Answer:
The mean of the sampling distribution
is 86
The standard deviation of the sampling distribution
is 2
The mean of the sampling distribution
is 71
The standard deviation of the sampling distribution
is 1
B) The distribution is approximately normal.
![\mathbf{P(\overline x >73) = 0.0228}](https://tex.z-dn.net/?f=%5Cmathbf%7BP%28%5Coverline%20x%20%3E73%29%20%3D%200.0228%7D)
Therefore, the probability that the sample mean is greater than 73 = 0.0228
![\mathbf{P(\overline x \leq 69) = 0.0228}](https://tex.z-dn.net/?f=%5Cmathbf%7BP%28%5Coverline%20x%20%5Cleq%2069%29%20%3D%200.0228%7D)
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}](https://tex.z-dn.net/?f=%5Cmathbf%7BP%28%2069.8%20%5Cleq%20%5Coverline%20x%20%5Cleq%2072.5%29%20%3D%200.8181%7D)
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
and
from the given parameters of a population and sample size.
Given that :
population mean
= 86
population standard deviation
= 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
is equal to the population mean
∴
=
= 86
The standard deviation of the sampling distribution can be computed by using the formula:
![\sigma_{\overline x } = \dfrac{\sigma }{\sqrt{n}}](https://tex.z-dn.net/?f=%5Csigma_%7B%5Coverline%20x%20%7D%20%3D%20%5Cdfrac%7B%5Csigma%20%7D%7B%5Csqrt%7Bn%7D%7D)
![\sigma_{\overline x } = \dfrac{16 }{\sqrt{64}}](https://tex.z-dn.net/?f=%5Csigma_%7B%5Coverline%20x%20%7D%20%3D%20%5Cdfrac%7B16%20%7D%7B%5Csqrt%7B64%7D%7D)
![\sigma_{\overline x }= \dfrac{16 }{8}](https://tex.z-dn.net/?f=%5Csigma_%7B%5Coverline%20x%20%7D%3D%20%5Cdfrac%7B16%20%7D%7B8%7D)
![\mathbf{\sigma_{\overline x } = 2}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Csigma_%7B%5Coverline%20x%20%7D%20%3D%202%7D)
∴
The mean of the sampling distribution
is 86
The standard deviation of the sampling distribution
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
= 71
standard deviation
= 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
is equal to the population mean
∴
=
= 71
The standard deviation of this sampling distribution
can be estimated as :
![\sigma_{\overline x }= \dfrac{\sigma }{\sqrt{n}}](https://tex.z-dn.net/?f=%5Csigma_%7B%5Coverline%20x%20%7D%3D%20%5Cdfrac%7B%5Csigma%20%7D%7B%5Csqrt%7Bn%7D%7D)
![\sigma_{\overline x }= \dfrac{6 }{\sqrt{36}}](https://tex.z-dn.net/?f=%5Csigma_%7B%5Coverline%20x%20%7D%3D%20%5Cdfrac%7B6%20%7D%7B%5Csqrt%7B36%7D%7D)
![\sigma_{\overline x } = \dfrac{6 }{6}](https://tex.z-dn.net/?f=%5Csigma_%7B%5Coverline%20x%20%7D%20%3D%20%5Cdfrac%7B6%20%7D%7B6%7D)
![\mathbf{\sigma_{\overline x } = 1}](https://tex.z-dn.net/?f=%5Cmathbf%7B%5Csigma_%7B%5Coverline%20x%20%7D%20%3D%201%7D)
∴
The mean of the sampling distribution
is 71
The standard deviation of the sampling distribution
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}](https://tex.z-dn.net/?f=P%28%5Coverline%20x%20%3E73%29%20%3D%20P%20%5Cbegin%20%7Bpmatrix%7D%20%5Cdfrac%7B%5Coverline%20x%20-%20%5Cmu%20%20%7D%7B%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D%20%3E%20%20%5Cdfrac%7B73%20-71%20%20%7D%7B%5Cdfrac%7B6%7D%7B%5Csqrt%7B36%7D%7D%7D%20%20%5Cend%20%7Bpmatrix%7D)
![P(\overline x >73) = P \begin {pmatrix} Z > \dfrac{73 -71 }{\dfrac{6}{\sqrt{36}}} \end {pmatrix}](https://tex.z-dn.net/?f=P%28%5Coverline%20x%20%3E73%29%20%3D%20P%20%5Cbegin%20%7Bpmatrix%7D%20Z%20%3E%20%20%5Cdfrac%7B73%20-71%20%20%7D%7B%5Cdfrac%7B6%7D%7B%5Csqrt%7B36%7D%7D%7D%20%20%5Cend%20%7Bpmatrix%7D)
