Question 1

Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 C3 150 0 69.912 0.0798 0.977 67.625 69.175 69.948
Variable Maximum
C3 72.339
If the data set presented here were normal, which value would most likely fall into the third quartile?
A) 72.84
B) 70.15
C) 69.76
D) 70.45

Accepted Answer

The answer of Variable N N* Mean SE Mean StDev...

Question 2

Metro buses are scheduled to arrive at each stop every 30 minutes.If the time a person waits at a bus stop is uniformly distributed and the maximum possible wait time for a bus is 30 minutes, what is the probability that you will wait more than 25 minutes for a bus?
A) 0.1667
B) 0.8333
C) 0.2000
D) 0.1333

Accepted Answer

The probability of waiting more than 25 minutes in a 30-minute interval is calculated by finding the proportion of the interval that exceeds 25 minutes. Since the wait time is uniformly distributed, the probability is $ \frac{30 - 25}{30} = \frac{5}{30} = 0.1667$.

Question 3

Spinal fusion surgery has become a popular and effective option for patients suffering from nerve pain caused by irregularities in the spinal disks.The effectiveness of the surgery varies from patient to patient.If the pain improvement post-surgery follows a normal distribution with a mean improvement of 2.25 and a standard deviation of 1.2, what is the probability that a randomly selected patient who undergoes fusion experiences a pain improvement between 0 and 2?
A) 0.3867
B) 0.4168
C) 0.2100
D) 0.0301

Accepted Answer

The answer of Spinal fusion surgery has become a popular...

Question 4

If X is a normal random variable with mean μ, standard deviation 0.15, and P(X < 2.10) = 0.025, what is the value of μ?
A) 2.394
B) 2.104
C) 2.096
D) 1.806

Accepted Answer

The answer of If X is a normal random variable...

Question 5

Large electrical generation companies use so-called peaking generators to carry additional load when demand for electrical power is particularly high.Running these generators at very low loads is bad for them for a number of reasons.A particular peaking generator carries an average load of 20 MW, with standard deviation of 4.23 MW.If this generator's low load threshold is 10 MW, what percentage of time is this generator being run at a low load condition?
A) 0.91%
B) 2.36%
C) 99.09%
D) 1%

Accepted Answer

The z-score for a load of 10 MW is (10 - 20) / 4.23 = -2.36. The percentage of time the generator is being run at a low load condition is the area under the normal distribution curve to the left of z = -2.36, which is 0.91%.

Question 6

Let X be a continuous random variable and let c be a constant.Which of the following statements is true?
A) The probability that X assumes a value in the interval X1 to X2 is the area under the probability density function between X1 and X2.
B) P(X = c) = 0.
C) P(X ≤ c) = P(X < c).
D) All of these statements are true.

Accepted Answer

A) This is true by the definition of probability density function for continuous random variables.
B) This is true since the probability of an exact value for a continuous random variable is always zero.
C) This is true since the probability that X equals to c is zero, so P(X ≤ c) = P(X < c).

Question 7

A sample of recorded dog barks is analyzed for decibel intensity.The mean intensity is 62.25 decibels, with a variance of 4.45.If this data set is normally distributed, we would expect approximately 95% of the barks to be within what range?
A) (58.03, 66.47)
B) (60.16, 64.36)
C) (55.92, 66.47)
D) (53.35, 71.15)

Accepted Answer

For a normally distributed data set, approximately 95% of the data falls within 2 standard deviations of the mean. The standard deviation is the square root of the variance, which is sqrt(4.45) ≈ 2.11. Therefore, 2 standard deviations is 2 * 2.11 = 4.22. Adding and subtracting this from the mean (62.25 ± 4.22) gives us the range (58.03, 66.47).

Question 8

Heights of children entering kindergarten are normally distributed with a mean height of 103 cm and a standard deviation of 1.27 cm.Sixty-seven percent of the children entering kindergarten are taller than:
A) 103 cm.
B) 0.44 cm.
C) 103.56 cm.
D) 102.44 cm.

Accepted Answer

The answer of Heights of children entering kindergarten are normally...

Question 9

The return on investment for a particular security has historically followed a normal distribution with a mean profit of 4.75% and standard deviation of 1.5%.If the historical pattern holds, what is the probability of investing in this security and seeing greater than a 6% return?
A) 0.7967
B) 0.8333
C) 0.2033
D) 0.1666

Accepted Answer

We need to find the probability of getting a return greater than 6%. 
Using the formula for standardizing values in a normal distribution, we get:

z = (6 - 4.75) / 1.5 = 0.833

Looking up the area under the standard normal curve for this z-value using a table or calculator, we get:

P(Z > 0.833) = 1 - P(Z ≤ 0.833) = 1 - 0.7967 = 0.2033

Therefore, the probability of investing in this security and seeing a return greater than 6% is 0.2033 or 20.33%, which is option C.

