Question 1

Suppose a population has a mean of 400 and a standard deviation of 24.If a random sample of size 144 is drawn from the population, the probability of drawing a sample with a mean less than 402 is ___.
A)0.3413
B)0.6826
C)0.8413
D)0.1587
E)0.9875

Accepted Answer

We know that the population mean (μ) = 400, the population standard deviation (σ) = 24, and the sample size (n) = 144.

The standard error of the mean (SEM) is given by the formula: SEM = σ/√n
= 24/√144
= 24/12
= 2

To find the probability of drawing a sample with a mean less than 402, we need to standardize this value using the formula:

z = (x̄ - μ) / (SEM)
= (402 - 400) / (2)
= 1

Using a standard normal distribution table or calculator, we can find the probability of z being less than 1 to be 0.8413.

Therefore, the probability of drawing a sample with a mean less than 402 is 0.8413. The mean of the sampling distribution of the mean is equal to the population mean which is 400.
The standard deviation of the sampling distribution of the mean is equal to the population standard deviation divided by the square root of the sample size which is 24/√144= 2.
We want to find the probability of drawing a sample with a mean less than 402 which can be represented as:
P(x̄ < 402)
To find this probability, we need to standardize the sample mean value using the formula:
z= (x̄ - μ) / (σ/√n)
Substituting the given values, we get:
z= (402-400) / (2)
z= 1
Now, we need to find the probability of z-score being less than 1 which can be obtained from the standard normal distribution table or calculator as 0.8413.
Therefore, the probability of drawing a sample with a mean less than 402 is 0.8413 or 84.13%. Answer: C.

Question 2

Albert Abbasi, VP of Operations at Ingleside International Bank, is evaluating the service level provided to walk-in customers.Accordingly, he plans a sample of waiting times for walk-in customers.If the population of waiting times has a mean of 15 minutes and a standard deviation of 4 minutes, the probability that Albert's sample of 64 will have a mean less than 15 minutes is ___.
A)0.5000
B)0.0228
C)0.9072
D)0.9544
E)1.0000

Accepted Answer

Given the symmetry of the normal distribution, the probability that the sample mean is less than the population mean (15 minutes) is 0.5000, as the distribution of sample means is centered around the population mean.

Question 3

Albert Abbasi, VP of Operations at Ingleside International Bank, is evaluating the service level provided to walk-in customers.Accordingly, he plans a sample of waiting times for walk-in customers.If the population of waiting times has a mean of 15 minutes and a standard deviation of 4 minutes, the probability that Albert's sample of 64 will have a mean less than 16 minutes is ___.
A)0.4772
B)0.0228
C)0.9072
D)0.9544
E)0.9772

Accepted Answer

The answer of Albert Abbasi, VP of Operations at Ingleside...

Question 4

Suppose a population has a mean of 90 and a standard deviation of 28.If a random sample of size 49 is drawn from the population, the probability of drawing a sample with a mean between 85 and 95 is ___.
A)0.1056
B)0.3944
C)0.7888
D)0.2112
E)0.5000

Accepted Answer

First, we need to convert the problem into standard normal distribution using the formula:

z = (x - μ) / (σ / sqrt(n))

where:
z = standard normal distribution (z-score)
x = sample mean
μ = population mean (given as 90)
σ = population standard deviation (given as 28)
n = sample size (given as 49)

Plugging in the values:

z1 = (85 - 90) / (28 / sqrt(49)) = -1.25
z2 = (95 - 90) / (28 / sqrt(49)) = 1.25

Next, we look up the probabilities associated with the z-scores using the standard normal distribution table or calculator. The probability of getting a z-score between -1.25 and 1.25 is:

P(-1.25 < z < 1.25) = P(z < 1.25) - P(z < -1.25)
= 0.8944 - 0.1056
= 0.7888

Therefore, the probability of drawing a sample with a mean between 85 and 95 is 0.7888 or approximately 0.79. The best choice is C.

