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 standard deviation of assembly time (in minutes) is approximately

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. In this case, a=6 and b=10. Therefore, the standard deviation is (10-6)/sqrt(12) = 1.1547 (rounded to four decimal places). Therefore, the answer is B.

Question 2

Exhibit 6-4
f(x) =(1/10) eS1U1P1-x/10S1S1P0 x  0
-Refer to Exhibit 6-4. The probability that x is between 3 and 6 is

Accepted Answer

The answer of Exhibit 6-4
f(x) =(1/10) eS1U1P1-x/10S1S1P0 x  0
-Refer...

Question 3

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 Z-scores for 180 and 220 pounds are -0.8 and 0.8, respectively. The area between these Z-scores in a standard normal distribution is approximately 57.62%.

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 that X will take on a value of at least 26 is

Accepted Answer

The probability that X will take on a value of at least 26 is equal to the area under the probability density function from 26 to 28. Since X has a uniform distribution, the probability density function is constant over the interval from 20 to 28 and equal to 1/(28-20) = 1/8. Therefore, the probability that X will take on a value of at least 26 is:
P(X ≥ 26) = ∫26 to 28 (1/8)dx = (1/8)∫26 to 28 dx = (1/8)(28-26) = 0.25 or 1/4 or 25%.

Hence, the answer is C (0.250).

Question 5

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 find the area under the normal curve to the left of 250 pounds. We can use a standard normal distribution table or a calculator to find this area. Using a calculator, we can standardize the value of 250 pounds as follows:

z = (x - μ) / σ 
z = (250 - 200) / 25 
z = 2

The area to the left of z = 2 in a standard normal distribution table is 0.9772. Therefore, the probability of a player weighing less than 250 pounds is 0.9772.

Question 6

Exhibit 6-6
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 6-6. What is the probability that a randomly selected individual with an MBA degree will get a starting salary of at least $30,000?

Accepted Answer

Using the normal distribution calculator with a mean of $40,000 and a standard deviation of $5,000, we can find the area to the left of $30,000.
Z-score = (30,000 - 40,000) / 5,000 = -2
Looking at the z-table, we find the probability of a z-score being less than -2 is 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.

Question 7

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

For a uniform distribution, the probability density function (pdf) is constant over the interval. The pdf is calculated as 1/(b-a), where a and b are the endpoints of the interval. Here, it is 1/(28-20) = 1/8 = 0.125.

Question 8

Exhibit 6-7
The weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces.
-Refer to Exhibit 6-7. What is the probability that a randomly selected item will weigh more than 10 ounces?

Accepted Answer

To find the probability that a randomly selected item weighs more than 10 ounces, calculate the z-score: z = (10 - 8) / 2 = 1. Using the standard normal distribution table, the probability of z being less than 1 is 0.8413. Therefore, the probability of z being more than 1 is 1 - 0.8413 = 0.1587.

Question 9

Exhibit 6-6
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 6-6. What is the probability that a randomly selected individual with an MBA degree will get a starting salary of at least $47,500?

Accepted Answer

To find the probability that a randomly selected individual will get a starting salary of at least $47,500, we calculate the z-score: z = (47,500 - 40,000) / 5,000 = 1.5. Using the standard normal distribution table, the probability of a z-score less than 1.5 is 0.9332. Therefore, the probability of a z-score greater than 1.5 is 1 - 0.9332 = 0.0668.

Question 10

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

Question 11

Exhibit 6-5
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 6-5. The probability that she will finish her trip in 80 minutes or less is

Accepted Answer

Since the travel time is uniformly distributed, the probability of finishing the trip in 80 minutes or less is simply the proportion of the total range of travel times that fall at or below 80 minutes: (80-40)/(90-40) = 0.8.

Question 12

Exhibit 6-4
f(x) =(1/10) eS1U1P1-x/10S1S1P0 x  0
-Refer to Exhibit 6-4. The probability that x is less than 5 is

Accepted Answer

To find the probability that x is less than 5, we need to evaluate the integral of f(x) from negative infinity to 5. This can be done using a calculator or software, and the result is approximately 0.3935. Therefore, the answer is C.

Question 13

Exhibit 6-5
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 6-5. The probability that her trip will take exactly 50 minutes is

Accepted Answer

In a continuous uniform distribution, the probability of the random variable taking on any single, exact value is zero.

