A random sample of size n = 122 is taken from a population of size N = 2,700 with mean μ = −56 and variance σ2 = 90. [You may find it useful to reference the z table.]
a-1. Is it necessary to apply the finite population correction factor?
multiple choice
Yes
No
a-2. Calculate the expected value and the standard error of the sample mean. (Negative values should be indicated by a minus sign. Round “expected value” to a whole number and "standard error" to 2 decimal places.)
b. What is the probability that the sample mean is between −58 and −54? (Round final answer to 4 decimal places.)
c. What is the probability that the sample mean is greater than −55? (Round final answer to 4 decimal places.)
a-1. No
a-2. The expected value of the sample mean E(x̄) is equal to the population mean μ:
E(x̄) = μ = -56
The standard error of the sample mean σ(x̄) is calculated using the formula:
σ(x̄) = σ / √n
σ(x̄) = √90 / √122
σ(x̄) ≈ 2.43 (rounded to 2 decimal places)
b. To find the probability that the sample mean is between -58 and -54, we need to calculate the z-scores for these values and use the z-table.
Find the z-scores:
z1 = (x1 - μ) / σ(x̄) = (-58 - (-56)) / 2.43
z2 = (x2 - μ) / σ(x̄) = (-54 - (-56)) / 2.43
P(-58 < x < -54) = P(z1 < Z < z2)
Using the z-table, find the probabilities corresponding to the z-scores z1 and z2, and subtract the smaller probability from the larger one:
P(-58 < x < -54) = P(z1 < Z < z2) = P(Z < z2) - P(Z < z1)
Round the final answer to 4 decimal places.
c. To find the probability that the sample mean is greater than -55, we need to find the z-score for -55 and use the z-table.
Find the z-score:
z = (x - μ) / σ(x̄) = (-55 - (-56)) / 2.43
P(x > -55) = P(Z > z)
Using the z-table, find the probability corresponding to the z-score z.
Round the final answer to 4 decimal places.
a-1. To determine if it is necessary to apply the finite population correction factor, we need to check if the sample size is a large proportion of the population size. The formula to calculate this is:
\[ \frac{n}{N} \leq 0.05 \]
In this case:
\[ \frac{122}{2700} = 0.045 \]
Since the result is less than 0.05, it is not necessary to apply the finite population correction factor.
Answer: No
a-2. The expected value of the sample mean (E[x̄]) is equal to the population mean (μ):
\[ E[x̄] = μ = -56 \]
The standard error (SE) of the sample mean is calculated using the formula:
\[ SE = \frac{σ}{\sqrt{n}} \]
where σ is the population standard deviation and n is the sample size.
In this case:
\[ SE = \frac{\sqrt{90}}{\sqrt{122}} = 0.778 \]
The expected value of the sample mean is -56 and the standard error is 0.778.
Answer:
Expected value of the sample mean: -56
Standard error of the sample mean: 0.78 (rounded to 2 decimal places)
b. To calculate the probability that the sample mean is between -58 and -54, we need to standardize the values using the formula:
\[ P(-58 < x̄ < -54) = P\left(\frac{{-58 - (-56)}}{0.778} < Z < \frac{{-54 - (-56)}}{0.778}\right) \]
Z is the standard score (Z-score) and can be found using the Z-table.
First, we calculate the standard scores:
\[ Z_1 = \frac{{-58 - (-56)}}{0.778} = -2.57 \]
\[ Z_2 = \frac{{-54 - (-56)}}{0.778} = 2.57 \]
Next, we look up the cumulative probabilities from the Z-table:
\[ P(-2.57 < Z < 2.57) \]
\[ = P(Z < 2.57) - P(Z < -2.57) \]
\[ = 0.9951 - 0.0049 \]
\[ = 0.9902 \]
The probability that the sample mean is between -58 and -54 is approximately 0.9902.
Answer: 0.9902
c. To calculate the probability that the sample mean is greater than -55, we first need to standardize the value using the formula:
\[ P(x̄ > -55) = P\left(Z > \frac{{-55 - (-56)}}{0.778}\right) \]
The standard score is:
\[ Z = \frac{{-55 - (-56)}}{0.778} = 1.29 \]
Next, we look up the cumulative probability from the Z-table:
\[ P(Z > 1.29) = 1 - P(Z < 1.29) \]
\[ = 1 - 0.9015 \]
\[ = 0.0985 \]
The probability that the sample mean is greater than -55 is approximately 0.0985.
Answer: 0.0985 (rounded to 4 decimal places)
a-1. To determine whether it is necessary to apply the finite population correction factor, we need to check if the sample size (n) is a significant portion of the population size (N).
If the sample size is less than 5% of the population size (n < 0.05N), then we can assume it is a large enough sample and not apply the finite population correction factor.
In this case, we have n = 122 and N = 2,700.
To check if n < 0.05N:
0.05N = 0.05 * 2,700 = 135
Since n = 122 is less than 135, it is necessary to apply the finite population correction factor.
So the answer is: Yes.
a-2. The expected value (mean) of the sample mean can be calculated using the formula:
Expected value of sample mean = population mean (μ)
In this case, the population mean (μ) is given as -56.
So the expected value of the sample mean is -56.
The standard error of the sample mean can be calculated using the formula:
Standard error of sample mean = square root(variance of population) / square root(sample size)
In this case, the variance of population (σ^2) is given as 90 and the sample size (n) is given as 122.
Standard error of sample mean = sqrt(90) / sqrt(122) ≈ 2.57
So the expected value of the sample mean is -56 and the standard error is approximately 2.57.
b. To find the probability that the sample mean is between -58 and -54, we can use the z-score formula:
z = (x - μ) / (σ / sqrt(n))
Where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.
In this case, x = -58, μ = -56, σ = sqrt(90) ≈ 9.49, and n = 122.
For -58:
z1 = (-58 - (-56)) / (9.49 / sqrt(122)) ≈ -1.085
For -54:
z2 = (-54 - (-56)) / (9.49 / sqrt(122)) ≈ 0.748
We can consult the z table or use a statistical calculator to find the corresponding probabilities for z1 and z2.
From the z table or calculator, we find that the probability for z1 is approximately 0.1389 and the probability for z2 is approximately 0.7734.
To find the probability between -58 and -54, we subtract the probability for z1 from the probability for z2:
Probability = 0.7734 - 0.1389 ≈ 0.6345
So the probability that the sample mean is between -58 and -54 is approximately 0.6345.
c. To find the probability that the sample mean is greater than -55, we can again use the z-score formula.
In this case, x = -55, μ = -56, σ = sqrt(90) ≈ 9.49, and n = 122.
z = (-55 - (-56)) / (9.49 / sqrt(122)) ≈ 0.265
We can consult the z table or use a statistical calculator to find the corresponding probability for z.
From the z table or calculator, we find that the probability for z ≈ 0.6030.
So the probability that the sample mean is greater than -55 is approximately 0.6030.