Question

Determine ux and sigma x from the given parameters of e population and sample size.

Answer 1

Answer:

The mean of the sampling distribution  $\mu_{\overline x}$ is 86

The standard deviation of the sampling distribution  $\sigma_x$  is 2

The mean of the sampling distribution  $\mu_{\overline x}$ is 71

The standard deviation of the sampling distribution  $\sigma_x$  is 1

B) The distribution is approximately normal.

$\mathbf{P(\overline x >73) = 0.0228}$

Therefore, the probability that the sample mean is greater than 73 = 0.0228

$\mathbf{P(\overline x \leq 69) = 0.0228}$

The probability that the sample mean is less than or equal to 69 is 0.0228

$\mathbf{P( 69.8 \leq \overline x \leq 72.5) = 0.8181}$

Thus, the probability that the sample mean is between 69.8 and 72 is 0.8181.

Step-by-step explanation:

We are to determine the $\mu_{\overline x}$ and $\sigma_x$ from the given parameters of a population and sample size.

Given that :

population mean $\mu$ = 86

population standard deviation $\sigma$ = 16

sample size n = 64

From the central limit theorem's knowledge, we know that as the sample distribution approximates a normal distribution, the sample size gets larger. Thus, the mean of the sampling distribution $\mu_{\overline x}$  is equal to the population mean $\mu$

∴

$\mu_{\overline x}$  = $\mu$ = 86

The standard deviation of the sampling distribution can be computed by using the formula:

$\sigma_{\overline x } = \dfrac{\sigma }{\sqrt{n}}$

$\sigma_{\overline x } = \dfrac{16 }{\sqrt{64}}$

$\sigma_{\overline x }= \dfrac{16 }{8}$

$\mathbf{\sigma_{\overline x } = 2}$

∴

The mean of the sampling distribution  $\mu_{\overline x}$ is 86

The standard deviation of the sampling distribution  $\sigma_x$  is 2

Suppose a simple random sample of size n = 36  is obtained from a population with mu = 71 and sigma = 6.

i.e

sample size n = 36

population mean $\mu$ = 71

standard deviation $\sigma$ = 6

From the central limit theorem's knowledge, we know that as the sample distribution approximates a normal distribution, the sample size gets larger. Thus, the mean of the sampling distribution $\mu_{\overline x}$  is equal to the population mean $\mu$

∴

$\mu_{\overline x}$  = $\mu$ = 71

The standard deviation of this sampling distribution $\sigma_{\overline x}$ can be estimated as :

$\sigma_{\overline x }= \dfrac{\sigma }{\sqrt{n}}$

$\sigma_{\overline x }= \dfrac{6 }{\sqrt{36}}$

$\sigma_{\overline x } = \dfrac{6 }{6}$

$\mathbf{\sigma_{\overline x } = 1}$

∴

The mean of the sampling distribution  $\mu_{\overline x}$ is 71

The standard deviation of the sampling distribution  $\sigma_x$  is 1

A)

The correct option from the  given question is:

B) The distribution is approximately normal.

B) What is P (xbar > 73)?

i.e

$P(\overline x >73) = P \begin {pmatrix} \dfrac{\overline x - \mu }{\dfrac{\sigma}{\sqrt{n}}} > \dfrac{73 -71 }{\dfrac{6}{\sqrt{36}}} \end {pmatrix}$

$P(\overline x >73) = P \begin {pmatrix} Z > \dfrac{73 -71 }{\dfrac{6}{\sqrt{36}}} \end {pmatrix}$

$P(\overline x >73) = P \begin {pmatrix} Z > \dfrac{2 }{\dfrac{6}{6}} \end {pmatrix}$

$P(\overline x >73) = P \begin {pmatrix} Z > \dfrac{2 \times 6 }{6} \end {pmatrix}$

$P(\overline x >73) = P \begin {pmatrix} Z > \dfrac{12 }{6} \end {pmatrix}$

$P(\overline x >73) = P \begin {pmatrix} Z > 2 \end {pmatrix}$

$P(\overline x >73) = 1- P \begin {pmatrix} Z < 2 \end {pmatrix}$

Using the Excel Function ( =NORMDIST(2) )

$P(\overline x >73) = 1- 0.9772$

$\mathbf{P(\overline x >73) = 0.0228}$

Therefore, the probability that the sample mean is greater than 73 = 0.0228

C) What is P (xbar ≤ 69)?

i.e

$P(\overline x \leq 69) = P \begin {pmatrix} \dfrac{\overline x - \mu }{\dfrac{\sigma}{\sqrt{n}}} \leq \dfrac{69 -71 }{\dfrac{6}{\sqrt{36}}} \end {pmatrix}$

$P(\overline x \leq 69) = P \begin {pmatrix}Z \leq \dfrac{-2 }{\dfrac{6}{6}} \end {pmatrix}$

$P(\overline x \leq 69) = P \begin {pmatrix} Z \leq \dfrac{-2 \times 6 }{6} \end {pmatrix}$

$P(\overline x \leq 69) = P \begin {pmatrix} Z \leq \dfrac{-12 }{6} \end {pmatrix}$

$P(\overline x \leq 69) = P \begin {pmatrix} Z \leq -2 \end {pmatrix}$

Using the EXCEL FUNCTION ( = NORMSDIST (-2) )

$\mathbf{P(\overline x \leq 69) = 0.0228}$

The probability that the sample mean is less than or equal to 69 is 0.0228

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