Question 1

Exhibit 6-7 
-Refer to Exhibit 6-7. The probability that x is less than 5 is

Accepted Answer

The area under the curve to the left of x=5 represents the probability that x is less than 5. From the graph, we can see that this area is approximately 0.3935.

Question 2

Exhibit 6-4
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 6-4. 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 from the mean ($40,000 ± $5,000). In a normal distribution, approximately 68% of the data falls within one standard deviation of the mean. However, since the range is from $34,000 to $46,000, which covers more than one standard deviation on one side, the percentage is higher, approximately 76.98%.

Question 3

Exhibit 6-7 
-The skewness measure for the exponential distributions is

Accepted Answer

The skewness of an exponential distribution is 2, indicating a right-skewed distribution.

Question 4

Exhibit 6-6
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.
-Refer to Exhibit 6-6. What is the probability that a randomly selected tire will have a life of at least 47,500 miles?

Accepted Answer

To find the probability that a tire lasts at least 47,500 miles, calculate the z-score: (47,500 - 40,000) / 5,000 = 1.5. The probability of a z-score greater than 1.5 is 0.0668.

Question 5

Exhibit 6-6
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.
-Refer to Exhibit 6-6. What is the probability that a randomly selected tire will have a life of exactly 47,500 miles?

Accepted Answer

The probability of a tire having an exact mileage is zero since the tire's lifespan is a continuous variable and the probability at one specific point is infinitesimally small.

Question 6

Exhibit 6-7 
-Exponential distributions

Accepted Answer

Exponential distributions are skewed to the right, with a long tail towards higher values.

Question 7

Exhibit 6-7 
-Refer to Exhibit 6-7. The mean of x is

Accepted Answer

The mean of x is found by adding up all the values of x and dividing by the number of values (10 in this case). Therefore, the mean of x is (100+10+1000+1+0.1+10+0.01+1000+10+1)/10 = 222.111, which is closest to 10 (choice B).

Question 8

Exhibit 6-4
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 6-4. 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

To find the probability that a randomly selected individual with an MBA degree will get a starting salary of at least $30,000, we need to find the area under the normal distribution curve to the right of $30,000.

Using the z-score formula: 
z = (x - μ) / σ 
where x = 30,000, μ = 40,000, and σ = 5,000

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

Looking up the z-score in the standard normal distribution table, we find the probability to the right of z = -2 is 0.9772. Therefore, the probability that a randomly selected individual with an MBA degree will get a starting salary of at least $30,000 is 0.9772 or 97.72%.

The best choice is B.

Question 9

Exhibit 6-5
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 6-5. What is the probability that a randomly selected item will weigh between 11 and 12 ounces?

Accepted Answer

First, we need to standardize the values:
$z_{11}=\frac{11-8}{2}=1.5$
$z_{12}=\frac{12-8}{2}=2$

Then, using a standard normal distribution table, we can find the probabilities:
$P(1.5<z<2)=P(z<2)-P(z<1.5)=0.9772-0.9332=0.0440$

Therefore, the probability that a randomly selected item will weigh between 11 and 12 ounces is 0.0440, which corresponds to answer choice D.

Question 10

Exhibit 6-6
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.
-Refer to Exhibit 6-6. What percentage of tires will have a life of 34,000 to 46,000 miles?

Accepted Answer

To find the percentage of tires that will have a life of 34,000 to 46,000 miles, we need to find the z-scores for both values:

z1 = (34,000 - 40,000) / 5,000 = -1.2
z2 = (46,000 - 40,000) / 5,000 = 1.2

Using a standard normal distribution table or calculator, we can find the area between these z-scores:
P(-1.2 < Z < 1.2) = 0.7698

This means that approximately 76.98% of tires will have a life of 34,000 to 46,000 miles. Therefore, the best choice is B.

Question 11

Exhibit 6-5
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 6-5. What is the probability that a randomly selected item will weigh more than 10 ounces?

Accepted Answer

We need to find the probability of an item weighing more than 10 ounces, which can be calculated using z-score: z = (X - mean) / standard deviation z = (10 - 8) / 2 z = 1 Now, we need to find the area under the normal distribution curve to the right of z = 1, which is the probability of an item weighing more than 10 ounces: P(z > 1) = 1 - P(z < 1) Using a standard normal distribution table or calculator, P(z < 1) = 0.8413 Therefore, P(z > 1) = 1 - 0.8413 = 0.1587 Thus, the probability that a randomly selected item will weigh more than 10 ounces is 0.1587 or 15.87%, which corresponds to choice C.

Question 12

Exhibit 6-4
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 6-4. 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 of a salary of at least $47,500, calculate the z-score: (47500 - 40000) / 5000 = 1.5. The probability of a z-score greater than 1.5 is 0.0668.

