Skip to main content

Sampling distribution: population 4,8,12,16,20

Short answer

Population mean is 12 and standard deviation is 5.6569. All 10 samples without replacement have means summing to 120, so mean of sample means is 12, equal to μ. Standard error is (5.6569/√2)√(3/4)=3.4641.

A population consists of five numbers 4, 8, 12, 16 and 20. If a random sample of size 2 is drawn without replacement. a) Find the population mean and population standard deviation. b) List all possible sample and find their sample mean. c) Show the mean of sample mean is equal to the population mean. d) Find the standard error of sample mean.

Population: 4, 8, 12, 16, 20 (N = 5); sample size n = 2, without replacement. Number of samples = ⁵C₂ = 10.

a) Population mean and standard deviation

μ = Σx/N = 60/5 = 12

σ² = Σ(x − μ)²/N = (64 + 16 + 0 + 16 + 64)/5 = 160/5 = 32

σ = √32 = 5.6569

b) All possible samples and their means

No.SampleSample mean
1(4, 8)6
2(4, 12)8
3(4, 16)10
4(4, 20)12
5(8, 12)10
6(8, 16)12
7(8, 20)14
8(12, 16)14
9(12, 20)16
10(16, 20)18

c) Mean of sample means equals population mean

Sum of sample means = 6 + 8 + 10 + 12 + 10 + 12 + 14 + 14 + 16 + 18 = 120

Mean of sample means = 120/10 = 12

This is equal to the population mean μ = 12. Hence E(x̄) = μ is verified.

d) Standard error of the sample mean

With sampling without replacement from a finite population:

SE = (σ/√n) × √((N − n)/(N − 1)) = (5.6569/√2) × √((5 − 2)/(5 − 1))

= 4.0000 × 0.8660 = 3.4641

(Direct check: the standard deviation of the 10 sample means is 3.4641.)

Answer: μ = 12, σ = 5.6569; mean of sample means = 12; SE = 3.4641.

✓ Answer checked by Claude Sonnet 5.5

Related questions

All Probability and Statistics chapter 4 questions →