4. A professor has records of the scores of all students that have taken one of

Question

4. A professor has records of the scores of all students that have taken one of his courses over the years. The average score is 70 and the corresponding population variance 82. The dataset of these scores can be approximated by the Logistic distribution with parameters = 70 and = 8 (Ex. 6 in HW 5). You can assume the total number of students that have ever taken this class is very large. (a) What is the (smallest) score of a student in the top 10% of this course? (b) If a student that has taken this class is selected at random, what is the probability that his/her score is between 65 and 75? Suppose that the professor samples at random from the population of students who have taken this class until she finds 2 students whose scores were between 65 and 7t5. (c) What is the probability that she won't have to sample more than 3 students? (d) If sampling hasn't been terminated after 3 students have been selected, what is the prob- ability that she won't have to sample more than 3 additional students?

Explanation / Answer

Let X be the random variable that score of student.

Here X ~ N(mu = 70, sigma = 8)

Now we have to find score of a student in the top 10% of this course.

So we will find score using formula,

x = mu + z*sigma

where z is z-score for probability = 1 - 0.1 = 0.9

z-score we can find in excel.

syntax :

=NORMSINV(probability)

where probability = 0.9

z = 1.28

x = 70 + (1.28 * 8) = 80.25

b) Now we have to find P(65 < X < 75).

Convert x = 65 and x = 75 into z-score.

z-score is,

z = (x - mu) / sigma

z-scores for x = 65 and x = 75 are,

z = (65 - 70) / 8 = -0.63

z = (75 - 70) / 8 = 0.63

Now we have to find P(-0.63 < Z < 0.63)

P(-0.63 < Z < 0.63) = P(Z < 0.63) - P(Z < -0.63)

This probability we can find in excel.

syntax :

=NORMSDIST(z)

where z is z-score.

P(-0.63 < Z < 0.63) = 0.7340 - 0.2660 = 0.4680

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