Question 5

Albert Abbasi, VP of Operations at Ingleside International Bank, is evaluating the service level provided to walk-in customers.Accordingly, he plans a sample of waiting times for walk-in customers.If the population of waiting times has a mean of 15 minutes and a standard deviation of 4 minutes, the probability that Albert's sample of 64 will have a mean between 16.5 and 18.5 minutes is ___.
A)0.1865
B)0.2667
C)0.7333
D)0.5468
E)1.0000

Accepted Answer

The sample mean follows a normal distribution with mean 15 and standard deviation 0.5 (4/sqrt(64)). The probability of the sample mean being between 16.5 and 18.5 can be found using the standard normal distribution, resulting in approximately 0.1865.

Question 6

Albert Abbasi, VP of Operations at Ingleside International Bank, is evaluating the service level provided to walk-in customers.Accordingly, he plans a sample of waiting times for walk-in customers.If the population of waiting times has a mean of 15 minutes and a standard deviation of 4 minutes, the probability that Albert's sample of 64 will have a mean less than 14 minutes is ___.
A)0.4772
B)0.0228
C)0.9772
D)0.9544
E)1.0000

Accepted Answer

We need to find the z-score for a sample mean of 14 minutes:
z = (14-15)/(4/sqrt(64)) = -2
And then find the probability of z being less than -2:
P(z < -2) = 0.0228 (from standard normal distribution table)
Therefore, the probability that Albert's sample of 64 will have a mean less than 14 minutes is 0.0228 or approximately 2.28%. The answer is B. We need to find the probability of having a sample mean less than 14 minutes when taking a sample of 64 waiting times. 
- The population mean (μ) is 15 minutes.
- The population standard deviation (σ) is 4 minutes.
- The sample size (n) is 64.
- The sample mean (x̄) will follow a normal distribution with a mean of μ and a standard error of σ/√n.

z = (x̄ - μ) / (σ/√n) = (14 - 15) / (4/√64) = -2
P(z < -2) = 0.0228 (from standard normal distribution table)

Therefore, the probability of Albert's sample of 64 having a mean less than 14 minutes is 0.0228, which is option B.

Question 7

Suppose a population has a mean of 90 and a standard deviation of 28.If a random sample of size 49 is drawn from the population, the probability of drawing a sample with a mean of less than 84 is ___.
A)0.9332
B)0.0668
C)0.4332
D)0.8664
E)1.0000

Accepted Answer

First we need to calculate the standard error of the mean (SEM) using the following formula:
SEM = σ/√n
where σ is the population standard deviation, n is the sample size.
SEM = 28/√49 = 4

Next, we need to standardize the sample mean using the following formula:
z = (x - μ)/SEM
where x is the sample mean, μ is the population mean, SEM is the standard error of the mean.
z = (84 - 90)/4 = -1.5

We need to find the probability of drawing a sample with a mean of less than 84, which is the same as finding the probability of z being less than -1.5. Using a standard normal distribution table or calculator, we find that the probability is approximately 0.0668. Therefore, the answer is B.

Question 8

Suppose the population of all public Universities shows the annual parking fee per student is normally distributed with a mean of $110 with a standard deviation of $18.If a random sample of size 49 is drawn from the population, the probability of drawing a sample with a sample mean between $100 and $115 is _______.
A)0.9738
B)0.4738
C)0.0262
D)0.6103
E)0.1103

Accepted Answer

The standard error of the mean is $18/sqrt(49) = $2.57. The z-scores for $100 and $115 are (100-110)/2.57 = -3.89 and (115-110)/2.57 = 1.95, respectively. The probability of a z-score between -3.89 and 1.95 is approximately 0.9738.

Question 9

A carload of steel rods has arrived at Cybermatic Construction Company.The car contains 50,000 rods.Claude Ong, Manager of Quality Assurance, directs his crew to measure the lengths of 100 randomly selected rods.If the population of rods have a mean length of 120 centimetres and a standard deviation of 0.05 centimetres, the probability that Claude's sample has a mean less than 119.985 centimetres is ___.
A)0.9974
B)0.0026
C)0.4987
D)0.0013
E)0.0030

Accepted Answer

The probability can be found using the standard normal distribution (Z-score). The sample mean's standard deviation (standard error) is 0.05/sqrt(100) = 0.005. The Z-score for 119.985 is (119.985 - 120) / 0.005 = -3. The probability of a Z-score less than -3 is 0.0013.

