Question 1

The life expectancy of a particular brand of tire is normally distributed with a mean of 40,000 and a standard deviation of 5,000 miles.What is the probability that a randomly selected tire will have a life of exactly 47,500 miles?
A) 0.0000
B) 0.9332
C) 0.0668
D) 0.4993

Accepted Answer

The probability of a continuous variable having an exact value is always 0. Therefore, the probability of a randomly selected tire having a life of exactly 47,500 miles is 0.

Question 2

A professor at a local university noted that the exam grades of her students were normally distributed with a mean of 73 and a standard deviation of 11.The professor has informed us that 7.93 percent of her students received grades of A.What is the minimum score needed to receive a grade of A?
A) 90.51
B) 93.2
C) 88.51
D) 100.00

Accepted Answer

The answer of A professor at a local university noted...

Question 3

What is the mean of x, given the function below f(x) =(1/10) e-x^/10 x $\geq$ 0 A) 0.10 B) 10 C) 100 D) 1,000

Accepted Answer

The given function is an exponential probability density function with a rate parameter (λ) of 1/10. The mean of an exponential distribution is 1/λ, which is 10 in this case.

Question 4

The travel time for a college student traveling between her home and her college is uniformly distributed between 40 and 90 minutes.The probability that her trip will take exactly 50 minutes is
A) zero
B) 0.02
C) 0.06
D) 0.20

Accepted Answer

Since the distribution is uniform, there is an equal probability of the student's trip taking any value between 40 and 90 minutes. Therefore, the probability of the trip taking exactly 50 minutes is zero, since it is just one value out of an infinite number of possible values in that range.

Question 5

What is the probability that x is between 3 and 6, given the function below f(x) =(1/10) e^-^x^/10 x $\geq$ 0 A) 0.4512 B) 0.1920 C) 0.2592 D) 0.6065

Accepted Answer

The answer of What is the probability that x is...

Question 6

The life expectancy of a particular brand of tire is normally distributed with a mean of 40,000 and a standard deviation of 5,000 miles.What percentage of tires will have a life of 34,000 to 46,000 miles?
A) 38.49%
B) 76.98%
C) 50%
D) 88.49%

Accepted Answer

To find the percentage of tires that will have a life of 34,000 to 46,000 miles, we need to calculate the area under the normal distribution curve between these two values. We can do this by standardizing the values using the formula:
z = (x - mu) / sigma
where x is the value we want to standardize, mu is the mean, and sigma is the standard deviation. 
For x = 34,000:
z = (34,000 - 40,000) / 5,000 = -1.2
For x = 46,000:
z = (46,000 - 40,000) / 5,000 = 1.2
Using a standard normal distribution table or calculator, we can find that the area between z = -1.2 and z = 1.2 is approximately 0.7698 or 76.98%. Therefore, the percentage of tires that will have a life of 34,000 to 46,000 miles is 76.98%.

Question 7

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.What percentage of MBA's will have starting salaries of $34,000 to $46,000?
A) 38.49%
B) 38.59%
C) 50%
D) 76.98%

Accepted Answer

To find the percentage of MBA's with starting salaries between $34,000 and $46,000, we need to find the area under the normal distribution curve between these two values. First, we need to standardize these values by subtracting the mean and dividing by the standard deviation:

$34,000 - $40,000 = -$6,000
-$6,000 / $5,000 = -1.2

$46,000 - $40,000 = $6,000
$6,000 / $5,000 = 1.2

Using a standard normal distribution table, we can find the area between -1.2 and 1.2 to be approximately 0.7698 or 76.98%. Therefore, the answer is D.

Question 8

The life expectancy of a particular brand of tire is normally distributed with a mean of 40,000 and a standard deviation of 5,000 miles.What is the probability that a randomly selected tire will have a life of at least 47,500 miles?
A) 0.4332
B) 0.9332
C) 0.0668
D) 0.4993

Accepted Answer

To find the probability that a tire lasts at least 47,500 miles, we calculate the Z-score: Z = (47,500 - 40,000) / 5,000 = 1.5. Using the standard normal distribution table, the probability of Z being less than 1.5 is approximately 0.9332. Therefore, the probability of a tire lasting at least 47,500 miles is 1 - 0.9332 = 0.0668.

