A random sample of size n = 72 is taken from a finite population of size N = 644 with mean μ = 239 and variance σ2 = 325. Use Table 1.
a-1.
Is it necessary to apply the finite population correction factor?
Yes
No
a-2.
Calculate the expected value and the standard error of the sample mean. (Do not round intermediate calculations. Round "standard error" to 3 decimal places.)
Expected value
Standard error
b.
What is the probability that the sample mean is less than 229?
Probability
c.
What is the probability that the sample mean lies between 234 and 249? (Use rounded standard deviation. Round "z" value to 2 decimal places and final answer to 4 decimal places.)
Probability
This might help:
http://davidmlane.com/hyperstat/z_table.html
Have no idea of contents of Table 1.
1. No
2a. SEm = SD/√n = (√variance)/n
2b. Z = (score-mean)/SEm
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
2c. Use same equation and Table.
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 more than 5% of the population size (N).
In this case, n = 72 and N = 644.
To calculate the percentage that n is of N, we divide n by N and multiply by 100:
(72 / 644) * 100 = 11.18%
Since 11.18% is less than 5%, it is not necessary to apply the finite population correction factor.
a-2.
To calculate the expected value (mean) of the sample mean, we use the formula:
Expected value = μ (population mean)
In this case, the population mean is given as μ = 239.
The standard error of the sample mean can be calculated using the formula:
Standard error = σ / sqrt(n)
Where σ is the population standard deviation and n is the sample size.
In this case, the population variance (σ^2) is given as 325, so the population standard deviation (σ) is the square root of 325:
σ = sqrt(325)
Now we can calculate the standard error:
Standard error = σ / sqrt(n) = sqrt(325) / sqrt(72)
Calculate the square root of 325:
sqrt(325) ≈ 18.028
Calculate the square root of 72:
sqrt(72) ≈ 8.485
Now divide the two values to get the standard error:
Standard error = 18.028 / 8.485 ≈ 2.126
Round the standard error to 3 decimal places:
Standard error ≈ 2.126
b.
To find the probability that the sample mean is less than 229, we need to convert the sample mean into a z-score and then look up the corresponding probability in Table 1.
The formula to calculate the z-score is:
z = (x - μ) / (σ / sqrt(n))
Where x is the value we want to find the probability for, μ is the population mean, σ is the population standard deviation, and n is the sample size.
In this case, we want to find the probability that the sample mean is less than 229, so x = 229.
Calculate the z-score:
z = (229 - 239) / (sqrt(325) / sqrt(72))
Substitute the values:
z = -10 / (sqrt(325) / sqrt(72)) ≈ -2.557
Round the z-score to 2 decimal places:
z ≈ -2.56
Now, use Table 1 to find the probability corresponding to a z-score of -2.56. This will give you the probability that the sample mean is less than 229.
c.
To find the probability that the sample mean lies between 234 and 249, we need to calculate the z-scores for both values and then look up the corresponding probabilities in Table 1.
First, calculate the z-score for 234 using the formula:
z = (x - μ) / (σ / sqrt(n))
x = 234
z = (234 - 239) / (sqrt(325) / sqrt(72))
Calculate the z-score:
z = -5 / (sqrt(325) / sqrt(72)) ≈ -1.268
Round the z-score to 2 decimal places:
z ≈ -1.27
Next, calculate the z-score for 249 using the same formula:
z = (x - μ) / (σ / sqrt(n))
x = 249
z = (249 - 239) / (sqrt(325) / sqrt(72))
Calculate the z-score:
z = 10 / (sqrt(325) / sqrt(72)) ≈ 2.557
Round the z-score to 2 decimal places:
z ≈ 2.56
Now, use Table 1 to find the probabilities corresponding to z-scores of -1.27 and 2.56. Subtract the probability for -1.27 from the probability for 2.56 to get the probability that the sample mean lies between 234 and 249.