Question 10

Suppose a population has a mean of 90 and a standard deviation of 28.If a random sample of size 49 is drawn from the population, the probability of drawing a sample with a mean of more than 95 is ___.
A)0.1056
B)0.3944
C)0.4286
D)0.8944
E)1.0000

Accepted Answer

The standard error of the mean (SEM) is calculated as the population standard deviation divided by the square root of the sample size: SEM = 28 / √49 = 4. The z-score for a sample mean of 95 is (95 - 90) / 4 = 1.25. Using the standard normal distribution table, the probability of a z-score greater than 1.25 is approximately 0.1056.

Question 11

Suppose a population has a mean of 400 and a standard deviation of 24.If a random sample of size 144 is drawn from the population, the probability of drawing a sample with a mean between 393.5 and 396.5 is ___.
A)0.0802
B)0.0012
C)0.0407
D)0.0395
E)1.0000

Accepted Answer

The answer of Suppose a population has a mean of...

Question 12

Suppose a population has a mean of 400 and a standard deviation of 24.If a random sample of size 144 is drawn from the population, the probability of drawing a sample with a mean of more than 404.5 is ___.
A)0.0139
B)0.4861
C)0.4878
D)0.0122
E)0.5000

Accepted Answer

The answer of Suppose a population has a mean of...

Question 13

Suppose the population of all public Universities shows the annual parking fee per student is normally distributed with a mean of $110 with a standard deviation of $18.If a random sample of size 49 is drawn from the population, the probability of drawing a sample with a sample mean between $112 and $115 is closest to _______.
A)0.9738
B)0.7777
C)0.7823
D)0.1915
E)1.7561

Accepted Answer

The answer of Suppose the population of all public Universities...

Question 14

Suppose a population has a mean of 870 and a variance of 1,600.If a random sample is size 64 is drawn from the population, the probability that the sample mean is between 860 and 875 is ___.
A)0.9544
B)0.6826
C)0.8785
D)0.5899
E)0.8185

Accepted Answer

The answer of Suppose a population has a mean of...

Question 15

Suppose a population has a mean of 90 and a standard deviation of 28.If a random sample of size 49 is drawn from the population, the probability of drawing a sample with a mean between 80 and 100 is ___.
A)0.9876
B)0.0124
C)0.4938
D)0.0062
E)1.0000

Accepted Answer

We can use the central limit theorem to approximate the distribution of sample means as a normal distribution, where the mean of the sample means is equal to the mean of the population (90) and the standard deviation of the sample means (also known as the standard error) is equal to the standard deviation of the population (28) divided by the square root of the sample size (√49 or 7).

z-score for a sample mean of 80:

z = (80-90) / (28/7) = -2.5

z-score for a sample mean of 100:

z = (100-90) / (28/7) = 2.5

Using a standard normal distribution table or calculator, the probability of getting a sample mean between -2.5 and 2.5 is approximately 0.9876. Therefore, the answer is A.

Question 16

Suppose a population has a mean of 870 and a variance of 8,100.If a random sample is size 36 is drawn from the population, the probability that the sample mean is between 840 and 900 is ___.
A)0.9544
B)0.6826
C)0.8185
D)0.5899
E)0.0897

Accepted Answer

The standard deviation of the sample mean is the population standard deviation divided by the square root of the sample size, which is 90/6 = 15. Using the standard normal distribution, the probability that the sample mean is between 840 and 900 is approximately 0.9544.

Question 17

Suppose the population of all public Universities shows the annual parking fee per student is normally distributed with a mean of $110 with a standard deviation of $18.If a random sample of size 49 is drawn from the population, the probability of drawing a sample with a mean of less than $92 is _______.
A)0.3400
B)0.1600
C)0.0000
D)1.0000
E)0.7000

Accepted Answer

The standard error of the mean is $18/sqrt(49) = $18/7 = $2.57. The z-score for a sample mean of $92 is (92-110)/2.57 = -7.00. A z-score of -7 is extremely low, making the probability essentially 0.