Question 10

Quality control engineers create acceptable boundaries for variability within the processes they monitor.They determine that the acceptable boundary for a certain heat exchange process at a gasoline refinery must be set such that under normal operating conditions the threshold would be exceeded only 1% of the time.Under normal operating conditions, the process varies according to a normal distribution, with mean temperature of 231°C and standard deviation of 2.5°C.What upper temperature should they set so that the process will normally exceed the temperature 0.51% of the time?
A) 257.00°C
B) 237.68°C
C) 224.58°C
D) 237.43°C

Accepted Answer

Given that the process should exceed the temperature 0.51% of the time, we look for the z-score corresponding to the upper tail probability of 0.51% (or 0.0051 in decimal form) in the standard normal distribution. This z-score is approximately 2.67. Using the z-score formula $z = \frac{X - \mu}{\sigma}$, where $X$ is the value we're solving for, $\mu = 231°C$ is the mean, and $\sigma = 2.5°C$ is the standard deviation, we can solve for $X$. Rearranging the formula gives $X = \mu + z\sigma = 231 + 2.67(2.5) = 231 + 6.675 = 237.675°C$, which rounds to 237.68°C, making option B the correct answer.

Question 11

If the heights of NBA players are normally distributed with a mean height of 73.5 inches and a standard deviation of 2.2 inches, what is the probability of randomly selecting an NBA player who is between 72 and 73 inches tall?
A) 0.1624
B) 0.5910
C) 0.4500
D) 0.4090

Accepted Answer

To find the probability of a player being between 72 and 73 inches tall, we use the Z-score formula: Z = (X - μ) / σ, where X is the value, μ is the mean, and σ is the standard deviation. For 72 inches, Z = (72 - 73.5) / 2.2 = -0.6818. For 73 inches, Z = (73 - 73.5) / 2.2 = -0.2273. Looking these Z-scores up in a standard normal distribution table or using a calculator, we find the probabilities for each and subtract the smaller from the larger to find the probability of being between these two heights. The calculation gives us approximately 0.1624, indicating the probability is 16.24%.

Question 12

Assume that the lifetime (in hours) of a particular car fan can be modeled with an exponential distribution with parameter 0.0003.Find the proportion of fans that will last for at least 10,000 hours.
A) 0.0302
B) 0.0498
C) 0.0357
D) 0.95012

Accepted Answer

The answer of Assume that the lifetime (in hours) of...

Question 13

The descriptive statistics for the age differences (husband age - wife age) from a sample of 100 married couples are shown here.  Based on these statistics, do the data appear to be normally distributed?
A) Yes, because the mean and the median are roughly the same.
B) Yes, because the value of IQR/s is roughly 1.3.
C) No, because the value of IQR/s is less than 1.3.
D) No, because the mean and the median are the same.

Accepted Answer

The answer of The descriptive statistics for the age differences...

Question 14

For any continuous random variable X, if a is any real fixed number, then P(X ≥ a) is always:
A) 1 - P(X< a).
B) 1 - P(X ≤ a).
C) P(X< a).
D) Both A and B are correct.

Accepted Answer

The answer of For any continuous random variable X, if...

Question 15

The smiling times (in seconds) of an 8-week-old baby can be modeled with a uniform distribution with a = 0 and b = 23 seconds.What is the probability that a randomly chosen 8-week-old baby smiles between 2 and 18 seconds?
A) 0.7826
B) 0.6956
C) 0.8696
D) 0.0870

Accepted Answer

To find the probability of a randomly chosen baby smiling between 2 and 18 seconds, we need to find the area under the uniform distribution curve between 2 and 18 seconds. The total area under the curve is (b-a) = (23-0) = 23 seconds. The area between 2 and 18 seconds is (18-2) = 16 seconds. Therefore, the probability is 16/23 = 0.6956. The closest answer choice is B.

Question 16

Assume Z is a standard normal random variable with P(-z < Z < z) = 0.8764.What is z?
A) 0.3764
B) 1.16
C) 1.54
D) 3.08

Accepted Answer

The probability P(-z < Z < z) = 0.8764 corresponds to the cumulative probability between -z and z for a standard normal distribution. Using a standard normal distribution table or calculator, z = 1.54 gives approximately this probability.

