Question 1

Exhibit 6-3
Consider the continuous random variable X, which has a uniform distribution over the interval from 20 to 28.
-Refer to Exhibit 6-3. The probability that X will take on a value between 21 and 25 is

Accepted Answer

The answer of Exhibit 6-3
Consider the continuous random variable X,...

Question 2

Exhibit 6-1
The assembly time for a product is uniformly distributed between 6 to 10 minutes.
-Refer to Exhibit 6-1. The probability of assembling the product between 7 to 9 minutes is

Accepted Answer

The answer of Exhibit 6-1
The assembly time for a product...

Question 3

Exhibit 6-3
Consider the continuous random variable X, which has a uniform distribution over the interval from 20 to 28.
-Refer to Exhibit 6-3. The mean of X is

Accepted Answer

The answer of Exhibit 6-3
Consider the continuous random variable X,...

Question 4

Exhibit 6-3
Consider the continuous random variable X, which has a uniform distribution over the interval from 20 to 28.
-Refer to Exhibit 6-3. The probability density function has what value in the interval between 20 and 28?

Accepted Answer

The answer of Exhibit 6-3
Consider the continuous random variable X,...

Question 5

Given that Z is a standard normal random variable. What is the value of Z if the area to the right of Z is 0.1401?

Accepted Answer

The answer of Given that Z is a standard normal...

Question 6

Exhibit 6-2
The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds.
-Refer to Exhibit 6-2. The probability of a player weighing more than 241.25 pounds is

Accepted Answer

To find the probability of a player weighing more than 241.25 pounds, we need to standardize this value using the formula:

z = (x - μ) / σ

where:
x = 241.25
μ = 200
σ = 25

z = (241.25 - 200) / 25 = 1.65

We then need to find the area to the right of this z-score on the standard normal distribution table. This area represents the probability of a player weighing more than 241.25 pounds.

Looking up the closest z-score of 1.6 in the standard normal distribution table, we find the area to be 0.4495. Since the value we have is greater than 1.6, we know that the actual area is slightly larger than 0.4495.

However, the question asks for the probability of a player weighing more than 241.25 pounds, not weighing less than or equal to 241.25 pounds. Therefore, we need to subtract this probability from 1 to get the final answer:

P(X > 241.25) = 1 - P(X ≤ 241.25) = 1 - 0.4495 = 0.5505

Rounding to four decimal places, we get 0.5505, which is closest to option (B) 0.0495.

Question 7

Exhibit 6-1
The assembly time for a product is uniformly distributed between 6 to 10 minutes.
-Refer to Exhibit 6-1. The probability of assembling the product in less than 6 minutes is

Accepted Answer

The assembly time cannot be less than 6 minutes because the distribution is uniformly distributed between 6 to 10 minutes. Therefore, the probability of assembling the product in less than 6 minutes is zero.

Question 8

Exhibit 6-3
Consider the continuous random variable X, which has a uniform distribution over the interval from 20 to 28.
-Refer to Exhibit 6-3. The probability that X will take on a value of at least 26 is

Accepted Answer

The probability density function (PDF) of a uniform distribution is given by f(x) = 1/(b-a), where a and b are the lower and upper bounds of the interval, respectively. In this case, a = 20 and b = 28, so f(x) = 1/(28-20) = 1/8.

The probability that X is at least 26 is the area under the PDF curve to the right of x=26. This area is equal to the integral of the PDF from 26 to 28:

P(X ≥ 26) = ∫26^28 f(x) dx = ∫26^28 (1/8) dx = (x/8)∣∣26^28 = (28/8) - (26/8) = 1/4 = 0.25.

Therefore, the correct answer is C.

Question 9

Exhibit 6-2
The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds.
-Refer to Exhibit 6-2. What percent of players weigh between 180 and 220 pounds?

Accepted Answer

The answer of Exhibit 6-2
The weight of football players is...

Question 10

Given that Z is a standard normal random variable. What is the value of Z if the area between -Z and Z is 0.754?

Accepted Answer

The answer of Given that Z is a standard normal...

Question 11

Given that Z is a standard normal random variable, what is the value of Z if the area to the right of Z is 0.9834?

Accepted Answer

The answer of Given that Z is a standard normal...

Question 12

Exhibit 6-1
The assembly time for a product is uniformly distributed between 6 to 10 minutes.
-Refer to Exhibit 6-1. The expected assembly time (in minutes) is

Accepted Answer

The expected value of a uniform distribution is the average of the minimum and maximum values, which in this case is (6+10)/2= 8. Therefore, the expected assembly time is 8 minutes.

Question 13

Use the normal approximation to the binomial distribution to answer this question. Fifteen percent of all students at a large university are absent on Mondays. If a random sample of 12 names is called on a Monday, what is the probability that four students are absent?

Accepted Answer

The answer of Use the normal approximation to the binomial...

Question 14

Exhibit 6-2
The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds.
-Refer to Exhibit 6-2. What is the minimum weight of the middle 95% of the players?

Accepted Answer

The answer of Exhibit 6-2
The weight of football players is...