Question 18

Suppose a population has a mean of 450 and a variance of 900.If a random sample is size 100 is drawn from the population, the probability that the sample mean is between 448 and 453 is ___.
A)0.4972
B)0.6826
C)0.4101
D)0.5899
E)0.9878

Accepted Answer

The answer of Suppose a population has a mean of...

Question 19

A carload of steel rods has arrived at Cybermatic Construction Company.The car contains 50,000 rods.Claude Ong, Manager of Quality Assurance, directs his crew to measure the lengths of 100 randomly selected rods.If the population of rods has a mean length of 120 centimetres and a standard deviation of 0.05 centimetres, the probability that Claude's sample has a mean greater than 120.0125 centimetres is ___.
A)0.0124
B)0.0062
C)0.4938
D)0.9752
E)1.0000

Accepted Answer

We can use the central limit theorem because the sample size is sufficiently large (n=100) and the population standard deviation is known.

The z-score for a sample mean of 120.0125 is:

z = (120.0125 - 120) / (0.05 / √100) = 2.5

Using a standard normal distribution table, we can find the probability that a z-score is greater than 2.5 to be approximately 0.0062. Therefore, the probability that Claude's sample has a mean greater than 120.0125 centimetres is 0.0062.

Question 20

Suppose the population of all public Universities shows the annual parking fee per student is normally distributed with a mean of $110 with a standard deviation of $18.If a random sample of size 49 is drawn from the population, the probability of drawing a sample with a mean of more than $115 is _______.
A)0.9738
B)0.4738
C)0.0262
D)0.6103
E)0.1103

Accepted Answer

We can use the central limit theorem to approximate the sampling distribution of the sample mean:

μx̄ = μ = $110
σx̄ = σ/√n = $18/√49 = $2.57

We want to find the probability of drawing a sample with a mean of more than $115:

P(x̄ > $115) = P(z > (115 - 110)/2.57) = P(z > 1.95)

Using a standard normal table or calculator, we find:

P(z > 1.95) = 0.0262

Therefore, the probability of drawing a sample with a mean of more than $115 is 0.0262 or 2.62%, which corresponds to choice (C).

Quiz 7: Sampling and Sampling Distributions

Suppose a population has a mean of 400 and a standard deviation of 24.If a random sample of size 144 is drawn from the population, the probability of drawing a sample with a mean less than 402 is ___.

Suppose a population has a mean of 90 and a standard deviation of 28.If a random sample of size 49 is drawn from the population, the probability of drawing a sample with a mean between 85 and 95 is ___.

Suppose a population has a mean of 90 and a standard deviation of 28.If a random sample of size 49 is drawn from the population, the probability of drawing a sample with a mean of less than 84 is ___.

Suppose a population has a mean of 90 and a standard deviation of 28.If a random sample of size 49 is drawn from the population, the probability of drawing a sample with a mean of more than 95 is ___.

Suppose a population has a mean of 400 and a standard deviation of 24.If a random sample of size 144 is drawn from the population, the probability of drawing a sample with a mean between 393.5 and 396.5 is ___.

Suppose a population has a mean of 400 and a standard deviation of 24.If a random sample of size 144 is drawn from the population, the probability of drawing a sample with a mean of more than 404.5 is ___.

Suppose a population has a mean of 870 and a variance of 1,600.If a random sample is size 64 is drawn from the population, the probability that the sample mean is between 860 and 875 is ___.

Suppose a population has a mean of 90 and a standard deviation of 28.If a random sample of size 49 is drawn from the population, the probability of drawing a sample with a mean between 80 and 100 is ___.

Suppose a population has a mean of 870 and a variance of 8,100.If a random sample is size 36 is drawn from the population, the probability that the sample mean is between 840 and 900 is ___.

Suppose a population has a mean of 450 and a variance of 900.If a random sample is size 100 is drawn from the population, the probability that the sample mean is between 448 and 453 is ___.