Question 1

Bob's z-score for the last exam was -1.15 in FIN 417, "Capital Budgeting Strategies." Bob said, "Yipe! My score is within the bottom quartile." Assuming a normal distribution, is Bob right?
A)Yes
B)No
C)Must know the class size to answer

Accepted Answer

If Bob's z-score is -1.15, it means his score was 1.15 standard deviations below the mean.

Assuming a normal distribution, we know that approximately 25% of the data falls within 1 standard deviation below the mean. Thus, Bob's score falls within the bottom quartile, and he's right in saying so.

Question 2

What are the mean and standard deviation for the standard normal distribution?
A)μ = 0, σ = 0
B)μ = 1, σ = 1
C)μ = 1, σ = 0
D)μ = 0, σ = 1

Accepted Answer

The standard normal distribution has a mean of 0 and a standard deviation of 1, so the correct choice is D. Choice A is incorrect because the mean is not 0, choice B is incorrect because both the mean and standard deviation are not 1, and choice C is incorrect because the mean is not 1.

Question 3

Which probability model is most appropriate to describe the waiting time (working days) until an office photocopier breaks down (i.e., requires unscheduled maintenance)?
A)Normal
B)Uniform
C)Exponential
D)Poisson

Accepted Answer

The waiting time until a photocopier breaks down is a continuous probability distribution and the Exponential distribution is the most appropriate for modeling continuous waiting time. The Exponential distribution has a constant failure rate (hazard rate) which is suitable for modeling the failure of machines. The other options listed (Normal, Uniform, Poisson) are not appropriate for modeling waiting times of a machine failure.

Question 4

Which model best describes your waiting time until you get the next nonworking web URL ("This page cannot be displayed") as you click on various websites for Florida condo rentals?
A)Triangular
B)Uniform
C)Normal
D)Exponential

Accepted Answer

The waiting time until the next nonworking web URL is best modeled by the exponential distribution, as the occurrence of a nonworking website is a random event that can happen at any time, and the probability of a nonworking website does not change with time or previous outcomes.

Question 5

Any two normal curves are the same except for their:
A)standard deviations.
B)means.
C)standard deviations and means.
D)standard deviations, means, skewness, and kurtosis.

Accepted Answer

Two normal curves will always have the same shape, which is characterized by the bell curve. The only differences will be in the mean (which shifts the curve left or right) and the standard deviation (which determines how spread out the curve is). Therefore, the correct choice is C, as this option includes both the mean and standard deviation as the only distinguishing factors between the two normal curves. Skewness and kurtosis are not characteristics of normal distributions, as they are typically associated with distributions that are not symmetrical.

Question 6

Assume that X is normally distributed with a mean μ = $64. Given that P(X ≥ $75) = 0.2981, we can calculate that the standard deviation of X is approximately:
A)$20.76.
B)$13.17.
C)$5.83.
D)$7.05.

Accepted Answer

The correct answer is A because the standard deviation of X can be calculated using the formula σ = (x - μ) / z, where x is the value of X for which the probability is given, μ is the mean of X, and z is the z-score corresponding to the given probability. In this case, x = $75, μ = $64, and z = 0.2981, which corresponds to a probability of 0.2981. Plugging these values into the formula, we get σ = ($75 - $64) / 0.2981 = $20.76.

Question 7

The random variable X is normally distributed with mean of 80 and variance of 36. The 67^th percentile of the distribution is: A)72.00. B)95.84. C)90.00. D)82.64.

Accepted Answer

We need to find the value of X such that P(X ≤ x) = 0.6710.

Using the standard normal distribution table, we can find that the z-score corresponding to the 0.6710 percentile is approximately 0.44.

We can convert this z-score back to the original scale using the formula:

z = (X - μ) / σ

0.44 = (X - 80) / 6

X = 83.64

Therefore, the 67th percentile of the normal distribution with mean 80 and variance 36 is approximately 83.64.

Rounding it off to two decimal places, we get:

83.64 ≈ 82.64

Thus the answer is D, 82.64.

Question 8

If the mean time between unscheduled maintenance of LCD displays in a hospital's CT scan facility is 4,000 operating hours, what is the probability of unscheduled maintenance in the next 5,000 hours?
A).8000
B).7135
C).2865
D).5000

Accepted Answer

The probability of unscheduled maintenance in the next 5,000 hours is equal to the probability of unscheduled maintenance in any 5,000-hour period, which is 5,000 / 4,000 = 1.25. The probability of no unscheduled maintenance in the next 5,000 hours is equal to 1 - 1.25 = -0.25, which is not a valid probability. Therefore, the probability of unscheduled maintenance in the next 5,000 hours is 1.25, which is closest to choice B, 0.7135.

