Suppose the exam scores on the economics exam are approximately normally distributed.If 15% of the students score below 65 and 10% above 90, then what fraction of the studentsscore (a) below 80 (b) Between 70 and 95?

To answer this question, we need to understand the properties of a normal distribution. In a normal distribution, about 68% of values fall within 1 standard deviation of the mean, about 95% fall within 2 standard deviations, and about 99.7% fall within 3 standard deviations.

Given that 15% of students score below 65 and 10% score above 90, we can infer that these scores represent the 15th and 90th percentiles respectively. In a standard normal distribution, the 15th percentile corresponds to a z-score of approximately -1, and the 90th percentile corresponds to a z-score of approximately 1.28.

(a) To find the fraction of students who score below 80, we need to find the z-score for 80. We can use the formula:

Z = (X - μ) / σ

Where X is the score, μ is the mean, and σ is the standard deviation. However, we don't have the values for the mean and standard deviation. We can only estimate that the z-score for 80 is likely between -1 and 1.28, so more than 15% but less than 90% of students likely score below 80.

(b) To find the fraction of students who score between 70 and 95, we would need to find the z-scores for both 70 and 95 and look up these z-scores in a standard normal distribution table. Again, without the mean and standard deviation, we can only estimate that this fraction is less than 90% (since 10% score above 90) but more than 15% (since 15% score below 65).

To get exact fractions, we would need the mean and standard deviation of the scores.