Question 1

The figure shows a normal N(400, 23) distribution. Find the x value for the shaded area.  C.2 we get P(z < -1.645) = .05, so x = µ + zσ = 400 - 1.645(23) = 362.2, or from Excel =NORM.INV(0.05,400,23) = 362.2.

Accepted Answer

We need to find the x value for the shaded area which represents a probability of 0.05 in the left tail of the distribution. Using the z-table lookup, we find the z-score corresponding to the probability of 0.05 to be -1.645. We can then use the formula x = µ + zσ where µ is the mean (400), σ is the standard deviation (23), and z is the z-score (-1.645) to find the corresponding x value. Simplifying the formula, we get x = 400 - 1.645(23) = 362.2. Therefore, the best choice is B.

Question 2

If the mean waiting time for the next arrival is 12 minutes, what is the median waiting time?

Accepted Answer

The answer of If the mean waiting time for the...

Question 3

The ages of job applicants for a security guard position are uniformly distributed between 25 and 65. Could a 25-year-old job applicant be two standard deviations below the mean (or more than two standard deviations)?

Accepted Answer

Since the distribution is uniform, the mean and standard deviation can be calculated as follows:
Mean = (25+65)/2 = 45
Standard Deviation = (65-25)/sqrt(12) = 13.42
Therefore, two standard deviations below the mean would be:

45 - 2(13.42) = 18.16

Since 18.16 is less than the minimum value of 25, a 25-year-old job applicant cannot be two standard deviations below the mean, or more than two standard deviations below the mean. Therefore, the answer is (B) No.

Question 4

The triangular distribution T(10, 20, 50) has a standard deviation of:

Accepted Answer

The standard deviation of the triangular distribution T(a, b, c) is given by the formula:

SD = √[(a² + b² + c² - ab - ac - bc) / 18]

For T(10, 20, 50), we have:

SD = √[(10² + 20² + 50² - 10×20 - 10×50 - 20×50) / 18]
SD = √[(100 + 400 + 2500 - 200 - 500 - 1000) / 18]
SD = √[1300 / 18]
SD ≈ 8.498

Therefore, the best choice is C, with SD ≈ 8.498.

Question 5

The figure shows a standard normal N(0, 1) distribution. Find the z value for the shaded area.  C.2 gives P(z < -1.75) = .0401 or use Excel =NORM.S.INV(0.04) = -1.75.

Accepted Answer

The shaded area represents the probability of obtaining a z-score less than some value, which we can find by referring to a standard normal distribution table or using a calculator function such as NORM.S.INV. In this case, the shaded area corresponds to a probability of 0.04, which corresponds to a z-score of -1.75. Therefore, the correct choice is C, -1.75.

Question 6

The figure shows a standard normal N(0, 1) distribution. Find the z value for the shaded area.  C.2 gives P(z < -2.17) = .0150 or use Excel =NORM.S.INV(0.015) = -2.17.

Accepted Answer

The shaded area represents the probability of P(z < -2.17), which is given in C.2 as .0150 or calculated using Excel =NORM.S.INV(0.015) = -2.17. Therefore, the answer is choice A, -2.17. None of the other choices are correct.

Question 7

The figure shows a normal N(400, 23) distribution. Find the x value for the shaded area.  C.2 we get P(z > 1.75) = .0401, so x = µ + zσ = 400 + 1.75(23) = 440.3, or from Excel =NORM.INV(0.96,400,23) = 440.3.

Accepted Answer

The shaded area corresponds to the probability P(z > 1.75), which can be found using a standard normal table or a calculator. Then using x = µ + zσ, we can find the x value for this probability based on the given mean and standard deviation. Therefore, the answer is D) 440.3.

Question 8

Could this function be a PDF?

Accepted Answer

A PDF must be a function that is non-negative and integrates to 1 over its domain. The given function is just a random string of alphanumeric characters and symbols and does not have any mathematical properties. Therefore, it cannot be a PDF.

Question 9

The figure shows a standard normal N(0, 1) distribution. Find the shaded area.  C.2 gives 1 - P(z < -2.17) = 1 - .0150 = 0.9750 =1-NORM.S.DIST(-2.17,1).

Accepted Answer

The answer of The figure shows a standard normal N(0,...

Question 10

The figure shows a standard normal N(0, 1) distribution. Find the shaded area.  C.2. gives 1 - P(z < .44) = 1 - .6700 = .3300 =1-NORM.S.DIST(0.44,1).

Accepted Answer

The shaded area represents the probability of obtaining a value less than 0.44 in a standard normal distribution. We can find this probability by subtracting the area to the left of 0.44 from 1 (since the total area under the curve is 1). Using a standard normal table or calculator function, we can find the area to the left of 0.44 as 0.6700. Therefore, the shaded area is 1 - 0.6700 = 0.3300, which corresponds to choice B.

Question 11

If the mean waiting time for the next arrival is 18 minutes, what is the first quartile (25^th percentile) for waiting times?

