The total scores on the Medical College Admission Test (MCAT) follow a Normal distribution with mean 24.4 and standard deviation 6.2 . What are the median and the first and third quartiles of the MCA

For the first question, we can use the formula for converting a normal distribution to a standard normal distribution: z = (x - mean) / standard deviation Using the values given, we get: z = (x - 24.4) / 6.2 To find the quartiles and median, we need to find the corresponding z-scores. We can use a standard normal distribution table or a calculator to find these values. Q1 is the 25th percentile, which corresponds to a z-score of -0.674. Median is the 50th percentile, which corresponds to a z-score of 0. Q3 is the 75th percentile, which corresponds to a z-score of 0.674. To convert these z-scores back to MCAT scores, we use the formula: x = z * standard deviation + mean Substituting the values, we get: Q1 ? -0.674 * 6.2 + 24.4 ? 20.65 Median ? 0 * 6.2 + 24.4 ? 24.4 Q3 ? 0.674 * 6.2 + 24.4 ? 28.15 Therefore, the first quartile is \approx imately 20.65, the median is \approx imately 24.4, and the third quartile is \approx imately 28.15. For the second question, we want to find the range of values that 95% of all pregnancies fall within. Since the distribution is normal, we can use the empirical rule, which states that \approx imately 95% of observations lie within 2 standard deviations of the mean. So, we have: Lower limit = mean - 2 * standard deviation = 266 - 2 * 13 = 240 Upper limit = mean + 2 * standard deviation = 266 + 2 * 13 = 292 Therefore, 95% of all pregnancies last between 240 and 292 days. For the third question, we need to use a standard normal distribution table or a calculator to find the proportions of observations that fall within each region. (a) z ? -2.34: This corresponds to \approx imately 0.0099 or 0.99% of observations. (b) z ? -2.34: This corresponds to \approx imately 0.9901 or 99.01% of observations. (c) z > 1.74: This corresponds to \approx imately 0.0401 or 4.01% of observations. (d) -2.34 < z < 0: This corresponds to the area to the left of z = 0.674 (which is the absolute value of -2.34) minus the area to the left of z = -2.34. Using a standard normal distribution table or a calculator, we get: Area to the left of z = 0.674: 0.7495 Area to the left of z = -2.34: 0.0099 Therefore, the area between -2.34 and 0 is \approx imately 0.7495 - 0.0099 ? 0.7396 or 73.96% of observations.