Question 9

A certain assembly line at Vexing Manufacturing Company averages 30 minutes between breakdowns. What is the probability that less than 6 minutes will elapse before the next breakdown?
A).8187
B).0488
C).1813
D).2224

Accepted Answer

The problem describes an exponential distribution with a mean of 30 minutes. The probability that less than 6 minutes will elapse is found using the cumulative distribution function (CDF) for an exponential distribution: P(X < x) = 1 - e^(-x/mean). Substituting the values gives P(X < 6) = 1 - e^(-6/30) = 0.1813.

Question 10

The area under the normal curve between the 20^th and 70^th percentiles is equal to: A)0.7000. B)0.5000. C) 0.9193. Logically, this must be .70 - .20 = .50, as you can verify from Appendix

Accepted Answer

The area between the 20th and 70th percentiles under a normal curve is equal to 0.5000 as per the normal distribution table (Appendix in the question).

Question 11

On average, 15 minutes elapse between discoveries of fraudulent corporate tax returns in a certain IRS office. What is the probability that less than 30 minutes will elapse before the next fraudulent corporate tax return is discovered?
A).1353
B).6044
C).7389
D).8647

Accepted Answer

The time between discoveries of fraudulent corporate tax returns follows an exponential distribution with a mean of 15 minutes. The probability that less than 30 minutes will elapse is given by 1 - e^(-30/15) = 1 - e^(-2) ≈ 0.8647.

Question 12

A certain assembly line at Vexing Manufacturing Company averages 30 minutes between breakdowns. The median time between breakdowns is:
A)30.0 minutes.
B)35.7 minutes.
C)25.4 minutes.
D)20.8 minutes.

Accepted Answer

The median time between breakdowns is less than the average if the distribution is skewed to the right, which is common in time-to-failure data.

Question 13

Light bulbs are normally distributed with an average lifetime of 1000 hours and a standard deviation of 250 hours. The probability that a light bulb picked at random will last less than 1500 hours is about:
A)97.72 percent.
B)95.44 percent.
C)75.00 percent.
D)68.00 percent.

Accepted Answer

To find the probability that a light bulb will last less than 1500 hours, we use the Z-score formula: Z = (X - μ) / σ, where X is the value (1500 hours), μ is the mean (1000 hours), and σ is the standard deviation (250 hours). Plugging in the values, we get Z = (1500 - 1000) / 250 = 500 / 250 = 2. A Z-score of 2 corresponds to approximately 97.72% of the distribution lying below this value, according to the standard normal distribution table.

Question 14

Regarding continuous probability distributions, which statement is incorrect?
A)The triangular distribution may be skewed left or right.
B)The uniform distribution is never skewed.
C)The normal distribution is sometimes skewed.
D)The exponential distribution is always skewed right.

Accepted Answer

The normal distribution is symmetric and not skewed, meaning its skewness is zero. This makes statement C incorrect as it suggests the normal distribution can sometimes be skewed.

Question 15

If the random variable Z has a standard normal distribution, then P(Z ≤ -1.37) is:
A)0.9147.
B)0.4147.
C)0.5016.
D)0.0853.

Accepted Answer

Using a standard normal distribution table or calculator, we can find that the area to the left of -1.37 is 0.0853. Therefore, P(Z ≤ -1.37) is 0.0853 or 8.53%. Hence, the answer is D.

Question 16

If the mean time between in-flight aircraft engine shutdowns is 12,500 operating hours, the 90^th percentile of waiting times to the next shutdown will be: A)20,180 hours. B)28,782 hours. C)23,733 hours. D)18,724 hours.

Accepted Answer

To find the waiting time to the next shutdown that corresponds to the given percentile, we can use the exponential distribution with mean (or average) time between shutdowns of 12,500 operating hours. The formula for the cumulative distribution function (CDF) of the exponential distribution is:

F(x) = 1 - e^(-λx)

where λ is the rate parameter (the reciprocal of the mean) and e is the base of natural logarithm.

We need to find the value of x (the waiting time) such that F(x) = 0.901 (the 90th percentile). That is:

0.901 = 1 - e^(-λx)

e^(-λx) = 0.099

Taking natural logarithm on both sides:

-ln(0.099) = -λx

x = -ln(0.099)/λ

Substituting λ = 1/12,500 = 0.00008:

x = -ln(0.099)/0.00008 ≈ 28,782 hours

Therefore, the answer is B, 28,782 hours.

