PS questions about standard deviation.

Are you talking seriously???

1. [1,5,1,1] and 2. [1,5,5,5] have SD of 2.
3. [1,5,3,3] has a SD of 1.63.
4. [1,5,1,3] and 6. [1,5,3,5] have 1.91.
5. [1,5,1,5] has a SD of 2.31.

So there are altogather 4 different SDs.

No typo there: the question is from GmatPrep but basically it's the same as you wrote.

A score of 58 was 2 standard deviations below the mean --> 58=Mean-2*sd
A score of 98 was 3 standard deviations above the mean --> 98=Mean+3*sd

Subtract (1) from (2) --> 98-58=Mean+3*sd-(Mean-2*sd) --> 40=5sd --> sd=8 --> Mean=58+2*sd=58+16=74.


Hope it's clear.

Interesting Questions on Mean,Median and Mode.
(Plucked from Dr.Math) I believe these type of conceptual questions would help us all. Very fundamental but somewhat tricky for a person like me. If this is not the right place to include these type of questions or that they do not qualify the post take action accordingly. Thanks.

Q1) Find a set of five data values with modes 0 and 2, median 2, and
mean 2.

Soln:
Mode = values appearing most often,
Median = value with as many other values above as below,
Mean = average (sum of the values divided by the number of values).

There are five values. If all were different, there would be five modes, but there are only two. The two modes must appear at least twice. They cannot appear three times each, because then you would have at least six values, not five. Thus four of the values must be 0, 0, 2, and 2.

For 2 to be the median, the remaining value, call it x, must be greater than 2. If 0 < x < 2, then x would be the median, and if
x < 0, 0 would be the median.

Then the mean is (0+0+2+2+x)/5 = 2, which you can solve for x.

Q2) We are told that the median of five numbers is 5, the mode is
1, and the mean is 4. Find the five numbers?

Soln: We can start by drawing a blank for each of the values. Then we'll try to fill them in, putting them in ascending order as we go. Here are the five blanks:

__ __ __ __ __

Now, what do we know about the numbers? We know that "the median of five numbers is 5." The median is the middle number when arranged in ascending order, so let's put it there:

__ __ 5 __ __

What else do we know? We're told that "the mode is 1." That means that 1 has to appear more than any other number. Since 1 is less than 5, all 1's will have to go to the left of the 5. We know we need at least two of them (otherwise, 5 would be a mode as well), so both blanks on the left will have to be 1's. Putting them in, we have:

1 1 5 __ __

Now the last clue is, "the mean is 4." The mean is the "average" of the numbers. It is computed by taking the sum of the numbers and dividing it by the number of numbers. Algebraically, if we call our values A, B, C, D and E, we'd write:

M = (A+B+C+D+E) / 5

Since we already know the first three numbers, let's plug them in for A, B, and C. We also know that the mean is 4, so we'll plug that in too:

4 = (1+1+5+D+E) / 5

4 = (7+D+E) / 5

Now let's solve for D+E, the two numbers we don't know:

4 = (7+D+E) / 5

4 * 5 = (7+D+E)

20 = 7+D+E

20 - 7 = D+E

D+E = 13

So the sum of the last two numbers must be 13. We also know that each of those two numbers must be greater than 5. What two numbers will work? Only 6 and 7 (can you think of WHY only 6 and 7 work?) So our five numbers must be:

1 1 5 6 7

To check; median = 5 (check), mode = 1 (check), mean is:

M = (1+1+5+6+7) / 5

= 20 / 5

= 4 (check)

let me explain, it is 0,s-t,s-r, thus, ... if say we multiply this set by -....0,t-s,r-s....now add s in all the terms <adding/subtracting same number in the set does not change the SD>....the set becomes <s,t,r>...

for the third one, the set may |r|,|s|,|t|..now since...now since SD=sqrt {summation(x-mean)2}/n...u can see that SD will differ as x will differ...

