Question 1

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're interested in the probability of earning at least $47,500, we look for the area to the right, which is 1 - 0.9332 = 0.0668.

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.Students who made 59.99 or lower on the exam failed the course.What percent of students failed the course?
A) 11.85%
B) 88.15%
C) 1.80%
D) 98.20%

Accepted Answer

To find the percentage of students who failed the course, we calculate the z-score for a grade of 59.99. The z-score formula is $z = \frac{(X - \mu)}{\sigma}$, where $X$ is the value in question, $\mu$ is the mean, and $\sigma$ is the standard deviation. Plugging in the values, we get $z = \frac{(59.99 - 73)}{11} = -1.1827$. Looking up this z-score in a standard normal distribution table or using a calculator, we find that approximately 11.85% of the area under the curve falls below this z-score, indicating that about 11.85% of students failed the course.

Question 3

What is the probability that x is between 3 and 6, given the function below? f(x) = (1/10) e^-^x^/10 x 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 4

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 8.75 percent of her students received grades of A.What is the minimum score needed to receive a grade of A?
A) 85.00
B) 95.00
C) 87.92
D) 90.00

Accepted Answer

First, we need to find the z-score corresponding to the top 8.75 percent of the distribution. Using a standard normal distribution table or calculator, we find that the z-score is approximately 1.44.

Next, we can use the formula for a z-score: 
z = (x - mu) / sigma

Plugging in the values we know: 
1.44 = (x - 73) / 11

Solving for x: 
x = 87.92

Therefore, the minimum score needed to receive a grade of A is 87.92, which is closest to choice C, 87.92.

Question 5

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 60 minutes or less is
A) 0.8.
B) 0.4.
C) 0.2.
D) 1.00.

Accepted Answer

The total time range is 90-40=50 minutes. Since the uniform distribution is evenly spread, the probability of the student finishing the trip in 60 minutes or less is the proportion of the total time range that falls within that range, which is (60-40)/50=0.4. Therefore, the answer is B.

Question 6

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 80 minutes is
A) 1.00.
B) 0.40.
C) 0.20.
D) 0.80.

Accepted Answer

Since the travel time is uniformly distributed between 40 and 90 minutes, the probability of the trip taking longer than 80 minutes is equal to the proportion of the interval from 80 to 90 minutes to the full interval from 40 to 90 minutes. 
This can be calculated as: 
(P(Trip takes longer than 80 minutes)) = (Length of interval from 80 to 90)/(Length of full interval from 40 to 90) = (90-80)/(90-40) = 10/50 = 0.20. 
Therefore, the probability that the trip will take longer than 80 minutes is 0.20 or choice C.

Question 7

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 52,500 miles?
A) 0.0062
B) 0.9938
C) 0.0000
D) 1.0000

Accepted Answer

To find the probability that a randomly selected tire will have a life of at least 52,500 miles, we use the Z-score formula: Z = (X - μ) / σ, where X is the value of interest (52,500 miles), μ is the mean (40,000 miles), and σ is the standard deviation (5,000 miles). Plugging in the values, we get Z = (52,500 - 40,000) / 5,000 = 2.5. Looking up the Z-score of 2.5 in a standard normal distribution table or using a calculator, we find that the area to the right of Z = 2.5 (which represents the probability of a tire lasting at least 52,500 miles) is approximately 0.0062.

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 35,000 miles?
A) 0.1587
B) 0.8413
C) 0.0000
D) 1.0000

Accepted Answer

We need to find the probability that a randomly selected tire will have a life of at least 35,000 miles. This can be computed as follows:

P(X ≥ 35,000) = P(Z ≥ (35,000 - 40,000)/5,000) [Converting X to a standard normal variable Z]
= P(Z ≥ -1)
= 0.8413 [Using standard normal distribution table]

Therefore, the answer is (B) 0.8413.

Question 9

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

Accepted Answer

To find the mean of x, we need to calculate the integral of x times the given function over the entire real line and then divide by the integral of the function over the entire real line.

Since the integral of the function over the entire real line is just a constant (equal to 1/10), we only need to calculate the integral of x times the function.

Using some calculus tricks and integrating by parts, we can show that the integral of x times the function is equal to 10.

Therefore, the mean of x is 10.

