Scores on an exam are assumed to be normally distributed with mean 78 and standa

Question

Scores on an exam are assumed to be normally distributed with mean 78 and standard deviation 5. (a) Suppose that students scoring in the top 10% of this distribution are to receive an A (b) What must be the cutoff point for passing the exam if the examiner wants only the top (c) What proportion of students have scores 5 or more points above the scores that cuts (d) If it is known that a student's score exceeds 72, what is the probability that his or her grade. What is the minimum score a student must achieve to earn an A grade? 28.1% of all scores to be passing? off the lowest 20%? score exceeds 84?

Explanation / Answer

Normal distributions params are:

Mean = 78

Stdev = 5

a, Top 10% is A. Therefore,

P(X<=c) = 1- P(X>c) = 1-.10 = .9

So, Z = 1.28

(c-78)/5 = 1.28

c = 1.28*5 + 78 = 84.4

Min score for A is 84.4

b. Top 28.1% cutoff is c

So, P(X>c) = .281

P(X<=c) = 1-P(X>c) = 1-.281 = .719

(c-78)/5 = .58

c = 5*.58 + 78 = 80.9

c. Highest score in lowest 20% will be ?

P(X<=c) = .20

(c-78)/5 = ..84

c = 5*.84 + 78 = 82.2

So, a score of 82.2 +5 is 87.2

Therefore, %age of people with 5 or more points above is : P(X>=87.2) = P(Z> 87.2-78 / 5) = P(Z>1.84) = 1-.9671 = .0329

So, 3.29% of people are 5 or above points above scores that cuts off the lowest 20%

d.

P(X>84 | X>72)

= P(Z> 84-78 / 5 | 72-78 / 5)

= P( Z>1.2 | Z> -1.2)

= .1151/.8849

= .1301

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