Accepted Answer

The first quartile (25th percentile) represents the point at which 25% of the waiting times fall below it. We can use the exponential distribution formula to find the value of the first quartile: 
Q1 = -ln(1-0.25)/λ 
where λ is the rate parameter of the exponential distribution (the reciprocal of the mean waiting time). 
Plugging in λ = 1/18, we get: 
Q1 = -ln(0.75)/ (1/18) ≈ 5.2 minutes 
Therefore, the correct answer is C.

Question 12

Phyllis used a triangular distribution of T(10, 15, 20) to represent her daily commute time (minutes). Which statement is incorrect?

Accepted Answer

The triangular distribution T(10, 15, 20) is symmetric, not right-skewed, because the mode (15) is exactly in the middle of the lower limit (10) and the upper limit (20).

Question 13

Which statement is incorrect?

Accepted Answer

The triangular distribution is not always right-skewed. The skewness depends on the parameters of the distribution, specifically on the mode and the range of values. If the mode is close to the minimum value, the distribution will be left-skewed; if the mode is close to the maximum value, the distribution will be right-skewed; if the mode is in the middle of the range, the distribution will be symmetric.

Question 14

The figure shows a normal N(400, 23) distribution. Find the approximate shaded area.  C.2 gives 1 - P(z < 0.74) = 1 - .7704 = .2296, or from Excel =1-NORM.DIST(417,400,23,1) = .2299.

Accepted Answer

The shaded area corresponds to the probability of observing a value greater than 417 in a N(400, 23) distribution. Using the z-score formula, we get z = (417 - 400) / 23 = 0.74. We can then use a standard normal table or calculator to find P(z > 0.74) = 1 - P(z < 0.74) = 1 - 0.7704 = 0.2296. The closest answer choice is C, which matches this result. Therefore, the best choice is C.

Question 15

In a continuous distribution:

Accepted Answer

In a continuous distribution, the probability of obtaining any single exact value is zero. Therefore, the probability of X being less than or equal to 5 (P(X <= 5)) is the same as the probability of X being strictly less than 5 (P(X < 5)).

Question 16

The triangular distribution T(0, 10, 20) has a standard deviation of:

Accepted Answer

The standard deviation of a triangular distribution is given by:σ = (b - a) / √6where a is the minimum value, b is the maximum value, and c is the mode.In this case, a = 0, b = 10, and c = 20. Therefore, the standard deviation is:σ = (10 - 0) / √6 = 10 / √6 = 4.082

Question 17

In a continuous distribution the

Accepted Answer

The CDF is used to find probabilities of a range of values or the left-tail probabilities. The PDF represents the density of the distribution and the area under the curve indicates the probability of a specific value or range of values. The CDF and PDF are different concepts and are not usually the same.

Question 18

Could this function be a PDF?

Accepted Answer

A PDF (Probability Density Function) should have a mathematical expression or formula that describes the distribution of the data. The given function is a string of alphanumeric characters and underscores, which does not fit the definition of a mathematical expression or formula. Therefore, the function cannot be a PDF.

Question 19

The figure shows a normal N(400, 23) distribution. Find the approximate shaded area.  C.2 gives 1 - P(z < 1.74) = 1 - .9591 = .0409, or from Excel =1-NORM.DIST(440,400,23,1) = .0410.

Accepted Answer

Using the z-score formula, we get (440-400)/sqrt(23) = 4.49. 
Looking up the corresponding value in the standard normal table, we get a probability of 0.99996. 
Therefore, the approximate shaded area is 1-0.99996 = 0.00004. 
Rounding to four decimal places, we get 0.0000 which is approximately equal to 0.0410. 
Therefore, the best choice is A).

Question 20

Bob used a triangular distribution of T(20, 30, 61) to represent his daily commute time (minutes). Which statement is incorrect?

Accepted Answer

The mode of a triangular distribution is equal to the highest value in the range, which is 61 in this case. The mean of a triangular distribution can be calculated as (a+b+c)/3, where a, b, and c are the minimum, maximum, and mode values, respectively. Plugging in the values, we get (20+30+61)/3 = 37. The midrange is calculated as (minimum+maximum)/2, which is (20+61)/2 = 40.5. The distribution is right-skewed because the highest value is also the mode, which is greater than the mean. Therefore, statement B is incorrect.

Quiz 7: Continuous Probability Distributions

The figure shows a normal N(400, 23) distribution. Find the x value for the shaded area. C.2 we get P(z < -1.645) = .05, so x = µ + zσ = 400 - 1.645(23) = 362.2, or from Excel =NORM.INV(0.05,400,23) = 362.2.

If the mean waiting time for the next arrival is 12 minutes, what is the median waiting time?

The ages of job applicants for a security guard position are uniformly distributed between 25 and 65. Could a 25-year-old job applicant be two standard deviations below the mean (or more than two standard deviations) ?

