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A population of values has a normal distribution with [tex]\mu=36.9[/tex] and [tex]\sigma=42.3[/tex]. You intend to draw a random sample of size [tex]n=147[/tex]. Please show your answers as numbers accurate to 4 decimal places.

1. Find the probability that a single randomly selected value is between 36.6 and 41.8.
[tex]
P(36.6\ \textless \ x\ \textless \ 41.8) = \square
[/tex]

2. Find the probability that a sample of size [tex]n=147[/tex] is randomly selected with a mean between 36.6 and 41.8.
[tex]
P(36.6\ \textless \ \bar{x}\ \textless \ 41.8) = \square
[/tex]

Answer :

Sure! Here is a detailed, step-by-step solution to the question:

### Given:
- Population mean [tex]\(\mu = 36.9\)[/tex]
- Population standard deviation [tex]\(\sigma = 42.3\)[/tex]
- Sample size [tex]\(n = 147\)[/tex]
- We need to find:
1. The probability that a single randomly selected value is between 36.6 and 41.8.
2. The probability that the mean of a sample of size 147 is between 36.6 and 41.8.

### Step 1: Probability for a Single Randomly Selected Value

A single randomly selected value follows the normal distribution defined by the given population parameters.

1. Convert the values into z-scores:

The z-score formula for a single value [tex]\(x\)[/tex] is:
[tex]\[
z = \frac{x - \mu}{\sigma}
\][/tex]

- For [tex]\(x = 36.6\)[/tex]:
[tex]\[
z_{36.6} = \frac{36.6 - 36.9}{42.3} \approx -0.0071
\][/tex]

- For [tex]\(x = 41.8\)[/tex]:
[tex]\[
z_{41.8} = \frac{41.8 - 36.9}{42.3} \approx 0.1156
\][/tex]

2. Find the probabilities corresponding to these z-scores:

Using the standard normal distribution table or a calculator:
- [tex]\(P(Z < -0.0071) \approx 0.4972\)[/tex]
- [tex]\(P(Z < 0.1156) \approx 0.5461\)[/tex]

3. Calculate the probability between the z-scores:

[tex]\[
P(36.6 < x < 41.8) = P(Z < 0.1156) - P(Z < -0.0071) = 0.5461 - 0.4972 = 0.0489
\][/tex]

So, the probability that a single randomly selected value is between 36.6 and 41.8 is [tex]\(0.0489\)[/tex].

### Step 2: Probability for the Sample Mean

The sampling distribution of the sample mean will have the same mean [tex]\(\mu\)[/tex] but a different standard deviation, called the standard error of the mean, which is calculated as:

[tex]\[
\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}
\][/tex]

1. Calculate the standard error:

[tex]\[
\sigma_{\bar{x}} = \frac{42.3}{\sqrt{147}} \approx 3.4946
\][/tex]

2. Convert the values into z-scores for the sample mean:

- For [tex]\(x = 36.6\)[/tex]:
[tex]\[
z_{36.6} = \frac{36.6 - 36.9}{3.4946} \approx -0.0859
\][/tex]

- For [tex]\(x = 41.8\)[/tex]:
[tex]\[
z_{41.8} = \frac{41.8 - 36.9}{3.4946} \approx 1.4097
\][/tex]

3. Find the probabilities corresponding to these z-scores:

Using the standard normal distribution table or a calculator:
- [tex]\(P(Z < -0.0859) \approx 0.4657\)[/tex]
- [tex]\(P(Z < 1.4097) \approx 0.9199\)[/tex]

4. Calculate the probability between the z-scores:

[tex]\[
P(36.6 < \bar{x} < 41.8) = P(Z < 1.4097) - P(Z < -0.0859) = 0.9199 - 0.4657 = 0.4542
\][/tex]

So, the probability that the mean of a sample of size 147 is between 36.6 and 41.8 is [tex]\(0.4542\)[/tex].

### Final Answers

1. [tex]\(P(36.6 < x < 41.8) = 0.0489\)[/tex]
2. [tex]\(P(36.6 < \bar{x} < 41.8) = 0.4542\)[/tex]

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