Question

In: Economics

Given a population with a mean of μ= 103 and a variance of σ2 = 16,...

Given a population with a mean of μ= 103 and a variance of σ² = 16, the central limit applies when the sample size n >_25. A random sample of size n = 25 is obtained.

a. What are the mean and variance of the sampling distribution for the sample means?

b. What is the probability that X^{- >} 104 ?

c. Whta is the probability that 102 < X < 104 ?

d. What is the probability that X^- <_ 103.5 ?

Solutions

Expert Solution

It is given that the -

$\mu = 103,{\sigma ^2} = 16,{\sigma ^{}} = \sqrt {16} = 4,n = 25$ .

As the central limit theorem is applicable. Thus, Z-test is used to compute the required probability.

a) It is given that the central limit theorem is applicable. Thus, it states that the sample mean ( $\overline X$ ) and sample standard deviation ( ${\sigma _{_{\overline X }}}$ ) of the sampling distribution of means is computed as-

$\mu = \overline X = 103$ , and

${\sigma _{_{\overline X }}} = \frac{\sigma }{{\sqrt n }} = \frac{4}{{\sqrt {25} }} = \frac{4}{5} = 0.8$

b) The required probability is computed as -

$\begin{array}{c}P\left( {\overline X > 104} \right) = 1 - P\left( {\overline X < 104} \right)\\ = 1 - P\left( {\frac{{\overline X - \mu }}{{{\sigma _{\overline X }}}} < \frac{{104 - 103}}{{0.8}}} \right)\\ = 1 - \Phi \left( {1.25} \right)\\ = 1 - 0.894\\ = 0.106\end{array}$

The required value is computed using Z-table.

Therefore, the required probability is 0.106.

c) The required probability is computed as -

$\begin{array}{c}P\left( {102 < \overline X < 104} \right) = P\left( {102 < \overline X < 104} \right)\\ = P\left( {\frac{{102 - 103}}{{0.8}} < \frac{{\overline X - \mu }}{{{\sigma _{\overline X }}}} < \frac{{104 - 103}}{{0.8}}} \right)\\ = \Phi \left( {1.25} \right) - \Phi \left( { - 1.25} \right)\\ = 0.894 - 0.106\\ = 0.788\end{array}$

The required value is computed using Z-table.

Therefore, the required probability is 0.788.

d) The required probability is computed as -

$\begin{array}{c}P\left( {\overline X \le 103.5} \right) = P\left( {\frac{{\overline X - \mu }}{{{\sigma _{\overline X }}}} < \frac{{103.5 - 103}}{{0.8}}} \right)\\ = \Phi \left( {0.625} \right)\\ = 0.734\end{array}$

The required value is computed using Z-table.

Therefore, the required probability is 0.734.

Rahul Sunny answered 1 year ago

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