![P(\overline x >73) = P \begin {pmatrix} Z > \dfrac{2 }{\dfrac{6}{6}} \end {pmatrix}](https://tex.z-dn.net/?f=P%28%5Coverline%20x%20%3E73%29%20%3D%20P%20%5Cbegin%20%7Bpmatrix%7D%20Z%20%3E%20%20%5Cdfrac%7B2%20%7D%7B%5Cdfrac%7B6%7D%7B6%7D%7D%20%20%5Cend%20%7Bpmatrix%7D)
![P(\overline x >73) = P \begin {pmatrix} Z > \dfrac{2 \times 6 }{6} \end {pmatrix}](https://tex.z-dn.net/?f=P%28%5Coverline%20x%20%3E73%29%20%3D%20P%20%5Cbegin%20%7Bpmatrix%7D%20Z%20%3E%20%20%5Cdfrac%7B2%20%5Ctimes%206%20%7D%7B6%7D%20%20%5Cend%20%7Bpmatrix%7D)
![P(\overline x >73) = P \begin {pmatrix} Z > \dfrac{12 }{6} \end {pmatrix}](https://tex.z-dn.net/?f=P%28%5Coverline%20x%20%3E73%29%20%3D%20P%20%5Cbegin%20%7Bpmatrix%7D%20Z%20%3E%20%20%5Cdfrac%7B12%20%7D%7B6%7D%20%20%5Cend%20%7Bpmatrix%7D)
![P(\overline x >73) = P \begin {pmatrix} Z > 2 \end {pmatrix}](https://tex.z-dn.net/?f=P%28%5Coverline%20x%20%3E73%29%20%3D%20P%20%5Cbegin%20%7Bpmatrix%7D%20Z%20%3E%20%202%20%5Cend%20%7Bpmatrix%7D)
![P(\overline x >73) = 1- P \begin {pmatrix} Z < 2 \end {pmatrix}](https://tex.z-dn.net/?f=P%28%5Coverline%20x%20%3E73%29%20%3D%201-%20P%20%5Cbegin%20%7Bpmatrix%7D%20Z%20%3C%202%20%5Cend%20%7Bpmatrix%7D)
Using the Excel Function ( =NORMDIST(2) )
![P(\overline x >73) = 1- 0.9772](https://tex.z-dn.net/?f=P%28%5Coverline%20x%20%3E73%29%20%3D%201-%200.9772)
![\mathbf{P(\overline x >73) = 0.0228}](https://tex.z-dn.net/?f=%5Cmathbf%7BP%28%5Coverline%20x%20%3E73%29%20%3D%200.0228%7D)
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}](https://tex.z-dn.net/?f=P%28%5Coverline%20x%20%5Cleq%2069%29%20%3D%20P%20%5Cbegin%20%7Bpmatrix%7D%20%5Cdfrac%7B%5Coverline%20x%20-%20%5Cmu%20%20%7D%7B%5Cdfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%7D%20%5Cleq%20%20%5Cdfrac%7B69%20-71%20%20%7D%7B%5Cdfrac%7B6%7D%7B%5Csqrt%7B36%7D%7D%7D%20%20%5Cend%20%7Bpmatrix%7D)
![P(\overline x \leq 69) = P \begin {pmatrix}Z \leq \dfrac{-2 }{\dfrac{6}{6}} \end {pmatrix}](https://tex.z-dn.net/?f=P%28%5Coverline%20x%20%5Cleq%2069%29%20%3D%20P%20%5Cbegin%20%7Bpmatrix%7DZ%20%5Cleq%20%5Cdfrac%7B-2%20%20%7D%7B%5Cdfrac%7B6%7D%7B6%7D%7D%20%20%5Cend%20%7Bpmatrix%7D)
![P(\overline x \leq 69) = P \begin {pmatrix} Z \leq \dfrac{-2 \times 6 }{6} \end {pmatrix}](https://tex.z-dn.net/?f=P%28%5Coverline%20x%20%5Cleq%2069%29%20%3D%20P%20%5Cbegin%20%7Bpmatrix%7D%20Z%20%5Cleq%20%20%5Cdfrac%7B-2%20%5Ctimes%206%20%20%7D%7B6%7D%20%20%5Cend%20%7Bpmatrix%7D)
![P(\overline x \leq 69) = P \begin {pmatrix} Z \leq \dfrac{-12 }{6} \end {pmatrix}](https://tex.z-dn.net/?f=P%28%5Coverline%20x%20%5Cleq%2069%29%20%3D%20P%20%5Cbegin%20%7Bpmatrix%7D%20Z%20%5Cleq%20%20%5Cdfrac%7B-12%20%20%7D%7B6%7D%20%20%5Cend%20%7Bpmatrix%7D)
![P(\overline x \leq 69) = P \begin {pmatrix} Z \leq -2 \end {pmatrix}](https://tex.z-dn.net/?f=P%28%5Coverline%20x%20%5Cleq%2069%29%20%3D%20P%20%5Cbegin%20%7Bpmatrix%7D%20Z%20%5Cleq%20-2%20%5Cend%20%7Bpmatrix%7D)
Using the EXCEL FUNCTION ( = NORMSDIST (-2) )
![\mathbf{P(\overline x \leq 69) = 0.0228}](https://tex.z-dn.net/?f=%5Cmathbf%7BP%28%5Coverline%20x%20%5Cleq%2069%29%20%3D%200.0228%7D)
The probability that the sample mean is less than or equal to 69 is 0.0228
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