Respuesta :
a. Mean of \( \bar{x} \):
\[ \text{Mean of } \bar{x} = \mu = 48 \text{ minutes} \]
Standard deviation of \( \bar{x} \):
\[ \text{Standard deviation of } \bar{x} = \frac{\sigma}{\sqrt{n}} = \frac{20}{\sqrt{17}} \]
Z-scores for 45 minutes and 50 minutes:
\[ Z_{45} = \frac{45 - 48}{\frac{20}{\sqrt{17}}} \approx -0.67 \]
\[ Z_{50} = \frac{50 - 48}{\frac{20}{\sqrt{17}}} \approx 0.67 \]
Using a Z-table, we find:
\[ P(Z_{45} < Z < Z_{50}) \approx P(-0.67 < Z < 0.67) \]
\[ \approx P(Z < 0.67) - P(Z < -0.67) \]
\[ \approx 0.7486 - (1 - 0.7486) \]
\[ \approx 0.7486 - 0.2514 \]
\[ \approx 0.4972 \]
So, the probability that the average time studying for the 17 students is between 45 and 50 minutes is approximately 0.4972.
b. Z-scores for one randomly selected student:
\[ Z_{45} = \frac{45 - 48}{20} = -0.15 \]
\[ Z_{50} = \frac{50 - 48}{20} = 0.1 \]
Using a Z-table, we find:
\[ P(Z_{45} < Z < Z_{50}) \approx P(-0.15 < Z < 0.1) \]
\[ \approx P(Z < 0.1) - P(Z < -0.15) \]
\[ \approx 0.5398 - (1 - 0.5398) \]
\[ \approx 0.5398 - 0.4602 \]
\[ \approx 0.0796 \]
So, the probability that one randomly selected student's time will be between 45 and 50 minutes is approximately 0.0796.
c. Mean of the total time:
\[ \mu_{\text{total}} = n \times \mu = 17 \times 48 = 816 \text{ minutes} \]
Standard deviation of the total time:
\[ \sigma_{\text{total}} = \sqrt{n} \times \sigma = \sqrt{17} \times 20 \approx 65.53 \text{ minutes} \]
Z-score for 700 minutes:
\[ Z_{700} = \frac{700 - 816}{65.53} \approx -1.77 \]
Using a Z-table, we find:
\[ P(Z < -1.77) \approx 0.0384 \]
So, the probability that the 17 randomly selected students will have a total study time less than 700 minutes is approximately 0.0384.
d. To find the least total time for a group to receive a sticker, we'll find the 80th percentile of the distribution of the total time, which corresponds to a Z-score.
Using a Z-table, we find the Z-score corresponding to the 80th percentile is approximately 0.85.
Then, we use the Z-score formula to find the total time:
\[ Z = \frac{x - \mu_{\text{total}}}{\sigma_{\text{total}}} \]
\[ 0.85 = \frac{x - 816}{65.53} \]
Solving for \( x \), we get:
\[ x \approx 816 + 0.85 \times 65.53 \]
\[ x \approx 816 + 55.7 \]
\[ x \approx 871.7 \]
So, the least total time that a group can study and still receive a sticker is approximately 871.7 minutes.
\[ \text{Mean of } \bar{x} = \mu = 48 \text{ minutes} \]
Standard deviation of \( \bar{x} \):
\[ \text{Standard deviation of } \bar{x} = \frac{\sigma}{\sqrt{n}} = \frac{20}{\sqrt{17}} \]
Z-scores for 45 minutes and 50 minutes:
\[ Z_{45} = \frac{45 - 48}{\frac{20}{\sqrt{17}}} \approx -0.67 \]
\[ Z_{50} = \frac{50 - 48}{\frac{20}{\sqrt{17}}} \approx 0.67 \]
Using a Z-table, we find:
\[ P(Z_{45} < Z < Z_{50}) \approx P(-0.67 < Z < 0.67) \]
\[ \approx P(Z < 0.67) - P(Z < -0.67) \]
\[ \approx 0.7486 - (1 - 0.7486) \]
\[ \approx 0.7486 - 0.2514 \]
\[ \approx 0.4972 \]
So, the probability that the average time studying for the 17 students is between 45 and 50 minutes is approximately 0.4972.
b. Z-scores for one randomly selected student:
\[ Z_{45} = \frac{45 - 48}{20} = -0.15 \]
\[ Z_{50} = \frac{50 - 48}{20} = 0.1 \]
Using a Z-table, we find:
\[ P(Z_{45} < Z < Z_{50}) \approx P(-0.15 < Z < 0.1) \]
\[ \approx P(Z < 0.1) - P(Z < -0.15) \]
\[ \approx 0.5398 - (1 - 0.5398) \]
\[ \approx 0.5398 - 0.4602 \]
\[ \approx 0.0796 \]
So, the probability that one randomly selected student's time will be between 45 and 50 minutes is approximately 0.0796.
c. Mean of the total time:
\[ \mu_{\text{total}} = n \times \mu = 17 \times 48 = 816 \text{ minutes} \]
Standard deviation of the total time:
\[ \sigma_{\text{total}} = \sqrt{n} \times \sigma = \sqrt{17} \times 20 \approx 65.53 \text{ minutes} \]
Z-score for 700 minutes:
\[ Z_{700} = \frac{700 - 816}{65.53} \approx -1.77 \]
Using a Z-table, we find:
\[ P(Z < -1.77) \approx 0.0384 \]
So, the probability that the 17 randomly selected students will have a total study time less than 700 minutes is approximately 0.0384.
d. To find the least total time for a group to receive a sticker, we'll find the 80th percentile of the distribution of the total time, which corresponds to a Z-score.
Using a Z-table, we find the Z-score corresponding to the 80th percentile is approximately 0.85.
Then, we use the Z-score formula to find the total time:
\[ Z = \frac{x - \mu_{\text{total}}}{\sigma_{\text{total}}} \]
\[ 0.85 = \frac{x - 816}{65.53} \]
Solving for \( x \), we get:
\[ x \approx 816 + 0.85 \times 65.53 \]
\[ x \approx 816 + 55.7 \]
\[ x \approx 871.7 \]
So, the least total time that a group can study and still receive a sticker is approximately 871.7 minutes.