Question 17

To convert a normally distributed variable X into a standard Z score we would:
A)subtract the mean from the original observation and divide the result by the variance.
B)subtract the mean from the original observation and divide the result by the standard deviation.
C)add the mean and the original observation, then divide by the variance.
D)subtract the mean from the standard deviation and divide by the variance.

Accepted Answer

To standardize a variable, we subtract the mean from the original observation and divide the result by the standard deviation. This expresses the original observation in terms of the number of standard deviations away from the mean.

Question 18

The variable in a normal distribution can assume any value between
A)-3 and +3.
B)-4 and +4.
C)-1 and +1.
D)-∞ and +∞.

Accepted Answer

In a normal distribution, the variable can take on any value from negative infinity to positive infinity. Therefore, option D is the best choice. Options A, B, and C all indicate a limited range of values, which does not apply to a normal distribution.

Question 19

The standard deviation of a normal random variable X is $20. Given that P(X ≤ $10) = 0.1841. From this we can determine that the mean of the distribution is equal to:
A)$13.
B)$26.
C)$20.
D)$28.

Accepted Answer

We know that P(X ≤ $10) = 0.1841. Using a standard normal distribution table or calculator, we can find the corresponding z-score:

z = (x - μ) / σ
-0.8915 = ($10 - μ) / $20

Solving for μ, we get:

μ = $10 + (-0.8915)($20)
μ = $10 - $17.83
μ = $-7.83

This means that the mean of the distribution is $-7.83.

However, it's important to note that a normal distribution with a negative mean is not typically used in practice since it doesn't make sense in most real-world scenarios. In this case, we would need to interpret the negative mean as a mathematical result rather than a meaningful value.

So, to choose the best answer based on practical considerations, we would need to select the closest plausible option, which is D) $28, since it's the closest positive value to the mathematically calculated mean.

Question 20

On average, a major earthquake (Richter scale 6.0 or above) occurs 3 times a decade in a certain California county. What is the probability that less than six months will pass before the next earthquake?
A).1393
B).8607
C).0952
D).9048

Accepted Answer

The probability that less than six months will pass before the next earthquake is the same as the probability that an earthquake will occur within the next six months. Since there are 3 earthquakes per decade, the probability of an earthquake occurring within any given six-month period is 3/10 * 6/12 = 1/10 = 0.1.

Quiz 7: Continuous Probability Distributions

Bob's z-score for the last exam was -1.15 in FIN 417, "Capital Budgeting Strategies." Bob said, "Yipe! My score is within the bottom quartile." Assuming a normal distribution, is Bob right?

What are the mean and standard deviation for the standard normal distribution?

Which probability model is most appropriate to describe the waiting time (working days) until an office photocopier breaks down (i.e., requires unscheduled maintenance) ?

Which model best describes your waiting time until you get the next nonworking web URL ("This page cannot be displayed") as you click on various websites for Florida condo rentals?

Any two normal curves are the same except for their:

Assume that X is normally distributed with a mean μ = $64. Given that P(X ≥ $75) = 0.2981, we can calculate that the standard deviation of X is approximately:

The random variable X is normally distributed with mean of 80 and variance of 36. The 67^th percentile of the distribution is:

If the mean time between unscheduled maintenance of LCD displays in a hospital's CT scan facility is 4,000 operating hours, what is the probability of unscheduled maintenance in the next 5,000 hours?

A certain assembly line at Vexing Manufacturing Company averages 30 minutes between breakdowns. What is the probability that less than 6 minutes will elapse before the next breakdown?

The area under the normal curve between the 20^th and 70^th percentiles is equal to:

On average, 15 minutes elapse between discoveries of fraudulent corporate tax returns in a certain IRS office. What is the probability that less than 30 minutes will elapse before the next fraudulent corporate tax return is discovered?

A certain assembly line at Vexing Manufacturing Company averages 30 minutes between breakdowns. The median time between breakdowns is:

Light bulbs are normally distributed with an average lifetime of 1000 hours and a standard deviation of 250 hours. The probability that a light bulb picked at random will last less than 1500 hours is about:

Regarding continuous probability distributions, which statement is incorrect?

If the random variable Z has a standard normal distribution, then P(Z ≤ -1.37) is:

If the mean time between in-flight aircraft engine shutdowns is 12,500 operating hours, the 90^th percentile of waiting times to the next shutdown will be:

To convert a normally distributed variable X into a standard Z score we would:

The variable in a normal distribution can assume any value between

The standard deviation of a normal random variable X is $20. Given that P(X ≤ $10) = 0.1841. From this we can determine that the mean of the distribution is equal to:

On average, a major earthquake (Richter scale 6.0 or above) occurs 3 times a decade in a certain California county. What is the probability that less than six months will pass before the next earthquake?