Question 9

The travel time for a college student traveling between her home and her college is uniformly distributed between 40 and 90 minutes.The probability that she will finish her trip in 80 minutes or less is
A) 0.02
B) 0.8
C) 0.2
D) 1.00

Accepted Answer

The uniform distribution has a constant probability. The range is 40 to 90 minutes, so the total range is 50 minutes. The probability of finishing in 80 minutes or less is the range from 40 to 80 minutes, which is 40 minutes. The probability is 40/50 = 0.8.

Question 10

A professor at a local university noted that the exam grades of her students were normally distributed with a mean of 73 and a standard deviation of 11.If 69.5 percent of the students received grades of C or better, what is the minimum score of those who received C's?
A) 70.39
B) 67.39
C) 50.39
D) 65.39

Accepted Answer

The answer of A professor at a local university noted...

Question 11

The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds.What is the minimum weight of the middle 95% of the players?
A) 196
B) 151
C) 249
D) 190

Accepted Answer

The middle 95% of a normal distribution falls within approximately 2 standard deviations of the mean. The mean is 200 pounds, and the standard deviation is 25 pounds. To find the minimum weight of the middle 95%, subtract 2 standard deviations from the mean: 200 - 2(25) = 200 - 50 = 150. Since 151 is the closest option, it is the correct answer.

Question 12

What is the probability that x is less than 5, given the function below f(x) =(1/10) e^-^x^/10 x $\geq$ 0 A) 0.6065 B) 0.0606 C) 0.3935 D) 0.9393

Accepted Answer

The answer of What is the probability that x is...

Question 13

The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds.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 range 180 to 220 pounds is within one standard deviation from the mean (200 ± 25). Approximately 68% of data in a normal distribution falls within one standard deviation, so the percentage of players weighing between 180 and 220 pounds is 57.62% (half of 68% since it's one side of the mean).

Question 14

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.What is the probability that a randomly selected individual with an MBA degree will get a starting salary of at least $30,000?
A) 0.4772
B) 0.9772
C) 0.0228
D) 0.5000

Accepted Answer

We need to find the probability of a starting salary of at least $30,000, which is the same as finding the probability of a salary less than $40,000 (since the distribution is symmetric). Using the z-score formula, we have:

z = (30,000 - 40,000) / 5,000 = -2

Looking up the corresponding probability in the standard normal distribution table, we get:

P(z < -2) = 0.0228

Therefore, the probability of a randomly selected individual with an MBA degree getting a starting salary of at least $30,000 is 1 - 0.0228 = 0.9772, which is option B.

Question 15

The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds.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 find the probability of a player weighing less than 250 pounds, we need to find the area under the normal curve to the left of 250 pounds. First, we standardize 250 pounds using the formula z = (x - μ) / σ where x is the weight value, μ is the mean, and σ is the standard deviation.

z = (250 - 200) / 25 = 2

Next, we look up the probability associated with a z-score of 2 in the standard normal distribution table, which is 0.9772. Therefore, the probability of a player weighing less than 250 pounds is 0.9772.

Question 16

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.What is the probability that a randomly selected individual with an MBA degree will get a starting salary of at least $47,500?
A) 0.4332
B) 0.9332
C) 0.0668
D) 0.5000

Accepted Answer

To find the probability that a randomly selected individual with an MBA degree will get a starting salary of at least $47,500, we use the Z-score formula: Z = (X - μ) / σ, where X is the value of interest ($47,500), μ is the mean ($40,000), and σ is the standard deviation ($5,000). Plugging in the values, we get Z = ($47,500 - $40,000) / $5,000 = 1.5. Looking up the Z-score of 1.5 in a standard normal distribution table or using a calculator, we find that the area to the left of Z = 1.5 is approximately 0.9332. Since we are interested in salaries at least $47,500, we look for the area to the right, which is 1 - 0.9332 = 0.0668. This means there's a 6.68% chance a randomly selected individual with an MBA will have a starting salary of at least $47,500.

Question 17

The travel time for a college student traveling between her home and her college is uniformly distributed between 40 and 90 minutes.The probability that her trip will take longer than 60 minutes is
A) 1.00
B) 0.40
C) 0.02
D) 0.60

Accepted Answer

Since the travel time is uniformly distributed between 40 and 90 minutes, the total area under the curve equals 1. The probability that her trip will take longer than 60 minutes can be found by calculating the area under the curve to the right of 60 minutes. 
The length of the total interval is 90 - 40 = 50 minutes. Since the distribution is uniform, the probability density function is a flat line with height 1/50. The area of a rectangle with height 1/50 and width (90 - 60) = 30 minutes is (1/50) * 30 = 0.6.
Therefore, the probability that her trip will take longer than 60 minutes is 0.6 or 60%.

