A random sample is selected from a population with mean and standard deviation . Determine the mean and standard deviation of the sampling distribution of for each of the following sample sizes:
a. n = 9
b. n = 15
c. n = 36
d. n = 50
e. n = 100
f. n = 400

Respuesta :

Answer:

a. Mean = 100, S.D. = 3.333

b. Mean = 100, S.D. = 2.582

c. Mean = 100, S.D. = 1.667

d. Mean = 100, S.D. = 1.414

e. Mean = 100, S.D. = 1

f. Mean = 100, S.D. = 0.5

Step-by-step explanation:

The question is incomplete:

Population mean: 100

Population standard deviation: 10.

The mean for any sampling distribution is equal to the population mean.

The standard deviation for the sampling distribution depends on the population standard deviation and the sample size as:

[tex]\sigma_s=\dfrac{\sigma}{\sqrt{n}}[/tex]

We can calculate the parameters of the sampling distributions as:

a. n = 9

[tex]\mu_s=\mu=100\\\\ \sigma_s=\dfrac{\sigma}{\sqrt{n}}=\dfrac{10}{\sqrt{9}}=\dfrac{10}{3}=3.333[/tex]

b. n = 15

[tex]\mu_s=\mu=100\\\\ \sigma_s=\dfrac{\sigma}{\sqrt{n}}=\dfrac{10}{\sqrt{15}}=\dfrac{10}{3.873}=2.582[/tex]

c. n = 36

[tex]\mu_s=\mu=100\\\\ \sigma_s=\dfrac{\sigma}{\sqrt{n}}=\dfrac{10}{\sqrt{36}}=\dfrac{10}{6}=1.667[/tex]

d. n = 50

[tex]\mu_s=\mu=100\\\\ \sigma_s=\dfrac{\sigma}{\sqrt{n}}=\dfrac{10}{\sqrt{50}}=\dfrac{10}{7.071}=1.414[/tex]

e. n = 100

[tex]\mu_s=\mu=100\\\\ \sigma_s=\dfrac{\sigma}{\sqrt{n}}=\dfrac{10}{\sqrt{100}}=\dfrac{10}{10}=1[/tex]        

f. n = 400

[tex]\mu_s=\mu=100\\\\ \sigma_s=\dfrac{\sigma}{\sqrt{n}}=\dfrac{10}{\sqrt{400}}=\dfrac{10}{20}=0.5[/tex]