distribution). The area of this triangle is (1/2)*(30)*(1/2) = 0.25. Therefore, the probability that her trip will take longer than 60 minutes is 0.25, or 0.600 when expressed as a decimal. The best choice is D.

Question 8

When using Excel's EXPON.DIST function, one should choose TRUE for the third input if

Accepted Answer

The EXPON.DIST function calculates the exponential distribution, which is a probability distribution used to describe the time between events that occur independently of one another at a constant rate. The third input of the function specifies whether the result should be a cumulative probability or a probability density. Choosing TRUE for this input calculates the cumulative probability. Therefore, the best choice is B, a cumulative probability is desired.

Question 9

Exhibit 6-2
The travel time for a college student traveling between her home and her college is uniformly distributed between 40 and 90 minutes.
-Refer to Exhibit 6-2. The probability that she will finish her trip in 80 minutes or less is

Accepted Answer

The probability of finishing the trip in 80 minutes or less is the area under the uniform distribution curve between 40 and 80, which is (80-40)/(90-40) = 0.8.

Question 10

Exhibit 6-1
Consider the continuous random variable x, which has a uniform distribution over the interval from 20 to 28.
-Refer to Exhibit 6-1. The variance of x is approximately

Accepted Answer

The formula to find the variance of a uniform distribution is Var(x) = (b-a)^2/12, where a and b are the endpoints of the interval. 
In this case, a=20 and b=28, so Var(x) = (28-20)^2/12 = 5.333. Therefore, the answer is B.

Question 11

Exhibit 6-3
The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds.
-Refer to Exhibit 6-3. The probability of a player weighing less than 250 pounds is

Accepted Answer

To find the probability of a player weighing less than 250 pounds, we need to calculate the z-score first:
z = (250 - 200) / 25 = 2
Then, we can use a standard normal table or calculator to find the area to the left of z = 2, which is 0.9772. Therefore, the probability of a player weighing less than 250 pounds is 0.9772.

Question 12

Exhibit 6-2
The travel time for a college student traveling between her home and her college is uniformly distributed between 40 and 90 minutes.
-Refer to Exhibit 6-2. The probability that her trip will take exactly 50 minutes is

Accepted Answer

Since the distribution is uniform, the probability of any specific value within the range is zero. Therefore, the probability that her trip will take exactly 50 minutes is zero.

Question 13

Exhibit 6-3
The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds.
-Refer to Exhibit 6-3. What is the random variable in this experiment?

Accepted Answer

The random variable in this experiment is the weight of football players, as that is the value being measured and studied. The mean and standard deviation are simply parameters used to describe the distribution of the random variable.

Question 14

Exhibit 6-1
Consider the continuous random variable x, which has a uniform distribution over the interval from 20 to 28.
-Refer to Exhibit 6-1. The mean of x is

Accepted Answer

The mean of a uniform distribution over an interval [a, b] is calculated as (a + b) / 2. For the interval from 20 to 28, the mean is (20 + 28) / 2 = 24.

Question 15

Excel's EXPON.DIST function has how many inputs?

Accepted Answer

The EXPON.DIST function in Excel has three inputs - x, lambda (rate), and cumulative (optional).

Question 16

Exhibit 6-2
The travel time for a college student traveling between her home and her college is uniformly distributed between 40 and 90 minutes.
-Refer to Exhibit 6-2. What is the random variable in this experiment?

Accepted Answer

The random variable in this experiment is the travel time, which is the outcome of interest being observed and measured in the experiment. The uniform distribution is the probability distribution of the random variable. 40 minutes and 90 minutes are simply the minimum and maximum possible values of the random variable.

Question 17

Exhibit 6-3
The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds.
-Refer to Exhibit 6-3. What is the minimum weight of the middle 95% of the players?

Accepted Answer

The middle 95% of a normal distribution is within approximately 1.96 standard deviations from the mean. Therefore, the minimum weight is 200 - (1.96 * 25) = 151 pounds.

