Assume we draw a simple random sample from a population having a mean of 100 and a standard deviation of 16. What is the probability that a sample mean will be within plus-or-minus two of the population mean for each of the following sample sizes?
(a.) n = 50 ____________
(b.) n = 100
(c.) n = 200 ____________
(d.) n = 400 ____________
μ ± 2SE will always include approximately 95% of the possible means. However, the range of the possible means for the 95% will be smaller as the n increases.
SE (Standard Error of the mean) = SD/sq.root of n
(For smaller samples, you might want to use n-1.)
I hope this helps.
To find the probability that a sample mean will be within plus-or-minus two of the population mean, we can use the Central Limit Theorem.
The Central Limit Theorem states that if a random sample with a sample size greater than 30 is drawn from any population, the distribution of the sample means will be approximately normally distributed, regardless of the shape of the population.
Given that the population mean (μ) is 100 and the standard deviation (σ) is 16, we can calculate the standard error (SE) using the formula:
SE = σ / √n
where n is the sample size.
(a.) For n = 50:
SE = 16 / √50 ≈ 2.262
To find the probability that the sample mean will be within plus-or-minus two of the population mean, we need to calculate the z-scores for both the upper and lower values:
Upper z-score = (102 - 100) / 2.262 ≈ 0.884
Lower z-score = (98 - 100) / 2.262 ≈ -0.884
Using a standard normal distribution table or a calculator, we can find the probability corresponding to these z-scores:
P(Z < 0.884) ≈ 0.812
P(Z > -0.884) ≈ 0.812
To find the probability that the sample mean will be within plus-or-minus two, we can subtract the probability of being outside this interval from 1:
P(-2 < X < 2) = 1 - (0.812 + 0.812) ≈ 1 - 1.624 ≈ 0.376
Therefore, the probability that a sample mean will be within plus-or-minus two of the population mean for n = 50 is approximately 0.376.
(b.) For n = 100:
SE = 16 / √100 = 1.6
Using the same process as above, we can find the z-scores:
Upper z-score = (102 - 100) / 1.6 = 1.25
Lower z-score = (98 - 100) / 1.6 = -1.25
Using the standard normal distribution table or a calculator:
P(Z < 1.25) ≈ 0.894
P(Z > -1.25) ≈ 0.894
P(-2 < X < 2) = 1 - (0.894 + 0.894) ≈ 1 - 1.788 ≈ 0.212
Therefore, the probability that a sample mean will be within plus-or-minus two of the population mean for n = 100 is approximately 0.212.
(c.) For n = 200:
SE = 16 / √200 ≈ 1.131
Upper z-score = (102 - 100) / 1.131 ≈ 1.769
Lower z-score = (98 - 100) / 1.131 ≈ -1.769
P(Z < 1.769) ≈ 0.962
P(Z > -1.769) ≈ 0.962
P(-2 < X < 2) = 1 - (0.962 + 0.962) ≈ 1 - 1.924 ≈ 0.076
Therefore, the probability that a sample mean will be within plus-or-minus two of the population mean for n = 200 is approximately 0.076.
(d.) For n = 400:
SE = 16 / √400 = 0.8
Upper z-score = (102 - 100) / 0.8 = 2.5
Lower z-score = (98 - 100) / 0.8 = -2.5
P(Z < 2.5) ≈ 0.9938
P(Z > -2.5) ≈ 0.9938
P(-2 < X < 2) = 1 - (0.9938 + 0.9938) ≈ 1 - 1.9876 ≈ 0.0124
Therefore, the probability that a sample mean will be within plus-or-minus two of the population mean for n = 400 is approximately 0.0124.