I hope it is clear...

Pavan

I hope it clear.

We have been given
Set A - Elements with Median > Mean.
Set B - Elements with Median = Mean.

I. Z > N : We cannot compare standard deviations of two alien sets - False. [ we don't know what is A, what is B and what is the relationship between them] - So I Out.
II. R > M : We are comparing mean of the combined sets with mean of set B. We have no knowledge of number of elements (cardinality) of A or B. Consider this as a case of weighted average. Unless we know the number of elements of A, number of elements of B and mean values of Both the sets, we can not evaluate this.
For Instance, If A has 4 elements and mean of 5, B has 10 elements and mean of 8 .Then M =8, R = (4*5 + 10*8)/(4+10) = 100/14 = approx 7. So R < M
now, suppose Set A has a mean of 20 instead of 5. M remains unchanged. R = (4*20 + 10*8)/(4+10) = 160/14 = approx. 11. So R > M
So II is Out.
III. Q > R : Mean > Median. Again as above, since we do not know anything about composition of A and B. This might be true or false depending on distribution of A and B.
As there is no co-relationship provided between A and B, we cannot derive any conclusion.

So the answer in E. None.

The problem could be twisted, if instead of L-M=0, we would be given N=0.
In that case, Z is always >0. So answer would be A then.

7. Which of the following data sets has the third largest standard deviation?
(A) {1, 2, 3, 4, 5}
(B) {2, 3, 3, 3, 4}
(C) {2, 2, 2, 4, 5}
(D) {0, 2, 3, 4, 6}
(E) {-1, 1, 3, 5, 7}


(A) standard deviation = 1.58113883
(B) standard deviation = 0.707106781
(C) standard deviation = 1.414213562
(D) standard deviation = 2.236067977
(E) standard deviation = 3.16227766

Answer is A.

To go by logic:

All the sets got same number of elements; so, not considering any effect from that.

Set E has highest range and elements are more distributed towards edge. --> set has the highest standard deviation.
Set D has next highest range and but little concentric towards mean. Still the distribution looks better than remaining sets. Thus, this set has the second highest standard deviation.
Set B has lowest range and elements are mainly centered towards mean. Thus, this set has the lowest standard deviation.
Now we are left with options A and C. Range of set A is more than range of set C. Elements of A are evenly distributed (more spread) and elements of C are concentrated towards one side. From these two observations, we can speculate that option A has higher standard deviation than that of option C.

Note: All these sets have small sample size, very small range, and the numbers are very close to each other. This scenario say little without calculation and, hopefully, real GMAT will not come up with this sort of questions.

That's not correct.

SD of {0, 2, 4, 6, 8} = ~2.83 (3.16 is sample standard deviation and 2.83 is population standard deviation, which is tested on the GMAT).

C. SD of {0, 2, 4, 6, 8, 3, 5} = ~2.45 --> difference=0.38
D. SD of {0, 2, 4, 6, 8, 2, 6} = ~2.62 --> difference=0.21
E. SD of {0, 2, 4, 6, 8, 0, 8} = ~3.21 --> difference=0.38

The correct answer is D.

Official answer from Manhattan Prep.

If X – Y > 0, then X > Y and the median of A is greater than the mean of set A. If L – M = 0, then L = M and the median of set B is equal to the mean of set B.

I. NOT NECESSARILY: According to the table, Z > N means that the standard deviation of set A is greater than that of set B. Standard deviation is a measure of how close the terms of a given set are to the mean of the set. If a set has a high standard deviation, its terms are relatively far from the mean. If a set has a low standard deviation, its terms are relatively close to the mean.

Recall that a median separates the set into two as far as the number of terms. There is an equal number of terms both above and below the median. If the median of a set is greater than the mean, however, the terms below the median must collectively be farther from the median than the terms above the median. For example, in the set {1, 89, 90}, the median is 89 and the mean is 60. The median is much greater than the mean because 1 is much farther from 89 than 90 is.

