Questions: A simple random sample of size n=36 is obtained from a population that is skewed right with μ=87 and σ=12. Describe the sampling distribution of x̄. ) What is P(x>90)? ) What is P(x ≤ 83)? d) What is P(84.7<x̄<90)? a) Choose the correct description of the shape of the sampling distribution of x̄. A. The distribution is uniform. B. The distribution is approximately normal. C. The distribution is skewed right. D. The distribution is skewed left. E. The shape of the distribution is unknown. Find the mean and standard deviation of the sampling distribution of x̄. μx̄=□ σx=□ (Type integers or decimals. Do not round.)

A simple random sample of size n=36 is obtained from a population that is skewed right with μ=87 and σ=12.
Describe the sampling distribution of x̄.
) What is P(x>90)?
) What is P(x ≤ 83)?
d) What is P(84.7<x̄<90)?
a) Choose the correct description of the shape of the sampling distribution of x̄.
A. The distribution is uniform.
B. The distribution is approximately normal.
C. The distribution is skewed right.
D. The distribution is skewed left.
E. The shape of the distribution is unknown.

Find the mean and standard deviation of the sampling distribution of x̄.

μx̄=□
σx=□
(Type integers or decimals. Do not round.)

Transcript text: A simple random sample of size $n=36$ is obtained from a population that is skewed right with $\mu=87$ and $\sigma=12$. Describe the sampling distribution of $\bar{x}$. ) What is $P(x>90)$ ? ) What is $P(x \leq 83)$ ? d) What is $P(84.7<\bar{x}<90)$ ? a) Choose the correct description of the shape of the sampling distribution of $\bar{x}$. A. The distribution is uniform. B. The distribution is approximately normal. C. The distribution is skewed right. D. The distribution is skewed left. E. The shape of the distribution is unknown. Find the mean and standard deviation of the sampling distribution of $\bar{x}$. \[ \begin{array}{l} \mu_{x}^{-}=\square \\ \sigma_{x}=\square \end{array} \] (Type integers or decimals. Do not round.)

Solution

Solution Steps

Step 1: Shape of the Sampling Distribution

According to the Central Limit Theorem, the shape of the sampling distribution of the sample mean $ \bar{x} $ will be approximately normal when the sample size is sufficiently large. In this case, since $ n = 36 $, which is greater than 30, we conclude that the distribution is approximately normal.

Thus, the shape of the sampling distribution of $ \bar{x} $ is: \[ \text{B. The distribution is approximately normal.} \]

Step 2: Mean and Standard Deviation of the Sampling Distribution

The mean $ \mu_{\bar{x}} $ of the sampling distribution of $ \bar{x} $ is equal to the population mean $ \mu $: \[ \mu_{\bar{x}} = 87 \]

The standard deviation $ \sigma_{\bar{x}} $ of the sampling distribution of $ \bar{x} $ is calculated using the formula: \[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{36}} = 2.0 \]

Step 3: Probability Calculations

Part a: $ P(x > 90) $

To find $ P(x > 90) $, we calculate the Z-score for $ x = 90 $: \[ Z = \frac{x - \mu_{\bar{x}}}{\sigma_{\bar{x}}} = \frac{90 - 87}{2.0} = 1.5 \] Using the standard normal distribution, we find: \[ P(x > 90) = 1 - P(Z \leq 1.5) = \Phi(\infty) - \Phi(1.5) = 0.0668 \]

Part b: $ P(x \leq 83) $

To find $ P(x \leq 83) $, we calculate the Z-score for $ x = 83 $: \[ Z = \frac{x - \mu_{\bar{x}}}{\sigma_{\bar{x}}} = \frac{83 - 87}{2.0} = -2.0 \] Using the standard normal distribution, we find: \[ P(x \leq 83) = P(Z \leq -2.0) = \Phi(-2.0) - \Phi(-\infty) = 0.0228 \]

Final Answer

The shape of the sampling distribution of $ \bar{x} $ is approximately normal.
The mean of the sampling distribution $ \mu_{\bar{x}} $ is $ 87 $.
The standard deviation of the sampling distribution $ \sigma_{\bar{x}} $ is $ 2.0 $.
$ P(x > 90) = 0.0668 $
$ P(x \leq 83) = 0.0228 $

Thus, the final answers are: \[ \boxed{\text{Shape: B, } \mu_{\bar{x}} = 87, \sigma_{\bar{x}} = 2.0, P(x > 90) = 0.0668, P(x \leq 83) = 0.0228} \]

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