The average grade for an exam is 84, and the standard deviation is 4. The possib

Question

The average grade for an exam is 84, and the standard deviation is 4. The possible grades are 'A','B', 'C', ‘D', ‘F'. a. If 92% of the class is given at least 'C' grade or better and the grades are curved to follow a normal distribution, what is the lowest possible 'C' and the highest possible ‘D'? b. If the grades are curved to follow a normal distribution and the grades of 80% students in the class range is (84 - c, 84 + c), find the value of c The grades are curved to follow a normal distribution and any student whose score is less than 55 is given F' grade. 20 students are randomly selected from the class. What is probability that at most 2 of them have 'F' grade:? c.

Explanation / Answer

(a) Here we assume he normal distribution is curved where average merks = 84 and standard deviation = 4

so IF lowest possible 'C' and highest possible 'D' is lets say k

then

Pr(x > k) = 0.92

1 - Pr(x < k) = 0.92

Pr(x < k) = 0.08

Pr(x < k ; 84 ; 4) = 0.08

Z - value = -1.4051

(k - 84)/4 = -1.4051

k = 84 - 4 * 1.4051 = 78.38

so the value is 78.38

(b) Here as the distribution is normal, os

Pr(84 -c < x < 84 + c) = 0.80

Pr(x < 84 + c) - Pr(x < 84 - c) = 0.80

Here as c is equal both sides then

Pr(x < 84 + c) = 0..50 + 0.80/2 = 0.90

so Here z for p - value 0.90 is

Z = 1.282

so,

(84 + c - 84)/4 = 1.282

c = 4 * 1.282 = 5.13

(C) Pr(x < 55 ; 84 ; 4)

Z = (55 - 84)/4 = -7.25

Pr(X < 55 84; 4) = 2.083 x 10-13

Now n = 20 and p =  2.083 x 10-13 and y is the number of people out of 20 who get the 'F' grade

Pr(y <= 2 ; 20 ;  2.083 x 10-13 ) = BIN (x = 0,1,2 ; 20 ;  2.083 x 10-13 ) = 1

as there is full probability of that event to occur.

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