Knowing that the median of set A is greater than the mean of set A just tells us that the terms below set A’s median are further from the median than the terms above set A’s median. This does not necessarily imply that the terms, overall, are further away from the mean than in set B, where the terms below the median are the same distance from the median as the terms above it. In fact, a set in which the mean and median are equal can have a very high standard deviation if the terms are both far below the mean and far above it.

II. NOT NECESSARILY: According to the table, R > M implies that the mean of set [A + B] is greater than the mean of set B. This is not necessarily true. When two sets are combined to form a composite set, the mean of the composite set must either be between the means of the individual sets or be equal to the mean of both of the individual sets. To prove this, consider the simple example of one member sets: A = [3], B = [5], A + B = [3, 5]. In this case the mean of A + B is greater than the mean of A and less than the mean of B. We could easily have reversed this result by reversing the members of sets A and B.

III. NOT NECESSARILY: According to the table, Q > R implies that the median of the set [A + B] is greater than the mean of set [A + B]. We can extend the rule given in statement II to medians as well: when two sets are combined to form a composite set, the median of the composite set must either be between the medians of the individual sets or be equal to the median of one or both of the individual sets. While the median of set A is greater than the mean of set A and the median of set B is equal to the mean of set B, set [A + B] might have a median that is greater or less than the mean of set [A + B].

Therefore none of the statements are necessarily true and the correct answer is E.

I m a little confused between 32% and 84% but i think 84% makes more sense.

If it is symmetric about the mean, then data less than m is and data greater than m is both 50%.

Then, each of the area from m to m+d and m to m-d would be 68/2=34%.

Thus, for data to be less than m+d, we have

Data from m+d to m = 34%
Data less than m = 50%
Thus, total 84% (D)

I'll attempt an easy one

10) A
Mean = 20.0
S.D = 3
More than 2.5 SD from mean means either less than 20- (2.5x3=7.5) = 12.5 or greater than 20+7.5= 27.5

By The same logic
11) A
13.5-3=10.5

Trying the difficult one...

8. The table below represents three sets of numbers with their respective medians, means and standard deviations. The third set, Set [A+B], denotes the set that is formed by combining Set A and Set B.

Median Mean StandardDeviation
Set A: X, Y, Z.
Set B: L, M, N.
Set [A + B]: Q, R, S.
If X – Y > 0 and L – M = 0, then which of the following must be true?
I. Z > N
II. R > M
III. Q > R
(A) I only
(B) II only
(C) III only
(D) I and II only
(E) None

----------------------------------
(C) III only
I. SD of one set is greater than another. We cannot prove this as no information is given on the actual data of the individual sets
II. Mean of combined set may or may not be greater than one of the given sets.
Consider Set A [1, 3, 3, 3] Median-3, Mean-2.5
Set B [1, 4, 4, 4] Median-4, Mean-3.25
Set A+B [1, 1, 3, 3, 3, 4, 4, 4] Median-3, Mean-2.8
In this case R is not greater than M. But if you interchange set A & B, R>M.

Yes. Your second solution for question #9 is correct. Indeed there are 6 possibilities and 2 pairs have the same SD, so there would be 4 different SD.

As for #1: if it would be E {0,8}, wouldn't these two numbers stretch the set making SD bigger? As {0,8} are too far from the the mean.

And for #6: 2 answers were given (by others) to this question and one of them is incorrect.

So yes, I'm serious. But maybe I didn't get your question right...

Agree with your logic but did you check whether other choices could deviate more further than E?
So what is your OA and workout for #1.
Did you solve it or just guessing? If you solve it, you will find E as the closest.

Frankly speaking when I solved this for the first time I didn't calculated SD-s. Just trusted my logic, then compared my answer to the OA and as they matched I didn't double checked my own solution and OA.

But know I did it. And I can say it again E {0,8} is not correct.

Maybe there is some misunderstanding in stem?