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# Class A B Average Age 15 6 No. of Students 16 12 Is the

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Class A B Average Age 15 6 No. of Students 16 12 Is the [#permalink]  09 Jan 2006, 22:09
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Class A B
Average Age 15 6

No. of Students 16 12

Is the standard deviation of ages of students in class A greater than the Standard Deviation of the age of students in class B?

1. The difference between the ages of any two students in class A is always more than 1

2. No Student in class B is more than 6 months older than any other student
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C? Not sure if this is correct

Q: Is SD (A) > SD (B)?

1) Difference between ages of any two is always > 1. To get the smallest possible SD for A, we can assume:
8 students have age = 14.5
8 students have age = 15.5 (which still makes mean = 15)

Smallest possible SD(A)^2 = [16* (deviation of each number from mean)^2]/16
= 12 * 0.5^2/16 = 0.5^2
=>Smallest possible SD(A) = 0.5

No information on ages of ClassB. INSUFF.

2) Difference between ages of any two is always < 0.5 (6 months). To get the largest possible SD we can assume:
6 students have age = 11.75
6 students have age = 12.25 (mean is still 12)

=> The largest possible SD(B) = 0.25

Combining 1 & 2:
Since the smallest possible SD(A) is greater than the largest possible SD(B) it must be true that
Hence SD(A) > SD(B)
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The problem doesn't give us the distribution of numbers, we need more info.

I. No info about B..INSUFF. All we know is that diff. b/w ANY two students is greater than 1. Thus, it has to be some sort of progression.

2. No info about A..INSUFF. The diff. b/w the ages is less than 6 months..so the numbers are tightly packed. SD will be low.

From I and II, SD(A) > SD(B)...C would be my answer.
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1 more for C as explained earlier. We need both
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1. Class A has spread (variance) of aprrox > or = 15*1
no info of students in calss B.

2. Class B has spread < or equalt to 12* 0.6

no info of spread in A

Case 1 and 2

Therere SD(A) > SD(B)

Hence C.
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