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Answer :
Final answer:
The distribution of the sample mean for a sample size of 7 from the normally distributed population will be normal with a mean of 159 lb and a standard deviation of approximately 9.43 lb.
Explanation:
The distribution of the sample mean ([tex]\overline{x}[/tex]) for a sample size of 7 from a normally distributed population with a known mean and standard deviation will itself be normally distributed according to the Central Limit Theorem. Even though the sample size is small, the underlying population distribution is normal, which allows us to assume that the distribution of the sample means will also be normal. For the given parameters:
Population mean (μm): 159 lb
Population standard deviation (σ): 25 lb
Sample size (n): 7
The mean of the sample means ([tex]\overline{x}[/tex]) will be the same as the population mean, which is 159 lb. The standard deviation of the sample means ([tex]\overline{x}[/tex]), also known as the standard error, is calculated using the formula ([tex]\overline{x}[/tex]) = σ/√n. Plugging in our numbers:
([tex]\overline{x}[/tex]) = 25 / √7 ≈ 9.43 lb
Therefore, the distribution of the sample mean for samples of size 7 is normal with a mean (μm) of 159 lb and a standard deviation ([tex]\overline{x}[/tex]) of approximately 9.43 lb.
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Rewritten by : Barada
Answer:
Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(159,25)[/tex]
Where [tex]\mu=159[/tex] and [tex]\sigma=25[/tex]
Since the distribution of X is normal than we can conclude that we know the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
With the following parameters:
[tex] \mu_{\bar X} = 159 [/tex]
[tex]\sigma_{\bar X} =\frac{25}{\sqrt{7}} = 9.449[/tex]
Step-by-step explanation:
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
Solution to the problem
Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(159,25)[/tex]
Where [tex]\mu=159[/tex] and [tex]\sigma=25[/tex]
Since the distribution of X is normal than we can conclude that we know the distribution for the sample mean [tex]\bar X[/tex] is given by:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]
With the following parameters:
[tex] \mu_{\bar X} = 159 [/tex]
[tex]\sigma_{\bar X} =\frac{25}{\sqrt{7}} = 9.449[/tex]