Question 17

Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 C4 150 0 64.940 0.0825 1.010 62.722 65.679
Variable Maximum
C4 67.445
If this data set were normal, which value would most likely fall into the first quartile?
A) 63.93
B) 65.073
C) 64.366
D) 63.730

Accepted Answer

If the data set were normal, the first quartile (Q1) would be approximately 0.6745 standard deviations below the mean. Using the values provided, we can calculate:
Q1 = Mean - 0.6745(StDev/√N) = 64.940 - 0.6745(1.010/√150) ≈ 64.366

Therefore, the best choice is (C), which is closest to the expected value of the first quartile based on a normal distribution.

Question 18

Let X = the time (in minutes) that a postal clerk spends with her customers.This time is known to have an exponential distribution with an average of 4 minutes.What is the parameter λ for this random variable?
A) 4
B) 2
C) 0.5
D) 0.25

Accepted Answer

The parameter λ of an exponential distribution is the reciprocal of the mean. Since the mean is 4 minutes, λ = 1/4 = 0.25.

Question 19

The speed at which an automated assembly line produces a product follows a normal distribution, with a mean production time of 32.20 seconds and a standard deviation of 1.05 seconds.What is the probability that a product requires more than 34 seconds to produce?
A) 0.0436
B) 0.1710
C) 0.9564
D) 0.0038

Accepted Answer

The answer of The speed at which an automated assembly...

Question 20

X is a normally distributed random variable with μx = 125 and σx = 12.5.What is P(X ≥ 110)?
A) 0.1151
B) 0.3849
C) 0.8849
D) 0.6151

Accepted Answer

To find P(X ≥ 110) for a normally distributed variable, we use the Z-score formula: Z = (X - μ) / σ. Here, Z = (110 - 125) / 12.5 = -1.2. Looking up the Z-score in a standard normal distribution table or using a calculator, we find the probability to the left of Z = -1.2. This is approximately 0.1151. Since we want P(X ≥ 110), we subtract this from 1 to get 1 - 0.1151 = 0.8849.

Quiz 7: Continuous Probability Distributions

Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 C3 150 0 69.912 0.0798 0.977 67.625 69.175 69.948 Variable Maximum C3 72.339 If the data set presented here were normal, which value would most likely fall into the third quartile?

Metro buses are scheduled to arrive at each stop every 30 minutes.If the time a person waits at a bus stop is uniformly distributed and the maximum possible wait time for a bus is 30 minutes, what is the probability that you will wait more than 25 minutes for a bus?

If X is a normal random variable with mean μ, standard deviation 0.15, and P(X < 2.10) = 0.025, what is the value of μ?

Let X be a continuous random variable and let c be a constant.Which of the following statements is true?

A sample of recorded dog barks is analyzed for decibel intensity.The mean intensity is 62.25 decibels, with a variance of 4.45.If this data set is normally distributed, we would expect approximately 95% of the barks to be within what range?

Heights of children entering kindergarten are normally distributed with a mean height of 103 cm and a standard deviation of 1.27 cm.Sixty-seven percent of the children entering kindergarten are taller than:

The return on investment for a particular security has historically followed a normal distribution with a mean profit of 4.75% and standard deviation of 1.5%.If the historical pattern holds, what is the probability of investing in this security and seeing greater than a 6% return?

If the heights of NBA players are normally distributed with a mean height of 73.5 inches and a standard deviation of 2.2 inches, what is the probability of randomly selecting an NBA player who is between 72 and 73 inches tall?

Assume that the lifetime (in hours) of a particular car fan can be modeled with an exponential distribution with parameter 0.0003.Find the proportion of fans that will last for at least 10,000 hours.

The descriptive statistics for the age differences (husband age - wife age) from a sample of 100 married couples are shown here. Based on these statistics, do the data appear to be normally distributed?

For any continuous random variable X, if a is any real fixed number, then P(X ≥ a) is always:

The smiling times (in seconds) of an 8-week-old baby can be modeled with a uniform distribution with a = 0 and b = 23 seconds.What is the probability that a randomly chosen 8-week-old baby smiles between 2 and 18 seconds?

Assume Z is a standard normal random variable with P(-z < Z < z) = 0.8764.What is z?

Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 C4 150 0 64.940 0.0825 1.010 62.722 65.679 Variable Maximum C4 67.445 If this data set were normal, which value would most likely fall into the first quartile?

Let X = the time (in minutes) that a postal clerk spends with her customers.This time is known to have an exponential distribution with an average of 4 minutes.What is the parameter λ for this random variable?

The speed at which an automated assembly line produces a product follows a normal distribution, with a mean production time of 32.20 seconds and a standard deviation of 1.05 seconds.What is the probability that a product requires more than 34 seconds to produce?

X is a normally distributed random variable with μx = 125 and σx = 12.5.What is P(X ≥ 110) ?