A random sample of size n = 53 is taken from a finite population of size N = 617 with mean μ = 248 and variance σ2 = 334. [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.(Round “expected value” to a whole number and "standard error" to 4 decimal places.)
b. What is the probability that the sample mean is less than 238? (Round final answer to 4 decimal places.)
c. What is the probability that the sample mean lies between 243 and 258? (Round final answer to 4 decimal places.)
a-1. It is necessary to apply the finite population correction factor if the sample size is more than 5% of the population size.
The sample size is n = 53 and the population size is N = 617.
To check if it is necessary to apply the finite population correction factor, we can calculate the sampling fraction as n/N = 53/617 ≈ 0.0859. Since the sampling fraction is less than 0.05 (5%), it is not necessary to apply the finite population correction factor.
Answer: No
a-2. The expected value of the sample mean is equal to the population mean μ, as the random sample is an unbiased estimator of the population mean. So, the expected value is 248.
The standard error of the sample mean (SE) can be calculated using the formula:
SE = σ / sqrt(n), where σ is the population standard deviation and n is the sample size.
Given that σ^2 = 334 and n = 53:
SE = sqrt(334) / sqrt(53) ≈ 4.5015 (rounded to 4 decimal places)
Answer:
Expected value: 248
Standard error: 4.5015
b. To calculate the probability that the sample mean is less than 238, we need to standardize the sample mean using the formula:
z = (x - μ) / (σ / sqrt(n))
In this case, x = 238, μ = 248, σ^2 = 334, and n = 53.
z = (238 - 248) / (sqrt(334) / sqrt(53))
z = -10 / (sqrt(334) / sqrt(53))
z ≈ -2.8086
Using the z-table or a calculator, we can find the corresponding probability of z < -2.8086, which is approximately 0.0025.
Answer: 0.0025 (rounded to 4 decimal places)
c. To calculate the probability that the sample mean lies between 243 and 258, we need to standardize both values using the formula:
z = (x - μ) / (σ / sqrt(n))
In this case, x1 = 243, x2 = 258, μ = 248, σ^2 = 334, and n = 53.
For x1 = 243:
z1 = (243 - 248) / (sqrt(334) / sqrt(53))
z1 ≈ -1.4150
For x2 = 258:
z2 = (258 - 248) / (sqrt(334) / sqrt(53))
z2 ≈ 1.4150
Using the z-table or a calculator, we can find the corresponding probabilities of z1 < z < z2.
P(z1 < z < z2) ≈ P(-1.4150 < z < 1.4150) ≈ 0.8038 - 0.1699 ≈ 0.6339
Answer: 0.6339 (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 is small relative to the population size. The general rule is that if the sample size is less than 5% of the population size (n < 0.05N), then the finite population correction factor should be applied.
In this case, the sample size (n) is 53 and the population size (N) is 617.
Let's calculate the threshold for applying the finite population correction factor:
0.05 * 617 = 30.85
Since 53 (the sample size) is greater than 30.85, it is not necessary to apply the finite population correction factor.
Therefore, the answer is: No.
a-2. The expected value (mean) of the sample mean can be calculated using the formula:
E(X̄) = μ = 248
The standard error (SE) of the sample mean can be calculated using the formula:
SE = σ / sqrt(n)
= sqrt(334) / sqrt(53)
= 5.1501 (rounded to 4 decimal places)
Therefore, the expected value of the sample mean is 248 and the standard error is 5.1501.
b. To calculate the probability that the sample mean is less than 238, we need to use the standard normal distribution (z-score) and the formula:
P(X̄ < 238) = P(Z < (238 - μ) / (σ / sqrt(n)))
Plugging in the values:
P(Z < (238 - 248) / (sqrt(334) / sqrt(53)))
= P(Z < -1.9323)
Using a z-table or calculator, we find the probability to be approximately 0.0273 (rounded to 4 decimal places).
c. To calculate the probability that the sample mean lies between 243 and 258, we need to calculate two probabilities separately and then subtract them.
P(X̄ < 243) = P(Z < (243 - μ) / (σ / sqrt(n)))
P(X̄ < 258) = P(Z < (258 - μ) / (σ / sqrt(n)))
Plugging in the values:
P(Z < (243 - 248) / (sqrt(334) / sqrt(53))) = P(Z < -0.9623)
P(Z < (258 - 248) / (sqrt(334) / sqrt(53))) = P(Z < 1.9323)
Using a z-table or calculator:
P(Z < -0.9623) ≈ 0.1667
P(Z < 1.9323) ≈ 0.9731
Now, we subtract the two probabilities:
P(243 < X̄ < 258) = P(Z < 1.9323) - P(Z < -0.9623)
≈ 0.9731 - 0.1667
≈ 0.8064 (rounded to 4 decimal places)
Therefore, the probability that the sample mean lies between 243 and 258 is approximately 0.8064.
To answer these questions, we need to follow a step-by-step process.
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 relatively large compared to the population size (N).
The formula to determine whether the correction factor is needed is:
n/N ≤ 0.05
In this case, n = 53 and N = 617. Let's calculate n/N:
53/617 ≈ 0.0859
Since 0.0859 is greater than 0.05, it is NOT necessary to apply the finite population correction factor.
Therefore, the answer is: No (Choice: No)
a-2. To calculate the expected value (mean) and the standard error of the sample mean:
Expected value (mean):
The expected value of the sample mean is the same as the population mean (μ). Therefore, the expected value is 248.
Standard error of the sample mean:
The formula to calculate the standard error of the sample mean is:
SE = σ / √n
Where σ is the population standard deviation and n is the sample size. In this case, σ^2 = 334, so σ = √334.
Let's calculate the standard error:
SE = √(334/53) ≈ 3.075
Therefore, the expected value (mean) is 248 and the standard error is approximately 3.075.
b. To calculate the probability that the sample mean is less than 238, we will use the z-score and the z-table.
The formula to calculate the z-score is:
z = (x - μ) / (σ / √n)
Where x is the value we are interested in, μ is the population mean, σ is the population standard deviation, and n is the sample size.
Let's calculate the z-score:
z = (238 - 248) / (√(334/53)) ≈ -1.729
Now, let's use the z-table or a statistical software to find the probability corresponding to the z-score -1.729.
Using the z-table, we can find that the probability is approximately 0.0425.
Therefore, the probability that the sample mean is less than 238 is approximately 0.0425.
c. To calculate the probability that the sample mean lies between 243 and 258, we will again use the z-score and the z-table.
First, let's calculate the z-scores for both values:
z1 = (243 - 248) / (√(334/53))
z2 = (258 - 248) / (√(334/53))
z1 ≈ -0.866
z2 ≈ 1.732
Now, we need to find the probabilities corresponding to these z-scores.
Using the z-table, we can find the probabilities as follows:
P(z < -0.866) ≈ 0.1936
P(z < 1.732) ≈ 0.9588
To find the probability between these two z-scores, we subtract the probability corresponding to the smaller z-score from the larger z-score:
P(-0.866 < z < 1.732) = P(z < 1.732) - P(z < -0.866)
≈ 0.9588 - 0.1936 ≈ 0.7652
Therefore, the probability that the sample mean lies between 243 and 258 is approximately 0.7652.