Question 1

For any normally distributed random variable with mean μ and standard deviation σ, the proportion of the observations that fall outside the interval [μ - σ, μ + σ] is the closest to ________.
A) 0.0466
B) 0.3174
C) 0.8413
D) 0.1687

Accepted Answer

For a normal distribution, approximately 68% of the observations fall within one standard deviation of the mean. Therefore, the proportion of observations that fall outside the interval [μ - σ, μ + σ] is approximately 32%, which is closest to option B) 0.3174.

Question 2

Sarah's portfolio has an expected annual return at 8%, with an annual standard deviation at 12%. If her investment returns are normally distributed, then in any given year Sarah has an approximate ________.
A) 50% chance that the actual return will be greater than 8%
B) 68% chance that the actual return will fall within 4% and 20%
C) 68% chance that the actual return will fall within -20% and 20%
D) 95% chance that the actual return will fall within -4% and 28%

Accepted Answer

The expected return is the mean of the distribution, so there's a 50% chance the return will be greater than the mean (8%).

Question 3

A hedge fund returns on average 26% per year with a standard deviation of 12%. Using the empirical rule, approximate the probability the fund returns over 50% next year.
A) 0.5%
B) 1%
C) 2.5%
D) 5%

Accepted Answer

The empirical rule states that 95% of data falls within 2 standard deviations. A return of 50% is 2 standard deviations above the mean (26% + 2*12% = 50%). Thus, approximately 2.5% of the data lies beyond 2 standard deviations above the mean.

Question 4

Let X be normally distributed with mean µ = 25 and standard deviation σ = 5. Find the value x such that P(X ≥ x) = 0.1736.
A) −0.94
B) 0.94
C) 20.30
D) 29.70

Accepted Answer

Using a standard normal table or calculator, we find that the z-score corresponding to the right-tail probability of 0.1736 is approximately 0.96. Thus, we have:

z = (x - µ)/σ = 0.96

Substituting the given values, we get:

(x - 25)/5 = 0.96

Solving for x, we obtain:

x = 4.8 + 25 = 29.8

Therefore, the value x such that P(X ≥ x) = 0.1736 is approximately 29.8. The closest choice is D, 29.70.

Question 5

The stock price of a particular asset has a mean and standard deviation of $58.50 and $8.25, respectively. Use the normal distribution to compute the 95th percentile of this stock price.
A) −1.645
B) 1.645
C) 44.93
D) 72.07

Accepted Answer

To find the 95th percentile, we need to find the z-score that corresponds to an area of 0.95 to the left of it. Using a z-table, we can find that the z-score for an area of 0.95 is 1.645.

We can use the formula z = (x - μ) / σ to solve for x (the stock price) given the mean (μ), standard deviation (σ), and z-score.

Plugging in the values we know: 
1.645 = (x - 58.50) / 8.25

Solving for x: 
x = 1.645 * 8.25 + 58.50 = 72.07

Therefore, the 95th percentile of the stock price is $72.07.

Question 6

Sarah's portfolio has an expected annual return at 10%, with an annual standard deviation at 12%. If her investment returns are normally distributed, then in any given year Sarah has an approximate ________.
A) 50% chance that the actual return will be greater than 12%
B) 68% chance that the actual return will fall within -2% and 22%
C) 68% chance that the actual return will fall within -20% and 22%
D) 95% chance that the actual return will fall within -2% and 24%

Accepted Answer

The 68% chance that the actual return will fall within -2% and 22% is based on the empirical rule (or 68-95-99.7 rule) for normal distributions, which states that approximately 68% of the data falls within one standard deviation of the mean. Here, one standard deviation from the mean (10%) is -2% (10% - 12%) to 22% (10% + 12%).

Question 7

The time to complete the construction of a soapbox derby car is normally distributed with a mean of three hours and a standard deviation of one hour. Find the probability that it would take exactly 3.7 hours to construct a soapbox derby car.
A) 0.0000
B) 0.5000
C) 0.7580
D) 0.2420

Accepted Answer

In a continuous probability distribution, such as the normal distribution, the probability of observing any exact value (like exactly 3.7 hours) is 0 because there are infinitely many possible values.

