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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]
### 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]
Thank you for reading the article 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. We hope the information provided is useful and helps you understand this topic better. Feel free to explore more helpful content on our website!
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