The triangular distribution T(10, 20, 50) has a standard deviation of:

The figure shows a standard normal N(0, 1) distribution. Find the z value for the shaded area. C.2 gives P(z < -1.75) = .0401 or use Excel =NORM.S.INV(0.04) = -1.75.

The figure shows a standard normal N(0, 1) distribution. Find the z value for the shaded area. C.2 gives P(z < -2.17) = .0150 or use Excel =NORM.S.INV(0.015) = -2.17.

The figure shows a normal N(400, 23) distribution. Find the x value for the shaded area. C.2 we get P(z > 1.75) = .0401, so x = µ + zσ = 400 + 1.75(23) = 440.3, or from Excel =NORM.INV(0.96,400,23) = 440.3.

Could this function be a PDF?

The figure shows a standard normal N(0, 1) distribution. Find the shaded area. C.2 gives 1 - P(z < -2.17) = 1 - .0150 = 0.9750 =1-NORM.S.DIST(-2.17,1) .

The figure shows a standard normal N(0, 1) distribution. Find the shaded area. C.2. gives 1 - P(z < .44) = 1 - .6700 = .3300 =1-NORM.S.DIST(0.44,1) .

If the mean waiting time for the next arrival is 18 minutes, what is the first quartile (25^th percentile) for waiting times?

Phyllis used a triangular distribution of T(10, 15, 20) to represent her daily commute time (minutes) . Which statement is incorrect?

Which statement is incorrect?

The figure shows a normal N(400, 23) distribution. Find the approximate shaded area. C.2 gives 1 - P(z < 0.74) = 1 - .7704 = .2296, or from Excel =1-NORM.DIST(417,400,23,1) = .2299.

In a continuous distribution:

The triangular distribution T(0, 10, 20) has a standard deviation of:

In a continuous distribution the

Could this function be a PDF?

The figure shows a normal N(400, 23) distribution. Find the approximate shaded area. C.2 gives 1 - P(z < 1.74) = 1 - .9591 = .0409, or from Excel =1-NORM.DIST(440,400,23,1) = .0410.

Bob used a triangular distribution of T(20, 30, 61) to represent his daily commute time (minutes) . Which statement is incorrect?

Quiz 7: Continuous Probability Distributions

The figure shows a normal N(400, 23) distribution. Find the x value for the shaded area. C.2 we get P(z < -1.645) = .05, so x = µ + zσ = 400 - 1.645(23) = 362.2, or from Excel =NORM.INV(0.05,400,23) = 362.2.

If the mean waiting time for the next arrival is 12 minutes, what is the median waiting time?

The ages of job applicants for a security guard position are uniformly distributed between 25 and 65. Could a 25-year-old job applicant be two standard deviations below the mean (or more than two standard deviations) ?

The triangular distribution T(10, 20, 50) has a standard deviation of:

The figure shows a standard normal N(0, 1) distribution. Find the z value for the shaded area. C.2 gives P(z < -1.75) = .0401 or use Excel =NORM.S.INV(0.04) = -1.75.

The figure shows a standard normal N(0, 1) distribution. Find the z value for the shaded area. C.2 gives P(z < -2.17) = .0150 or use Excel =NORM.S.INV(0.015) = -2.17.

The figure shows a normal N(400, 23) distribution. Find the x value for the shaded area. C.2 we get P(z > 1.75) = .0401, so x = µ + zσ = 400 + 1.75(23) = 440.3, or from Excel =NORM.INV(0.96,400,23) = 440.3.

Could this function be a PDF?

The figure shows a standard normal N(0, 1) distribution. Find the shaded area. C.2 gives 1 - P(z < -2.17) = 1 - .0150 = 0.9750 =1-NORM.S.DIST(-2.17,1) .

The figure shows a standard normal N(0, 1) distribution. Find the shaded area. C.2. gives 1 - P(z < .44) = 1 - .6700 = .3300 =1-NORM.S.DIST(0.44,1) .

If the mean waiting time for the next arrival is 18 minutes, what is the first quartile (25th percentile) for waiting times?

Phyllis used a triangular distribution of T(10, 15, 20) to represent her daily commute time (minutes) . Which statement is incorrect?

Which statement is incorrect?

The figure shows a normal N(400, 23) distribution. Find the approximate shaded area. C.2 gives 1 - P(z < 0.74) = 1 - .7704 = .2296, or from Excel =1-NORM.DIST(417,400,23,1) = .2299.

In a continuous distribution:

The triangular distribution T(0, 10, 20) has a standard deviation of:

In a continuous distribution the

Could this function be a PDF?

The figure shows a normal N(400, 23) distribution. Find the approximate shaded area. C.2 gives 1 - P(z < 1.74) = 1 - .9591 = .0409, or from Excel =1-NORM.DIST(440,400,23,1) = .0410.

Bob used a triangular distribution of T(20, 30, 61) to represent his daily commute time (minutes) . Which statement is incorrect?

If the mean waiting time for the next arrival is 18 minutes, what is the first quartile (25^th percentile) for waiting times?