Question 1

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?
A)0.25
B)4.00
C)5.00
D)zero

Accepted Answer

The probability density function for a uniform distribution is simply the reciprocal of the interval length. Therefore, from 6 to 10, the interval length is 4, and the probability density function has a value of 1/4 = 0.25.

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 standard deviation of assembly time (in minutes)is approximately
A)1.3333
B)1.1547
C)0.1111
D)0.5773

Accepted Answer

The formula for the standard deviation of a uniform distribution is (b-a)/sqrt(12), where a is the lower limit and b is the upper limit. Plugging in the values given, we get (10-6)/sqrt(12) which simplifies to approximately 1.1547. Therefore, the answer is B.

Question 3

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
A)zero
B)0.50
C)0.15
D)1

Accepted Answer

The question states that the assembly time is uniformly distributed between 6 to 10 minutes, meaning that it cannot be less than 6 minutes. Therefore, the probability of assembling the product in less than 6 minutes is zero.

Question 4

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?
A)1.08
B)0.1401
C)2.16
D)-1.08

Accepted Answer

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

Question 5

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
A)0.000
B)0.125
C)23
D)24

Accepted Answer

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

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 less than 250 pounds is
A)0.4772
B)0.9772
C)0.0528
D)0.5000

Accepted Answer

To calculate the probability of a player weighing less than 250 pounds, we can use the standard normal distribution formula and convert the weight to a z-score:

z = (250 - 200) / 25 = 2

Using a standard normal distribution table or calculator, we can find that the probability of a z-score less than 2 is 0.9772. Therefore, the answer is B.

Question 7

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
A)0.4505
B)0.0495
C)0.9505
D)0.9010

Accepted Answer

We need to find the area to the right of 241.25 in a normal distribution with mean=200, σ=25. First, we need to standardize the value of 241.25:

Z = (241.25 - 200)/25 = 1.65

Then, using a standard normal distribution table, we can find the area to the right of 1.65:

P(Z > 1.65) = 0.0495

Therefore, the probability of a player weighing more than 241.25 pounds is 0.0495 or approximately 5%.

Question 8

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?
A)28.81%
B)0.5762%
C)0.281%
D)57.62%

Accepted Answer

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

Question 9

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
A)zero
B)0.50
C)0.20
D)1

Accepted Answer

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

Question 10

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?
A)0.4834
B)-2.13
C)+2.13
D)zero

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 left of Z is 0.119?
A)0.381
B)+1.18
C)-1.18
D)2.36

Accepted Answer

Using a standard normal distribution table, we can find the Z-score that corresponds to an area of 0.119 to the left of it. We look for the closest value to 0.119 in the table and find that it is 0.379. The corresponding Z-score is -1.18. Therefore, the answer is choice C (-1.18).

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
A)16
B)2
C)8
D)4

Accepted Answer

The expected value of a uniform distribution is the average of the minimum and maximum values. Therefore, the expected assembly time can be calculated as: 
Expected assembly time = (Minimum assembly time + Maximum assembly time) / 2 
Expected assembly time = (6 + 10) / 2 
Expected assembly time = 8 
Thus, the best choice is C.

Question 13

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
A)0.125
B)0.250
C)0.500
D)1.000

Accepted Answer

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

Question 14

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
A)0.25
B)0.75
C)zero
D)1

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 15

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?
A)0.0683
B)0.0213
C)0.0021
D)0.1329

Accepted Answer

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

Question 16

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
A)0.000
B)0.125
C)0.250
D)1.000

Accepted Answer

Since X has a uniform distribution over the interval from 20 to 28, the probability of X taking on a value of at least 26 is equal to the probability of X being greater than or equal to 26, which can be calculated as follows:
P(X ≥ 26) = (28 - 26)/(28 - 20) = 2/8 = 0.25
Therefore, the answer is choice C.

Question 17

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?
A)0
B)0.050
C)0.125
D)1.000

Accepted Answer

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

Question 18

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?
A)1.96
B)-1.96
C)0.4505
D)1.65

Accepted Answer

The area between -Z and Z being 0.901 implies that the cumulative probability from -Z to Z is 0.901. This corresponds to a Z value of approximately 1.65 for a standard normal distribution.

Question 19

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?
A)1.16
B)1.96
C)2.0
D)11.6

Accepted Answer

The area between -Z and Z being 0.754 implies that the cumulative probability from -Z to Z is 0.754. This corresponds to a Z-score of approximately 1.16 for each tail, as the standard normal distribution is symmetric.

Question 20

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?
A)196
B)151
C)249
D)190

Accepted Answer

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

Quiz 6: Continuous Probability Distributions

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?

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

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

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 6-3 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 6-2 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 6-2 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 6-2 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?

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

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?

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 expected assembly time (in 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 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 in 7 minutes or more 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 6-3 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 6-3 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 between -Z and Z is 0.901?

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?

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?