Question 13

Exhibit 6-5
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 6-5. What percentage of items will weigh at least 11.7 ounces?

Accepted Answer

To solve this problem, we need to find the area under the normal curve to the right of the value of 11.7 ounces. We can standardize the value of 11.7 by subtracting the mean and dividing by the standard deviation: 
z = (11.7 - 8) / 2 = 1.85

Using a standard normal distribution table or calculator, we can find that the area to the right of z = 1.85 is 0.0322, or 3.22%. Therefore, the percentage of items that will weigh at least 11.7 ounces is 3.22%.

Question 14

Exhibit 6-6
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.
-Refer to Exhibit 6-6. What is the random variable in this experiment?

Accepted Answer

The life expectancy of this particular brand of tire is the random variable being measured in this experiment.

Question 15

Exhibit 6-5
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 6-5. What is the random variable in this experiment?

Accepted Answer

The random variable is the weight of items produced by the machine, as this is the variable that is being measured and can vary from item to item. The mean and standard deviation are simply parameters of the distribution of this random variable. The normal distribution is not the random variable itself, but rather a description of the distribution of the random variable.

Question 16

Exhibit 6-5
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 6-5. What is the probability that a randomly selected item weighs exactly 8 ounces?

Accepted Answer

In a continuous probability distribution, the probability of a single exact value is zero.

Question 17

Exhibit 6-7 
-Refer to Exhibit 6-7. The probability that x is between 3 and 6 is

Accepted Answer

The answer of Exhibit 6-7 
-Refer to Exhibit 6-7. The...

Question 18

Exhibit 6-6
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.
-Refer to Exhibit 6-6. What is the probability that a randomly selected tire will have a life of at least 30,000 miles?

Accepted Answer

We need to find the area to the right of 30,000 in a normal distribution with mean=40,000 and standard deviation=5,000. Using the z-score formula, z=(x-μ)/σ, we get z=(30,000-40,000)/5,000=-2. Using a standard normal table or calculator, we find the probability of z being greater than -2 is 0.9772. Therefore, the probability that a randomly selected tire will have a life of at least 30,000 miles is 0.9772 or 97.72%.

Question 19

Exhibit 6-5
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 6-5. What percentage of items will weigh between 6.4 and 8.9 ounces?

Accepted Answer

To find the percentage of items weighing between 6.4 and 8.9 ounces, we calculate the z-scores for both values and find the area between them using the standard normal distribution. The z-score for 6.4 is (6.4-8)/2 = -0.8, and for 8.9 is (8.9-8)/2 = 0.45. The area between these z-scores corresponds to approximately 0.4617 or 46.17%.

Question 20

Exhibit 6-7 
-A property of the exponential distribution is that the mean equals the

Accepted Answer

The answer of Exhibit 6-7 
-A property of the exponential...

Quiz 6: Continuous Probability Distributions

Exhibit 6-7 -Refer to Exhibit 6-7. The probability that x is less than 5 is

Exhibit 6-4 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 6-4. What percentage of MBA's will have starting salaries of $34,000 to $46,000?

Exhibit 6-7 -The skewness measure for the exponential distributions is

Exhibit 6-6 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. -Refer to Exhibit 6-6. What is the probability that a randomly selected tire will have a life of at least 47,500 miles?

Exhibit 6-6 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. -Refer to Exhibit 6-6. What is the probability that a randomly selected tire will have a life of exactly 47,500 miles?

Exhibit 6-7 -Exponential distributions

Exhibit 6-7 -Refer to Exhibit 6-7. The mean of x is

Exhibit 6-5 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 6-5. What is the probability that a randomly selected item will weigh between 11 and 12 ounces?

Exhibit 6-6 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. -Refer to Exhibit 6-6. What percentage of tires will have a life of 34,000 to 46,000 miles?

Exhibit 6-5 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 6-5. What is the probability that a randomly selected item will weigh more than 10 ounces?

Exhibit 6-5 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 6-5. What percentage of items will weigh at least 11.7 ounces?

Exhibit 6-6 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. -Refer to Exhibit 6-6. What is the random variable in this experiment?

Exhibit 6-5 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 6-5. What is the random variable in this experiment?

Exhibit 6-5 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 6-5. What is the probability that a randomly selected item weighs exactly 8 ounces?

Exhibit 6-7 -Refer to Exhibit 6-7. The probability that x is between 3 and 6 is

Exhibit 6-6 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. -Refer to Exhibit 6-6. What is the probability that a randomly selected tire will have a life of at least 30,000 miles?

Exhibit 6-5 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 6-5. What percentage of items will weigh between 6.4 and 8.9 ounces?

Exhibit 6-7 -A property of the exponential distribution is that the mean equals the