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A population of values has a normal distribution with [tex]\(\mu=218.2\)[/tex] and [tex]\(\sigma=81.6\)[/tex].

a. Find the probability that a single randomly selected value is between 207.3 and 208.5. Round your answer to four decimal places.
[tex]\[ P(207.3 \ \textless \ X \ \textless \ 208.5) = \][/tex]
[tex]\(.0058 \ \square\)[/tex]

b. Find the probability that a randomly selected sample of size [tex]\(n=183\)[/tex] has a mean between 207.3 and 208.5. Round your answer to four decimal places.
[tex]\[ P(207.3 \ \textless \ M \ \textless \ 208.5) = \][/tex]
[tex]\(\square\)[/tex]

Sagot :

GinnyAnsweridn
To find the probabilities, we'll start by calculating the corresponding Z-scores for the given values.

### Part a

#### Given:
- Population mean ([tex]\(\mu\)[/tex]) = 218.2
- Population standard deviation ([tex]\(\sigma\)[/tex]) = 81.6
- Lower bound ([tex]\(X_{lower}\)[/tex]) = 207.3
- Upper bound ([tex]\(X_{upper}\)[/tex]) = 208.5

We need to find the Z-scores for both bounds using the formula:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]

#### Steps:
1. Calculate the Z-score for the lower bound ([tex]\(X_{lower} = 207.3\)[/tex]):
[tex]\[ Z_{lower} = \frac{207.3 - 218.2}{81.6} = -0.1336 \][/tex]

2. Calculate the Z-score for the upper bound ([tex]\(X_{upper} = 208.5\)[/tex]):
[tex]\[ Z_{upper} = \frac{208.5 - 218.2}{81.6} = -0.1189 \][/tex]

3. Use the standard normal distribution (Z-table) to find the probabilities corresponding to [tex]\( Z_{lower} \)[/tex] and [tex]\( Z_{upper} \)[/tex].

4. The desired probability is the difference between these probabilities:
[tex]\[ P(207.3 < X < 208.5) = P(Z_{upper}) - P(Z_{lower}) \][/tex]

After calculating, we get:
[tex]\[ P(207.3 < X < 208.5) = 0.0058 \][/tex]

### Part b

#### Given:
- Sample size ([tex]\(n\)[/tex]) = 183
- Population mean ([tex]\(\mu\)[/tex]) = 218.2
- Population standard deviation ([tex]\(\sigma\)[/tex]) = 81.6
- Lower bound for mean ([tex]\(M_{lower}\)[/tex]) = 207.3
- Upper bound for mean ([tex]\(M_{upper}\)[/tex]) = 208.5

Since we are dealing with a sample mean, we need to use the standard error of the mean ([tex]\(\text{SE}\)[/tex]) instead of the population standard deviation. The standard error is calculated as:
[tex]\[ \text{SE} = \frac{\sigma}{\sqrt{n}} \][/tex]

#### Steps:
1. Calculate the standard error:
[tex]\[ \text{SE} = \frac{81.6}{\sqrt{183}} \approx 6.026 \][/tex]

2. Convert the lower bound and upper bound of the sample mean to Z-scores:
[tex]\[ Z_{lower} = \frac{207.3 - 218.2}{6.026} = -1.807 \][/tex]
[tex]\[ Z_{upper} = \frac{208.5 - 218.2}{6.026} = -1.608 \][/tex]

3. Use the standard normal distribution (Z-table) to find the probabilities corresponding to [tex]\( Z_{lower} \)[/tex] and [tex]\( Z_{upper} \)[/tex].

4. The desired probability is the difference between these probabilities:
[tex]\[ P(207.3 < M < 208.5) = P(Z_{upper}) - P(Z_{lower}) \][/tex]

After calculating, we get:
[tex]\[ P(207.3 < M < 208.5) = 0.0185 \][/tex]

### Conclusion

- Part a: [tex]\( P(207.3 < X < 208.5) = 0.0058 \)[/tex]
- Part b: [tex]\( P(207.3 < M < 208.5) = 0.0185 \)[/tex]
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