Question 8

If X has a normal distribution with µ = 100 and σ = 5, then the probability P(90 ≤ X ≤ 95) can be expressed in terms of a standard normal variable Z as ________.
A) P(2 ≤ Z ≤ 1)
B) P(-2 ≤ Z ≤ 1)
C) P(-2 ≤ Z ≤ -1)
D) P(-2 ≤ Z ≤ -2)

Accepted Answer

To express the probability in terms of a standard normal variable Z, we use the transformation $Z = \frac{X - \mu}{\sigma}$. For $X = 90$, $Z = \frac{90 - 100}{5} = -2$, and for $X = 95$, $Z = \frac{95 - 100}{5} = -1$. Therefore, the probability $P(90 \leq X \leq 95)$ is equivalent to $P(-2 \leq Z \leq -1)$.

Question 9

Gold miners in Alaska have found, on average, 12 ounces of gold per 1,000 tons of dirt excavated with a standard deviation of 3 ounces. Assume the amount of gold found per 1,000 tons of dirt is normally distributed. What is the probability the miners find more than 16 ounces of gold in the next 1,000 tons of dirt excavated?
A) 0.0912
B) 0.4082
C) 0.5918
D) 0.9082

Accepted Answer

First, we need to standardize the value of 16 ounces to find the corresponding z-score:

z = (16 - 12) / 3 = 1.33

Next, we use a standard normal distribution table or calculator to find the probability of getting a z-score greater than 1.33:

P(z > 1.33) = 0.0912

Therefore, the probability the miners find more than 16 ounces of gold in the next 1,000 tons of dirt excavated is 0.0912 or approximately 9.12%, which corresponds to choice (A).

Question 10

Let X be normally distributed with mean µ = 250 and standard deviation σ = 80. Find the value x such that P(X ≤ x) = 0.0606.
A) −1.55
B) 1.55
C) 126
D) 374

Accepted Answer

We need to convert the probability of P(X ≤ x) = 0.0606 to a z-score using the standard normal distribution table.

First, we calculate the z-score:

z = (x - µ) / σ

0.0606 = P(X ≤ x) => 1 - P(X > x) = 0.0606
P(Z ≤ z) = 0.0606
From the standard normal distribution table, the closest value to 0.0606 is 0.0618, which corresponds to a z-score of -1.56.

-1.56 = (x - 250) / 80

x = 126.72

Therefore, the value x such that P(X ≤ x) = 0.0606 is x = 126.72, which is closest to answer C: 126.

Question 11

The time to complete the construction of a soapbox derby car is normally distributed with a mean of three hours and a standard deviation of one hour. Find the probability that it would take between 2.5 and 3.5 hours to construct a soapbox derby car.
A) 0.3085
B) 0.3830
C) 0.6170
D) 0.6915

Accepted Answer

Using the z-score formula, we get:

z1 = (2.5 - 3) / 1 = -0.5
z2 = (3.5 - 3) / 1 = 0.5

Then, using a standard normal distribution table, we find the area between z1 and z2:

P(-0.5 < Z < 0.5) = 0.3830

Therefore, the probability that it would take between 2.5 and 3.5 hours to construct a soapbox derby car is 0.3830, which corresponds to choice B).

Question 12

Let X be normally distributed with mean µ = 250 and standard deviation σ = 80. Find the value x such that P(X ≤ x) = 0.9394.
A) −1.55
B) 1.55
C) 126
D) 374

Accepted Answer

Using Z-score formula, 
( x - µ ) / σ = Z 
( x - 250 ) / 80 = 1.556 
x - 250 = 1.556 x 80 
x = 374.48 
Round the answer up to the nearest whole number 
x = 374 
Therefore, the answer is D.

Question 13

For any normally distributed random variable with mean μ and standard deviation σ, the percent of the observations that fall between [μ - 2σ, μ + 2σ] is the closest to ________.
A) 68%
B) 68.26%
C) 95%
D) 99.73%

Accepted Answer

According to the empirical rule, for any normally distributed random variable, approximately 95% of the observations fall within 2 standard deviations of the mean. Therefore, the percent of observations that fall between [μ - 2σ, μ + 2σ] is 95%.