Question 18

A professor at a local university noted that the exam grades of her students were normally distributed with a mean of 73 and a standard deviation of 11.Students who made 57.93 or lower on the exam failed the course.What percent of students failed the course?
A) 8.53%
B) 18.53%
C) 91.47%
D) 0.853%

Accepted Answer

To find the percentage of students who failed, we calculate the z-score for 57.93 using the formula: z = (X - μ) / σ, where X is 57.93, μ is 73, and σ is 11. The z-score is approximately -1.36. Using the standard normal distribution table, a z-score of -1.36 corresponds to about 8.53% of students scoring below 57.93.

Question 19

The weight of football players is normally distributed with a mean of 200 pounds and a standard deviation of 25 pounds.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

To find the probability of a player weighing more than 241.25 pounds, we need to find the area under the normal curve to the right of 241.25. Firstly, we need to standardize this value:

z = (x - μ) / σ
z = (241.25 - 200) / 25
z = 1.65

We can then use a standard normal distribution table or calculator to find the probability of a z-score of 1.65 or greater, which is 0.0495 or approximately 0.05. Therefore, the answer is B.

Question 20

The life expectancy of a particular brand of tire is normally distributed with a mean of 40,000 and a standard deviation of 5,000 miles.What is the probability that a randomly selected tire will have a life of at least 30,000 miles?
A) 0.4772
B) 0.9772
C) 0.0228
D) 0.5000

Accepted Answer

To find the probability that a randomly selected tire will have a life of at least 30,000 miles, we need to find the area under the normal distribution curve to the right of 30,000 miles. 
Using the standard normal distribution formula, Z= (x - μ) / σ, we get Z = (30,000 - 40,000) / 5,000 = -2. 
Looking up the area under the standard normal distribution curve for Z = -2, we find that it is 0.0228. 
Therefore, the probability that a randomly selected tire will have a life of at least 30,000 miles is 1 - 0.0228 = 0.9772 or approximately 97.72%. 
Thus, the best choice is B.

Quiz 6: Continuous Probability Distributions

The life expectancy of a particular brand of tire is normally distributed with a mean of 40,000 and a standard deviation of 5,000 miles.What is the probability that a randomly selected tire will have a life of exactly 47,500 miles?

A professor at a local university noted that the exam grades of her students were normally distributed with a mean of 73 and a standard deviation of 11.The professor has informed us that 7.93 percent of her students received grades of A.What is the minimum score needed to receive a grade of A?

What is the mean of x, given the function below f(x) =(1/10) e-x/10 x ≥\geq≥ 0

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

What is the probability that x is between 3 and 6, given the function below f(x) =(1/10) e-x/10 x ≥\geq≥ 0

The life expectancy of a particular brand of tire is normally distributed with a mean of 40,000 and a standard deviation of 5,000 miles.What percentage of tires will have a life of 34,000 to 46,000 miles?

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.What percentage of MBA's will have starting salaries of $34,000 to $46,000?

The life expectancy of a particular brand of tire is normally distributed with a mean of 40,000 and a standard deviation of 5,000 miles.What is the probability that a randomly selected tire will have a life of at least 47,500 miles?

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

A professor at a local university noted that the exam grades of her students were normally distributed with a mean of 73 and a standard deviation of 11.If 69.5 percent of the students received grades of C or better, what is the minimum score of those who received C's?

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

What is the probability that x is less than 5, given the function below f(x) =(1/10) e-x/10 x ≥\geq≥ 0

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

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.What is the probability that a randomly selected individual with an MBA degree will get a starting salary of at least $30,000?

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

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.What is the probability that a randomly selected individual with an MBA degree will get a starting salary of at least $47,500?

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

A professor at a local university noted that the exam grades of her students were normally distributed with a mean of 73 and a standard deviation of 11.Students who made 57.93 or lower on the exam failed the course.What percent of students failed the course?

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

The life expectancy of a particular brand of tire is normally distributed with a mean of 40,000 and a standard deviation of 5,000 miles.What is the probability that a randomly selected tire will have a life of at least 30,000 miles?

What is the mean of x, given the function below f(x) =(1/10) e-x^/10 x $\geq$ 0

What is the probability that x is between 3 and 6, given the function below f(x) =(1/10) e^-^x^/10 x $\geq$ 0

What is the probability that x is less than 5, given the function below f(x) =(1/10) e^-^x^/10 x $\geq$ 0