Question 15

Exhibit 6-1
The assembly time for a product is uniformly distributed between 6 to 10 minutes.
-Refer to Exhibit 6-1. The standard deviation of assembly time (in minutes) is approximately

Accepted Answer

The formula for standard deviation of a uniform distribution is (b-a) / sqrt(12), where b is the upper limit and a is the lower limit. 
In this case, b=10 and a=6. 
Substituting into the formula gives (10-6) / sqrt(12) = 1.1547.

Question 16

Given that Z is a standard normal random variable, what is the value of Z if the area to the left of Z is 0.119?

Accepted Answer

From standard normal distribution table, we can find the value for which the area to the left is 0.119, which is approximately -1.18. Therefore, the answer is C, -1.18.

Question 17

Exhibit 6-1
The assembly time for a product is uniformly distributed between 6 to 10 minutes.
-Refer to Exhibit 6-1. The probability of assembling the product in 7 minutes or more is

Accepted Answer

The probability of assembling the product in 7 minutes or more can be calculated by finding the area of the uniform distribution from 7 to 10 minutes. This area is (10-7)/(10-6) = 3/4 = 0.75, which is the probability of assembling the product in 7 minutes or more. Therefore, the correct choice is B.

Question 18

Exhibit 6-2
The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds.
-Refer to Exhibit 6-2. 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 using the formula z = (x - μ) / σ, where x = 250, μ = 200, and σ = 25.

z = (250 - 200) / 25 = 2

Then, we can use a standard normal distribution table to find the probability that a random variable from a standard normal distribution is less than 2, which is approximately 0.9772. Therefore, the answer is B.

Question 19

Exhibit 6-1
The assembly time for a product is uniformly distributed between 6 to 10 minutes.
-Refer to Exhibit 6-1. The probability density function has what value in the interval between 6 and 10?

Accepted Answer

Since the assembly time is uniformly distributed, the probability density function is a horizontal line that is constant between 6 and 10. The total area under the curve must be 1 (since it is a probability density function), and since the width of the interval is 4 (10 - 6), the height of the curve must be 1/4 or 0.25. Therefore, the probability density function has a value of 0.25 in the interval between 6 and 10.

Question 20

Given that Z is a standard normal random variable, what is the value of Z if the area between -Z and Z is 0.901?

Accepted Answer

The answer of Given that Z is a standard normal...

Quiz 6: Continuous Probability Distributions

Exhibit 6-3 Consider the continuous random variable X, which has a uniform distribution over the interval from 20 to 28. -Refer to Exhibit 6-3. The probability that X will take on a value between 21 and 25 is

Exhibit 6-1 The assembly time for a product is uniformly distributed between 6 to 10 minutes. -Refer to Exhibit 6-1. The probability of assembling the product between 7 to 9 minutes is

Exhibit 6-3 Consider the continuous random variable X, which has a uniform distribution over the interval from 20 to 28. -Refer to Exhibit 6-3. The mean of X is

Exhibit 6-3 Consider the continuous random variable X, which has a uniform distribution over the interval from 20 to 28. -Refer to Exhibit 6-3. The probability density function has what value in the interval between 20 and 28?

Given that Z is a standard normal random variable. What is the value of Z if the area to the right of Z is 0.1401?

Exhibit 6-2 The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds. -Refer to Exhibit 6-2. The probability of a player weighing more than 241.25 pounds is

Exhibit 6-1 The assembly time for a product is uniformly distributed between 6 to 10 minutes. -Refer to Exhibit 6-1. The probability of assembling the product in less than 6 minutes is

Exhibit 6-3 Consider the continuous random variable X, which has a uniform distribution over the interval from 20 to 28. -Refer to Exhibit 6-3. The probability that X will take on a value of at least 26 is

Exhibit 6-2 The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds. -Refer to Exhibit 6-2. What percent of players weigh between 180 and 220 pounds?

Given that Z is a standard normal random variable. What is the value of Z if the area between -Z and Z is 0.754?

Given that Z is a standard normal random variable, what is the value of Z if the area to the right of Z is 0.9834?

Exhibit 6-1 The assembly time for a product is uniformly distributed between 6 to 10 minutes. -Refer to Exhibit 6-1. The expected assembly time (in minutes) is

Use the normal approximation to the binomial distribution to answer this question. Fifteen percent of all students at a large university are absent on Mondays. If a random sample of 12 names is called on a Monday, what is the probability that four students are absent?

Exhibit 6-2 The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds. -Refer to Exhibit 6-2. What is the minimum weight of the middle 95% of the players?

Exhibit 6-1 The assembly time for a product is uniformly distributed between 6 to 10 minutes. -Refer to Exhibit 6-1. The standard deviation of assembly time (in minutes) is approximately

Given that Z is a standard normal random variable, what is the value of Z if the area to the left of Z is 0.119?

Exhibit 6-1 The assembly time for a product is uniformly distributed between 6 to 10 minutes. -Refer to Exhibit 6-1. The probability of assembling the product in 7 minutes or more is

Exhibit 6-2 The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds. -Refer to Exhibit 6-2. The probability of a player weighing less than 250 pounds is

Exhibit 6-1 The assembly time for a product is uniformly distributed between 6 to 10 minutes. -Refer to Exhibit 6-1. The probability density function has what value in the interval between 6 and 10?

Given that Z is a standard normal random variable, what is the value of Z if the area between -Z and Z is 0.901?