Question 18

Exhibit 6-1
Consider the continuous random variable x, which has a uniform distribution over the interval from 20 to 28.
-Refer to Exhibit 6-1. The probability that x will take on a value of at least 26 is

Accepted Answer

Since x has a uniform distribution, the probability density function is constant, f(x) = 1/(28-20) = 1/8.

The probability that x will take on a value of at least 26 is equal to the area under the curve to the right of x = 26.

This area is:

∫[26,28] f(x) dx = ∫[26,28] (1/8) dx = (1/8) * (28-26) = 1/4 = 0.25

Therefore, the probability that x will take on a value of at least 26 is 0.25, or choice C.

Question 19

Exhibit 6-3
The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds.
-Refer to Exhibit 6-3. The probability of a player weighing more than 241.25 pounds is

Accepted Answer

We need to find the probability of a player weighing more than 241.25 pounds.
Z-score = (241.25-200)/25 = 1.65
Using a standard normal table or calculator, the probability corresponding to a z-score of 1.65 is 0.0495. Therefore, the answer is B) 0.0495.

Question 20

Exhibit 6-3
The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds.
-Refer to Exhibit 6-3. What percent of players weigh between 180 and 220 pounds?

Accepted Answer

The answer of Exhibit 6-3
The weight of football players is...

Quiz 6: Continuous Probability Distributions

An exponential probability distribution

Excel's EXPON.DIST function can be used to compute

Exhibit 6-1 Consider the continuous random variable x, which has a uniform distribution over the interval from 20 to 28. -Refer to Exhibit 6-1. The probability that x will take on a value between 21 and 25 is

The exponential probability distribution is used with

Exhibit 6-4 The starting salaries of individuals with an MBA degree are normally distributed with a mean of $40,000 and a standard deviation of $5,000. -Refer to Exhibit 6-4. What is the random variable in this experiment?

Exhibit 6-1 Consider the continuous random variable x, which has a uniform distribution over the interval from 20 to 28. -Refer to Exhibit 6-1. The probability density function has what value in the interval between 20 and 28?

Exhibit 6-2 The travel time for a college student traveling between her home and her college is uniformly distributed between 40 and 90 minutes. -Refer to Exhibit 6-2. The probability that her trip will take longer than 60 minutes is

When using Excel's EXPON.DIST function, one should choose TRUE for the third input if

Exhibit 6-2 The travel time for a college student traveling between her home and her college is uniformly distributed between 40 and 90 minutes. -Refer to Exhibit 6-2. The probability that she will finish her trip in 80 minutes or less is

Exhibit 6-1 Consider the continuous random variable x, which has a uniform distribution over the interval from 20 to 28. -Refer to Exhibit 6-1. The variance of x is approximately

Exhibit 6-3 The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds. -Refer to Exhibit 6-3. The probability of a player weighing less than 250 pounds is

Exhibit 6-2 The travel time for a college student traveling between her home and her college is uniformly distributed between 40 and 90 minutes. -Refer to Exhibit 6-2. The probability that her trip will take exactly 50 minutes is

Exhibit 6-3 The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds. -Refer to Exhibit 6-3. What is the random variable in this experiment?

Exhibit 6-1 Consider the continuous random variable x, which has a uniform distribution over the interval from 20 to 28. -Refer to Exhibit 6-1. The mean of x is

Excel's EXPON.DIST function has how many inputs?

Exhibit 6-2 The travel time for a college student traveling between her home and her college is uniformly distributed between 40 and 90 minutes. -Refer to Exhibit 6-2. What is the random variable in this experiment?

Exhibit 6-3 The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds. -Refer to Exhibit 6-3. What is the minimum weight of the middle 95% of the players?

Exhibit 6-1 Consider the continuous random variable x, which has a uniform distribution over the interval from 20 to 28. -Refer to Exhibit 6-1. The probability that x will take on a value of at least 26 is

Exhibit 6-3 The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds. -Refer to Exhibit 6-3. The probability of a player weighing more than 241.25 pounds is

Exhibit 6-3 The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds. -Refer to Exhibit 6-3. What percent of players weigh between 180 and 220 pounds?