Question 10

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

Accepted Answer

We need to find the probability that x is less than 5, which can be written as P(x < 5). Using the given function, we can find the cumulative distribution function (CDF) as follows: F(x) = ∫f(t) dt from -∞ to x = (1/10) ∫e^(^-^{t) (1/10}) dt from -∞ to x = (-1/10) e^(^-^{t) from -∞ to x
= (-1/10}) (e^(^-^{x) - e^(-∞))
= (-1/10}) (e^(^-^{x))

Now, we can use the CDF to find the required probability:

P(x < 5) = F(5)
= (-1/10}) (e^(^-^{5))
≈ 0.3935

Therefore, the best choice is C.}

Question 11

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

To solve this problem, we need to use the z-score formula which tells us how many standard deviations a data point is from the mean.

z = (x - μ) / σ

where 
x = starting salary
μ = mean starting salary 
σ = standard deviation

We want to find the probability that a randomly selected individual with an MBA degree will get a starting salary of at least $30,000.

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

Using a standard normal distribution table, we can find that the probability of getting a z-score of -2 or lower is 0.9772.

Therefore, the probability of a randomly selected individual with an MBA degree getting a starting salary of at least $30,000 is 0.9772 or 97.72%.

The answer is B.

Question 12

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 35,000 to 45,000 miles?
A) 31.73%
B) 68.27%
C) 50%
D) 95%

Accepted Answer

We need to find the percentage of tires with a life of 35,000 to 45,000 miles, which is one standard deviation above and below the mean. Using the empirical rule, we know that about 68% of the tires will fall within this range. Therefore, the answer is (B) 68.27%.

Question 13

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 is captured within 2 standard deviations from the mean. The calculation is: mean - 2(standard deviation) = 200 - 2(25) = 150. However, since 150 is not an option and the question likely involves rounding, the closest value provided is 151.

Question 14

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 continuous and uniform, the probability of any specific value (e.g. 50 minutes) is zero.

Question 15

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 16

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 240 pounds is
A) 0.0548.
B) 0.9452.
C) 0.9956.
D) 0.0044.

Accepted Answer

We need to find the area under the normal distribution curve to the left of 240 pounds. Using a standard normal distribution table, we find the corresponding z-score to be 1.6. Therefore, the probability of a player weighing less than 240 pounds is the same as the probability of a standard normal variable being less than 1.6, which is 0.9452.

Question 17

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 175 and 225 pounds?
A) 50%
B) 68.27%
C) 31.73%
D) 95%

Accepted Answer

We can solve this problem using the empirical rule or z-score table.
Empirical rule:
About 68% of the data falls within one standard deviation of the mean.
About 95% of the data falls within two standard deviations of the mean.
About 99.7% of the data falls within three standard deviations of the mean.

Since 175 and 225 are each 1 standard deviation away from the mean of 200, we know that about 68% of the data falls between these two values.

Therefore, the best choice is B) 68.27%.

Question 18

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 250 pounds is
A) 0.228.
B) 0.9772.
C) 0.0022.
D) 0.9978.

Accepted Answer

The answer of The weight of football players is normally...

Question 19

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 42,500 miles?
A) 0.0000
B) 1.0000
C) 0.5000
D) 0.9500

Accepted Answer

The probability of a tire having an exact life of 42,500 miles is zero, as the normal distribution is a continuous probability distribution and the probability of any exact value is zero.

Question 20

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 calculate the Z-scores for both values and then find the percentage of values between those Z-scores in a standard normal distribution. The Z-score formula is $Z = \frac{X - \mu}{\sigma}$, where $X$ is the value, $\mu$ is the mean, and $\sigma$ is the standard deviation.For $34,000: $Z = \frac{34,000 - 40,000}{5,000} = -1.2$For $46,000: $Z = \frac{46,000 - 40,000}{5,000} = 1.2$Looking up these Z-scores in a standard normal distribution table or using a calculator, we find that the area between Z = -1.2 and Z = 1.2 is approximately 76.98%. This represents the percentage of MBA's with starting salaries between $34,000 and $46,000.

Quiz 6: Continuous Probability Distributions

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 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 59.99 or lower on the exam failed the course.What percent of students failed the course?

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

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 8.75 percent of her students received grades of A.What is the minimum score needed to receive a grade of A?

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 60 minutes or less is

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 80 minutes 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 52,500 miles?

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 35,000 miles?

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

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

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 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 35,000 to 45,000 miles?

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?

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 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.The probability of a player weighing less than 240 pounds is

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 175 and 225 pounds?

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 250 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 exactly 42,500 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?

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

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

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