Question 14

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 variance of X is approximately

Accepted Answer

The formula for calculating the variance of a uniform distribution is:

Var(X) = (b - a)^2 / 12

where a and b are the minimum and maximum values of the interval.

In this case, a = 20 and b = 28, so:

Var(X) = (28 - 20)^2 / 12 = 5.333

Therefore, the variance of X is approximately 5.333.

Option B is the correct answer.

Question 15

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

We need to standardize the value of 241.25 using the formula z = (x - mu) / sigma, where x is the value we're interested in, mu is the mean, and sigma is the standard deviation. So z = (241.25 - 200) / 25 = 1.65. Then we can use a standard normal distribution table to find the probability of a value being greater than 1.65, which is 0.0495. Therefore, the answer is choice B.

Question 16

Exhibit 6-5
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 6-5. The probability that her trip will take longer than 60 minutes is

Accepted Answer

Since the travel time is uniformly distributed, we know that the probability of any segment of time within the range is proportional to the length of that segment. Therefore, the probability of the trip taking longer than 60 minutes is equal to the length of the segment from 60 to 90 minutes, which is 30 minutes out of the total range of 50 minutes. Therefore, the probability is 30/50 = 0.6 or 60%. The answer is D.

Question 17

Exhibit 6-6
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 6-6. What percentage of MBA's will have starting salaries of $34,000 to $46,000?

Accepted Answer

The range $34,000 to $46,000 is within one standard deviation (±$5,000) from the mean ($40,000). In a normal distribution, approximately 68% of the data falls within one standard deviation of the mean. However, since $34,000 to $46,000 covers more than one standard deviation (from -1.2 to +1.2), the percentage is higher, approximately 76.98%.

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. 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 19

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 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 20

Exhibit 6-4
f(x) =(1/10) eS1U1P1-x/10S1S1P0 x  0
-Refer to Exhibit 6-4. The mean of x is

Accepted Answer

The mean of a continuous random variable with probability density function f(x) is given by the formula ∫xf(x)dx. Plugging in the given function f(x), we get ∫x(1/10) e^(S1U1P1-x/10S1S1P0)dx. This integral can be solved using integration by substitution, with u=S1U1P1-x/10S1S1P0 and du=-1/10S1S1P0 dx. The limits of integration are not given, so we assume they are from 0 to infinity (since the function is defined for all x ≥ 0). Therefore, the integral becomes ∫u (-1/10) e^u du, which simplifies to (-1/10) e^u (u+1)+C. Substituting back in for u, we get (-1/10) e^(S1U1P1-x/10S1S1P0) (S1U1P1-x/10S1S1P0/10+S1S1P0/10) from 0 to infinity. Evaluating the limit as x approaches infinity gives 0, so we only need to evaluate the expression at x=0. Plugging in x=0 and simplifying gives (-1/10) e^0 (S1U1P1/10+S1S1P0/10), which simplifies to (1/10) (S1U1P1+S1S1P0/10). This means that the mean of x is (1/10) (S1U1P1+S1S1P0/10). Since the answer choices are given in whole numbers, we need to estimate this value. Since both S1U1P1 and S1S1P0 are in the thousands place (based on the scientific notation used in the function), we can estimate the mean to be approximately 10. Note that if we had been given more information (such as the actual values of S1U1P1 and S1S1P0), we could have calculated the mean more precisely.

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 standard deviation of assembly time (in minutes) is approximately

Exhibit 6-4 f(x) =(1/10) e^-x/10 x 0 -Refer to Exhibit 6-4. The probability that x is between 3 and 6 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 6-3 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. The probability of a player weighing less than 250 pounds 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?

Exhibit 6-7 The weight of items produced by a machine is normally distributed with a mean of 8 ounces and a standard deviation of 2 ounces. -Refer to Exhibit 6-7. What is the probability that a randomly selected item will weigh more than 10 ounces?

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 6-5 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 6-5. The probability that she will finish her trip in 80 minutes or less is

Exhibit 6-4 f(x) =(1/10) e^-x/10 x 0 -Refer to Exhibit 6-4. The probability that x is less than 5 is

Exhibit 6-5 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 6-5. The probability that her trip will take exactly 50 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 variance of X is approximately

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 6-5 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 6-5. The probability that her trip will take longer than 60 minutes is

Exhibit 6-6 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 6-6. What percentage of MBA's will have starting salaries of $34,000 to $46,000?

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 6-3 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 6-4 f(x) =(1/10) e^-x/10 x 0 -Refer to Exhibit 6-4. The mean of x is