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Two Statistic Questions (Not too hard)

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60 bags of soup mix were weight. They found the mean of the 60 bags to be 27 grams. Suddenly they discovered that the first 20 bags were weight incorrectly and they needed to add 3 grams to the first 20 bags. Find the new mean of the 60 bags.

In a class of 35 students there were tests taken at different times. The entire class mean was 8.1. The mean grade of the first 15 students was 8.5. What was the mean of the other 20 students.



If you could explain this to me step by step I would appreciate it greatly. Thanks fellow nerds .
 Dec 16, 2012

2+0 Answers

 #1
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will solve in 2 (separate) replies.

prob. #1:

mean (original) = mo = (w1+w2+...+w59+w60)/60

mean (adjusted) = ma = [(w1+3)+(w2+3)+...(w20+3)+w21+...+w60]/60

using algebra to rearrange terms and using the properties of fractions

ma = [(w1+w2+...w59+w60)+(3+3+...+3)]/60 = [(w1+w2+...w59+w60)/60]+[(20*3)/60]

the first term is the original mean and has value 27g, the second term is a constant with value 1g

so the answer is 28g
 Dec 16, 2012
 #2
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prob #2

mean = sum/N or sum=N*mean
consider the two groups of students applying the latter formula:
group1: 8.5 = sum/15 or sum=8.5*15=127.5
group2: sum = 20*mean (and this mean is the unknown value that we are looking for)

the mean for the entire class is 8.1 and that is gotten by adding up all the values (sum of everybodies' scores) and dividng by 60:

8.1 = (sum group1 + sum group2)/35 or (127.5+20*mean)/35=8.1

using algebra, solve this equation for "mean"

mean = [(8.1*35)-127.5]/20 = 7.8
 Dec 16, 2012

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