Mathematics, 29.08.2020 01:01 AllanCasey12

Determine ux and sigma x from the given parameters of e population and sample size. mu = 86, sigma = 16, n = 64
Suppose a simple random sample of size n = 36 is obtained from a population with mu = 71 and sigma = 6.
A) Describe the sampling distribution of xbar.
B) What is P (xbar > 73)?
C) What is P (xbar 69)?
D) What is P (69.8 < xbar < 72.5)?
1) Choose the correct description of the shape of the sampling distribution of xbar.
A) The distribution is skewed left.
B) The distribution is approximately normal.
C) The distribution is uniform.
D) The distribution is skewed right.
E) The shape of the distribution is unknown.
Find the mean and standard deviation of the sampling distribution of x-.
2) P(x->73) =.
3) P (x- 69) =.
4) P (69.8 < xbar < 72.5) =.

Answers: 2

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Answer from: dozsyerra

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

∴

$\mu_{\overline x}$ = $\mu$ = 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

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

NOTE: This text editor can't contain more than 5000 characters so the last part of the question is attached in the image below.

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

mu = 86, sigma

Answer from: Quest

He was paid 23 dollars for each lawn

Answer from: Quest

answer: x=3 and y=4

step-by-step explanation:

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