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Mean comparison

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Solution Summary

Two tests were given, the tests were designed with different scales. The solution discusses the mean comparison.

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2.)
For test#1:
mean = 100; sd = 25; score = 130
Hence,
(score-mean)/sd = (130-100)/25 = 30/25 = 6/5 = 1.2

For test#2:
mean = 40; sd = 5; score = 52
Hence,
(score-mean)/sd = (52-40)/5 = 12/5 = 2.4 > 1.2

Because, in 1st case score is closer to the mean in comparison to second score, therfore 130 on test#1 is better score --Answer (A)

3.)
Sample space ={(BBB), (GBB), (BBG), (BGB), (BGG), (GBG), (GGB), (GGG)}
where G stand for girl and B stands for boy.

Hence, sample size = 8 --Answer

Probability of at least two girls P= n(2 girls or 3 girls)/n(total)
2 or 3 girls = {(BGG), (GBG), (GGB), (GGG)}

=> P = 4/8 = 1/2 --Answer (B)

4.)
P(Aisle or smoking)
= [n(Aisle) + n(smoking) - n(aisle and smoking)]/n(total)

n(aisle) = 15 + 80 = 95
n(smoking) = 30
n(smoking and aisle) = 15
n(total) = 230
Hence,
P(Aisle or smoking) = (95 + 30 - 15)/230

=> P = 0.478 --Answer (D)

5.)
probability of a eligible voter to vote = p = 48/100 = 0.48

Hence, probability that all 4 eligible voters vote
P = p^4 = (0.48)^4

=> P = 0.0531 --Answer (A)

6.)
Number of possible exams each comprised of 10 questions out of 20
n = 20C10 = 20!/(10!*10!)
=> n = 20*19*18*17*16*15*14*13*12*11/(10*9*8*7*6*5*4*3*2*1)

7.)
Let us take class size = ...

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Education
• MSc , Pune University, India
• PhD (IP), Pune University, India
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Measures of Central Tendency

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