Question

The weights of 1500 ball bearings are normally distributed with a mean of 22.40 oz and a standard deviation of 0.048 oz. If 300 random samples of size 36 are drawn from this population, determine the expected mean and standard deviation of the sampling distribution of means if sampling is done (a) with replacement, (b) without replacement. Solve Problem 5.51 if the population consists of 72 ball bearings.

Answer 1

Answer:

Step-by-step explanation:

Given that:

The mean $\mu$ = 22.40

The standard deviation = 0.048

Sample size n = 36

Then; the expected mean $\mu_{\overline x }$ = $\mu$ = 22.40

With Replacement;

The standard deviation $\sigma _{ \overline x} = \dfrac{\sigma}{\sqrt{n}}$

$\sigma _{ \overline x} = \dfrac{0.048}{\sqrt{36}}$

$\sigma _{ \overline x} = \dfrac{0.048}{6}$

$\sigma _{ \overline x} =0.008$

Without replacement;

The standard deviation $\sigma _{ \overline x} = \sqrt{\dfrac{{N-n}}{N-1}} \times \dfrac{\sigma}{\sqrt{n}}$

$\sigma _{ \overline x} = \sqrt{\dfrac{{300-36}}{300-1}} \times \dfrac{0.048}{\sqrt{36}}$

$\sigma _{ \overline x} = \sqrt{\dfrac{{264}}{299}} \times \dfrac{0.048}{6}$

$\sigma_{\overline x} = 0.0075$

Also;

For population consisting of 72 ball bearings.

the expected mean $\mu_{\overline x }$ = $\mu$ = 22.40

With Replacement;

The standard deviation $\sigma _{ \overline x} = \dfrac{\sigma}{\sqrt{n}}$

$\sigma _{ \overline x} = \dfrac{0.048}{\sqrt{72}}$

$\sigma _{ \overline x} = \dfrac{0.048}{8.485}$

$\sigma _{ \overline x} =0.0057$

Without replacement;

The standard deviation $\sigma _{ \overline x} = \sqrt{\dfrac{{N-n}}{N-1}} \times \dfrac{\sigma}{\sqrt{n}}$

$\sigma _{ \overline x} = \sqrt{\dfrac{{300-72}}{300-1}} \times \dfrac{0.048}{\sqrt{72}}$

$\sigma _{ \overline x} = \sqrt{\dfrac{{228}}{299}} \times \dfrac{0.048}{8.485}$

$\sigma_{\overline x} =0.0049$

Answer 2

A) The expected mean and standard deviation of the sampling distribution of means if sampling is done with replacement are;

μₓ = 22.4 oz and σₓ = 0.008 oz

B) The expected mean and standard deviation of the sampling distribution of means if sampling is done without replacement are;

μₓ = 22.4 oz and σₓ = 0.008 oz

A) Sample size; n = 36

Standard deviation; σ = 0.048

Mean; μ = 22.4

Sampling with replacement;

Expected mean is; μₓ = μ

Thus; μₓ = 22.4 oz

Expected standard deviation is gotten from the formula;

σₓ = σ/√n

σₓ = 0.0048/√36

σₓ = 0.008 oz

B) Sampling without replacement;

Expected mean is; μₓ = μ

Thus; μₓ = 22.4 oz

Expected standard deviation is gotten from the formula;

σₓ = (σ/√n) × √((N - n)/(N - 1))

The sample size of 36 is way less than 5% of the population size 1500. Thus, we would drop the factor √((N - n)/(N - 1)) according to general rule of fpc and our expected standard deviation is;

σₓ = (σ/√n)

σₓ = 0.0048/√36

σₓ = 0.008 oz

The population now consists of a sample size of 72 ball bearings. Thus;

C) With Replacement;

Expected mean remains; μₓ = 22.4 oz

Expected standard deviation is;

σₓ = (σ/√n)

σₓ = 0.048/√72

σₓ = 0.0057 oz

D) Without Replacement;

Expected mean remains; μₓ = 22.4 oz

Expected standard deviation is;

σₓ = (σ/√n) × √((N - n)/(N - 1))

We will maintain that formula because 72 as sample size is very close to 5% of the population size. Thus;

σₓ = (0.048/√72) × √((1500 - 72)/(1500 - 1))

σₓ = 0.0055 oz

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