Question

1. Imagine you have collected data on 1500 english students related to depression levels, on a 100-point scale. Your distribution of data is normal, with a mean of 52, a standard deviation of 7.
   1. What score is at the 95th percentile?
   2. how many students would have deviant depression levels (using the 5% criterion)

1. You are teaching a large section of CHE 100, with 117 students. Your class is part of a larger course enrollment totaling 3000 students. For the whole course (all 3000 students), the first exam average is 75, with a standard deviation of 7.
   1. what is the range of scores that encompases the middle 50% of the class?
   2. What are the deviant scores on both tails of the distribution?
   3. How many of your 117 students would you expect to earn a B (a grade of 80-89)?

Answer 1

Question 1:

1. What score is at the 95th percentile?

z = 1.645

1.645 = (x - 52)/7

x = 1.645*7 + 52 = 63.515

1. how many students would have deviant depression levels (using the 5% criterion)

0.05*1500 = 75

Question 2:

1. what is the range of scores that encompases the middle 50% of the class?

-0.67 = (x - 75)/7

x = -0.67*7 + 75 = 70.31

0.67 = (x - 75)/7

x = 0.67*7 + 75 =79.69

1. What are the deviant scores on both tails of the distribution?

-1.645 = (x - 75)/7

x = -1.645*7 + 75 = 63.485

1.645 = (x - 75)/7

x = 1.645*7 + 75 = 86.515

1. How many of your 117 students would you expect to earn a B (a grade of 80-89)?

z = (80 - 75)/7 = 0.71

P (z = 0.71) = 0.7625

z = (89 - 75)/7 = 2

P (z = 2) = 0.9772

Required probability = 0.9772 - 0.7625 = 0.2148

n = 0.2148*117 = 26

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