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 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.)
1 answer
2 answers