Is the standard deviation of ages of students in class A greater than

OA is C

My math concepts around statistics and standard deviation are weak. Can you share some resources for improvement?

If More detail is better! thanks!

Thanks you, you force me to think more carefully


fast (guessing) way: the more is difference between values, the more SD. In other words SD corresponds to dispersion of data. Taking both condition, we can see that dispersion of students of A class is obviously less than that of B class. So, C

usual way:

\(SD=\sqrt{\frac{\sum{(x-x_{av})^2}}{n}}\)

1) first condition says that for class A \(|x_j-x_i|>12\)

Additionally, we can states that minimum SD is (when \(x_{av}\) is evenly between \(x_i\) and \(x_j\))

\(SD_{Amin}>\sqrt{\frac{({x_j}-x_{av})^2+({x_i}-x_{av})^2}{2}}=\sqrt{\frac{6^2+6^2}{2}}=6\)

\(SD_{Amin}>6\)

2) second condition says that for class B \(|x_j-x_i|<=6\)

Additionally, we can states that maximum SD is (when \(x_{av}\) is close to one of \(x_i\) or \(x_j\))

\(SD_{Bmax}<\sqrt{\frac{({x_j}-x_{av})^2+({x_i}-x_{av})^2}{2}}=\sqrt{\frac{6^2+0^2}{2}}=\frac{6}{\sqrt{2}}\)

\(SD_{Bmin}<\frac{6}{\sqrt{2}}\)

1)&2) Combine two conditions:

\(SD_{A}=>SD_{Amin}>6>\frac{6}{\sqrt{2}}>SD_{Bmin}>=SD_{B}\)

\(SD_{A}>SD_{B}\)

Walker, one more: How do you come from "statement1: the difference between the ages of any two students in class A is always more than 1 year" to : , I mean |xj-xi|>12


One more: I chose E because I follow one rule from Gmatclub that if we dont know exactly specific each age, we can have no conclusion about SD!

We have to choose the same units for age to make SD compatible. I choose months and, therefore, |xj-xi|>12



Think about SD as an average difference between an average value and other data. The problem directly says about differences.

I think GMAT is far far away from statistics. For many people who did not study statistics SD seems to be a mysterious feature from very difficult mathematical jungle of statistics. But GMAT does not go so deeply. Think about SD as an average deviation of data from an average value and remember the formula.

(1) The difference between the ages of any two students in class A is always more than 1 year.
INSUFFICIENT. No mention about class B here.

(2) No student in class B is more than 6 months older than any other student.
INSUFFICIENT. No mention about class A here.

Combing both and assuming that both classes have at least 2 students;
We can see that the standard deviation will increase with every student added for class A and the the deviation will decrease with every student added to class B.

Maximum standard deviation for class B can be somewhere around 0.25 years or 3 months and the minimum standard deviation of class B will be somewhere around 0.5 years.

Thus Std Dev(A) > Std Dev(B)

Sufficient.

Ans: "C"

C.

Statement 1 alone is not enough. It doesn't give any indication what the differences of ages in class B are.
For example, everyone in class B could be exactly the same age, or everyone in class B could be 2 years apart.

Statement 2 alone is not enough. It doesn't give any indication what the difference of ages in class A are.
Everyone in class A could be exactly the same age, or everyone in class A could be 2 years apart.

If you combine them, you know that A has a significantly larger standard deviation in ages.

A - a1, a2, a3,a4,a5,a6

Total A = 90

a1-a2 > 12 months and so on, but no information about B, so insufficient.

B - b1,b2,b3,b4,b5... b12

Total B = 192

b1-b2 <= 6 month , but no information a bout A, so insufficient


From (1) and (2) together, the difference between elements of set A is > the difference between elements of Set B, and the denominator in A is < the denominator in set B for the Std Dev formula.

=> Std Dev A > Std Dev B

So answer - C.

standard deviation is the underroot of the (sum of squares of the difference b/w the quantity and the mean divided by the no of quantities).
now since the statement 1 doesnt mentions abt B and stat. 2 doesnt mentions abt A. hence they both are insufficient.

on combining the two statements do we get the full information to make a comparison?
YES.

also keep in mind that the diff b/w the mean and the quantity in B wud always be less than 1 and therefore its square wud be much more smaller.

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

Statement 1: No mention of students in class B. Thus INSUFFICIENT

Statement 2: No mention of students in class A. Thus INSUFFICIENT

Combining the two statements, we can see that the difference between the ages of two students in class A is considerably larger than the difference between the ages of any two students of class B.
As we know that SD is a measure of the compactness within the elements of a set, we can infer that the elements of Set A are more dispersed than are the elements of Set B. Thus, we know that SD (A) > SD (B). SUFFICIENT.

Answer: C

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