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Mean comparison

I have a poor understanding of this subject. Please assist me with the attachment and explain your steps so that I can try to understand this area. Please use the traditional method becasue I can follow it a little better. Thank you for your help.

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2.)
<br><br>For test#1:
<br><br>mean = 100; sd = 25; score = 130
<br><br>Hence,
<br><br>(score-mean)/sd = (130-100)/25 = 30/25 = 6/5 = 1.2
<br><br>
<br><br>For test#2:
<br><br>mean = 40; sd = 5; score = 52
<br><br>Hence,
<br><br>(score-mean)/sd = (52-40)/5 = 12/5 = 2.4 > 1.2
<br><br>
<br><br>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)
<br><br>
<br><br>
<br><br>3.)
<br><br>Sample space ={(BBB), (GBB), (BBG), (BGB), (BGG), (GBG), (GGB), (GGG)}
<br><br>where G stand for girl and B stands for boy.
<br><br>
<br><br>Hence, sample size = 8 --Answer
<br><br>
<br><br>Probability of at least two girls P= n(2 girls or 3 girls)/n(total)
<br><br>2 or 3 girls = {(BGG), (GBG), (GGB), (GGG)}
<br><br>
<br><br>=> P = 4/8 = 1/2 --Answer (B)
<br><br>
<br><br>
<br><br>4.)
<br><br>P(Aisle or smoking)
<br><br>= [n(Aisle) + n(smoking) - n(aisle and smoking)]/n(total)
<br><br>
<br><br>n(aisle) = 15 + 80 = 95
<br><br>n(smoking) = 30
<br><br>n(smoking and aisle) = 15
<br><br>n(total) = 230
<br><br>Hence,
<br><br>P(Aisle or smoking) = (95 + 30 - 15)/230
<br><br>
<br><br>=> P = 0.478 --Answer (D)
<br><br>
<br><br>
<br><br>5.)
<br><br>probability of a eligible voter to vote = p = 48/100 = 0.48
<br><br>
<br><br>Hence, probability that all 4 eligible voters vote
<br><br>P = p^4 = (0.48)^4
<br><br>
<br><br>=> P = 0.0531 --Answer (A)
<br><br>
<br><br>
<br><br>6.)
<br><br>Number of possible exams each comprised of 10 questions out of 20
<br><br>n = 20C10 = 20!/(10!*10!)
<br><br>=> n = 20*19*18*17*16*15*14*13*12*11/(10*9*8*7*6*5*4*3*2*1)
<br><br>
<br><br>=> n = 184756 --Answer
<br><br>
<br><br>
<br><br>7.)
<br><br>Let us take class size = ...

Solution Summary

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

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