Normal Distribution Study Dc3999

1. **Problem Statement:** We have a normal distribution for study times with mean $$\mu = 2068$$ and standard deviation $$\sigma = 126$$. We want to find: - The probability a single student's time is between 45 and 47 minutes. - The probability the average time of 44 students is between 45 and 47 minutes. - The probability the total study time of 44 students exceeds 1936 minutes. - Whether the normality assumption is necessary for parts e) and f). - The minimum total time for the top 10% of groups of 44 students. --- 2. **Distribution and Formulas:** - Single student's time: $$X \sim N(2068, 126^2)$$ - For sample size $$n=44$$, the sample mean $$\bar{X}$$ has distribution $$N\left(2068, \frac{126^2}{44}\right)$$ - Total time for 44 students: $$T = 44 \times \bar{X} \sim N(44 \times 2068, 44 \times 126^2) = N(90992, 699936)$$ Use the standard normal variable: $$Z = \frac{X - \mu}{\sigma}$$ --- 3. **Probability for a single student between 45 and 47 minutes:** Calculate $$P(45 < X < 47)$$. Convert to $$Z$$: $$Z_1 = \frac{45 - 2068}{126} = \frac{45 - 2068}{126} = -16.22$$ $$Z_2 = \frac{47 - 2068}{126} = -16.21$$ These values are extremely low, so $$P(45 < X < 47) \approx 0$$. --- 4. **Probability average time of 44 students between 45 and 47 minutes:** Sample mean $$\bar{X} \sim N(2068, \frac{126^2}{44})$$. Standard deviation of $$\bar{X}$$: $$\sigma_{\bar{X}} = \frac{126}{\sqrt{44}} \approx 19.01$$ Calculate $$P(45 < \bar{X} < 47)$$: $$Z_1 = \frac{45 - 2068}{19.01} = -107.9$$ $$Z_2 = \frac{47 - 2068}{19.01} = -107.8$$ Again, probability is essentially 0. --- 5. **Probability total time of 44 students more than 1936 minutes:** Total time $$T \sim N(90992, 699936)$$. Calculate $$P(T > 1936)$$: $$Z = \frac{1936 - 90992}{\sqrt{699936}} = \frac{1936 - 90992}{836.5} = -106.5$$ This is almost 1, so $$P(T > 1936) \approx 1$$. --- 6. **Is normality assumption necessary for parts e) and f)?** Yes, because we use normal distribution properties to calculate probabilities and percentiles. --- 7. **Top 10% total study time cutoff:** Find $$x$$ such that $$P(T > x) = 0.10$$. Find $$z$$ for 90th percentile: $$z = 1.28155$$ Calculate $$x$$: $$x = \mu_T + z \sigma_T = 90992 + 1.28155 \times 836.5 = 90992 + 1071.5 = 92063.5$$ So the least total time for top 10% is approximately $$92063.5$$ minutes. --- **Final answers:** - Probability single student between 45 and 47 minutes: approximately 0. - Probability average of 44 students between 45 and 47 minutes: approximately 0. - Probability total time of 44 students more than 1936 minutes: approximately 1. - Normality assumption is necessary. - Least total time for top 10%: approximately $$92063.5$$ minutes.