Question 14

The starting salary of an administrative assistant is normally distributed with a mean of $50,000 and a standard deviation of $2,500. We know that the probability of a randomly selected administrative assistant making a salary between μ − x and μ + x is 0.7416. Find the salary range referred to in this statement.
A) $42,825 to $52,825
B) $42,825 to $57,175
C) $47,175 to $52,825
D) $47,175 to $57,175

Accepted Answer

The probability of 0.7416 corresponds to approximately ±1 standard deviation from the mean in a normal distribution. Therefore, the range is $50,000 ± $2,500, which is $47,500 to $52,500. However, the closest option is $47,175 to $52,825, considering rounding or approximation.

Question 15

You are planning a May camping trip to Denali National Park in Alaska and want to make sure your sleeping bag is warm enough. The average low temperature in the park for May follows a normal distribution with a mean of 32°F and a standard deviation of 8°F. Above what temperature must the sleeping bag be suited such that the temperature will be too cold only 5% of the time?
A) −1.645
B) 1.645
C) 18.84
D) 45.16

Accepted Answer

We want to find the temperature that is too cold only 5% of the time. This means we want to find the z-score that corresponds to the 5th percentile of the distribution. Using a standard normal table or calculator, we can find this to be -1.645.

We can use the formula z = (x - mu) / sigma to find the corresponding temperature x. Solving for x, we get:

x = z * sigma + mu
x = -1.645 * 8 + 32
x = 18.84

Thus, the sleeping bag must be suited for temperatures above 18.84°F to ensure that it will be too cold only 5% of the time. Therefore, the answer is C.

Question 16

An investment consultant tells her client that the probability of making a positive return with her suggested portfolio is 0.90. What is the risk, measured by standard deviation that this investment manager has assumed in her calculation if it is known that returns from her suggested portfolio are normally distributed with a mean of 6%?
A) 1.28%
B) 4.69%
C) 6.00%
D) 10.0%

Accepted Answer

We can use the inverse normal distribution function to find the z-score corresponding to a probability of 0.90:
z = invNorm(0.90) = 1.2816
We can then use the z-score formula to solve for the standard deviation:
z = (x - mu) / sigma
1.2816 = (0 - 0.06) / sigma
sigma = (0.06) / 1.2816 = 0.0469 or 4.69%
Therefore, the investment manager has assumed a risk of 4.69% in her calculation.

Question 17

You are planning a May camping trip to Denali National Park in Alaska and want to make sure your sleeping bag is warm enough. The average low temperature in the park for May follows a normal distribution with a mean of 32°F and a standard deviation of 8°F. One sleeping bag you are considering advertises that it is good for temperatures down to 25°F. What is the probability that this bag will be warm enough on a randomly selected May night at the park?
A) 0.1894
B) 0.3106
C) 0.8092
D) 0.8800

Accepted Answer

To solve this problem, we need to standardize the value of 25°F using the formula z = (x - μ) / σ, where x is the value we want to standardize, μ is the mean, and σ is the standard deviation.

z = (25 - 32) / 8 = -0.875

We then look up the probability of z being less than or equal to -0.875 in a standard normal distribution table or by using a calculator.

Using a calculator, we find that P(z ≤ -0.875) = 0.1894.

Since the probability of the temperature being lower than 25°F is only 0.1894, the probability of the sleeping bag being warm enough is 1 - 0.1894 = 0.8106 or approximately 0.81.

Therefore, the best choice is C, with a probability of 0.8092.

Question 18

You are planning a May camping trip to Denali National Park in Alaska and want to make sure your sleeping bag is warm enough. The average low temperature in the park for May follows a normal distribution with a mean of 32°F and a standard deviation of 8°F. An inexpensive bag you are considering advertises to be good for temperatures down to 38°F. What is the probability that the bag will not be warm enough?
A) 0.2266
B) 0.2734
C) 0.7500
D) 0.7734

Accepted Answer

The answer of You are planning a May camping trip...

