Question

Suppose we have assigned grades for the 11 students in our data: Grade A for students who scored≥ 90; B for students who scored≥ 80 and < 90; C for students who scored ≥ 70and < 80; D for students who scored≥ 60and < 70; F for students who scored< 60.Following the above grade scheme, we observe that we have 8 students who received grade A, 1 student received grade B, 0 students received grade C, 1 student received grade D and 1 student received grade F. Using this, please answer the following questions: Data: 105, 91, 52, 86, 100, 96, 98, 109, 96, 106, 67 Q33. What is the probability that a student received grade A and grade B? Q34.What is the probability that a student received grade A, i.e., P(A) is Q35.What is the probability that a student received grade B, i.e., P(B)is Q36.What is the probability that a student received grade C, i.e., P(C) is Q37.What is the probability that a student received grade D, i.e., P(D) is Q38.What is the probability that a student received grade F, i.e., P(F) is Q39.What is the expected value of these grades? Q40.What is the variance of these grades?

Answer 1

grade	number of students
A	8
B	1
C	0
D	1
F	1
total	11

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33)

total students =11

total students = 9/11

P(A and B) = 9/11

34)

P(A) = 8/11

35)

P(B) = 1/11

36)

P(C) = 0

37)

P(F) = 1/11

38)

P(F) = 1/11

39) and 40)

grade A ≥90

so,

X	(X - X̄)²
105	23.766
91	83.266
100	0.016
96	17.016
98	4.516
109	78.766
96	17.016
106	34.5

	X	(X - X̄)²
total sum	801	258.875
n	8	8

mean of grade A = ΣX/n =    100.125

variance of grade A = Σ(X - X̄)²/(n-1)=   36.9821

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grade B who scored≥ 80 and < 90 i.e 86

mean = 86

variance=0

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for grade C

mean=0

variance = 0

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for grade D ,

mean=67

varince=0

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for grade F

mean=52

variance = 0