Question 19

The time to complete the construction of a soapbox derby car is normally distributed with a mean of three hours and a standard deviation of one hour. Find the probability that it would take more than five hours to construct a soapbox derby car.
A) 0
B) 0.0228
C) 0.4772
D) 0.9772

Accepted Answer

To find the probability that it takes more than five hours to construct a soapbox derby car, we use the Z-score formula: Z = (X - μ) / σ, where X is the value of interest (5 hours), μ is the mean (3 hours), and σ is the standard deviation (1 hour). Plugging in the values, we get Z = (5 - 3) / 1 = 2. Looking up a Z-score of 2 in the standard normal distribution table or using a calculator, we find the area to the left of Z=2 is approximately 0.9772. Since we're interested in the probability of taking more than 5 hours, we subtract this from 1 to get 1 - 0.9772 = 0.0228.

Question 20

The salary of teachers in a particular school district is normally distributed with a mean of $50,000 and a standard deviation of $2,500. Due to budget limitations, it has been decided that the teachers who are in the top 2.5% of the salaries would not get a raise. What is the salary level that divides the teachers into one group that gets a raise and one that doesn't?
A) −1.96
B) 1.96
C) 45,100
D) 54,900

Accepted Answer

To find the salary level that divides the teachers into one group that gets a raise and one that doesn't, we need to find the z-score that corresponds to the top 2.5% of the salaries.

Using a z-table, we can find that the z-score that corresponds to the top 2.5% is approximately 1.96.

We can then use this z-score to find the salary level that corresponds to it:

z-score = (x - mean) / standard deviation

1.96 = (x - 50,000) / 2,500

x - 50,000 = 1.96 * 2,500

x - 50,000 = 4,900

x = 54,900

Therefore, the salary level that divides the teachers into one group that gets a raise and one that doesn't is $54,900.

Quiz 6: Continuous Probability Distributions

For any normally distributed random variable with mean μ and standard deviation σ, the proportion of the observations that fall outside the interval [μ - σ, μ + σ] is the closest to ________.

Sarah's portfolio has an expected annual return at 8%, with an annual standard deviation at 12%. If her investment returns are normally distributed, then in any given year Sarah has an approximate ________.

A hedge fund returns on average 26% per year with a standard deviation of 12%. Using the empirical rule, approximate the probability the fund returns over 50% next year.

Let X be normally distributed with mean µ = 25 and standard deviation σ = 5. Find the value x such that P(X ≥ x) = 0.1736.

The stock price of a particular asset has a mean and standard deviation of $58.50 and $8.25, respectively. Use the normal distribution to compute the 95th percentile of this stock price.

Sarah's portfolio has an expected annual return at 10%, with an annual standard deviation at 12%. If her investment returns are normally distributed, then in any given year Sarah has an approximate ________.

The time to complete the construction of a soapbox derby car is normally distributed with a mean of three hours and a standard deviation of one hour. Find the probability that it would take exactly 3.7 hours to construct a soapbox derby car.

If X has a normal distribution with µ = 100 and σ = 5, then the probability P(90 ≤ X ≤ 95) can be expressed in terms of a standard normal variable Z as ________.

Let X be normally distributed with mean µ = 250 and standard deviation σ = 80. Find the value x such that P(X ≤ x) = 0.0606.

The time to complete the construction of a soapbox derby car is normally distributed with a mean of three hours and a standard deviation of one hour. Find the probability that it would take between 2.5 and 3.5 hours to construct a soapbox derby car.

Let X be normally distributed with mean µ = 250 and standard deviation σ = 80. Find the value x such that P(X ≤ x) = 0.9394.

For any normally distributed random variable with mean μ and standard deviation σ, the percent of the observations that fall between [μ - 2σ, μ + 2σ] is the closest to ________.

The time to complete the construction of a soapbox derby car is normally distributed with a mean of three hours and a standard deviation of one hour. Find the probability that it would take more than five hours to construct a soapbox derby car.