PS questions about standard deviation.

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?

lol.

Probably the difference is from how the SDs are calculated. I used (n-1) instead of n while dividing the sum of the sqare dev. If n is used, then it is (2, 6).

I don't think Answer will be E, here i'll prove it and you can also check with the calculations that E cannot be true.

With the given set the mean = 4 (as it is set of consecutive even integers therefore, 8+0/2 = 4 )

For all the options the mean remains 4 (as they add 8 to the total it will be divided by 7)

As the values are close to the mean then SD will be lower than the original SD and if the values are far from the mean than SD will be higher.

so D is the option which is not closer and not far from the mean.

Convincing explanation for question 1.

Mean = 4
Var = (16+4+0+4+16)/5 = \(40/5 = 8\)
After addition of 2 numbers, New Var = \((40+x)/7\)
The question is What x will pitch New Var closest to 8 so that 56/7 = 8
OR Which of the options will give a value of x that is closest to 16
So from the 5 options find out which (deviation^2) from 4 is closest to 16
Naked eye will tell you that (A), (B) are a long shot.
(C) 1^2 + 1^2 = 2
(D) 2^2 + 2^2 = 8 ==> |16-8| = 8
(E) 4^2 + 4^2 = 32 ==> |16-32| = 16
So option (D) gives an SD that is closest to the original SD.

Bunuel,

I think the statement in the Q10 is incorrect. Some typo error, IMO:

10. If a certain sample of data has a mean of 20.0 and a standard deviation of 3.0, which of the following values is more than 2.5 standard deviations AWAY from the mean?
(A) 12.0
(B) 13.5
(C) 17.0
(D) 23.5
(E) 26.5
Answer: A.

Please confirm.

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

I found the Q11 in GMATPrep and based on the explaination for it, I asked you.

As per the question:
10. If a certain sample of data has a mean of 20.0 and a standard deviation of 3.0, which of the following values is more than 2.5 standard deviations AWAY from the mean?
(A) 12.0
(B) 13.5
(C) 17.0
(D) 23.5
(E) 26.5
Answer: A.

So, the answer would be > 20+2.5x3 or >27.5, while none of the options say this.
But, if we are given that the required value is 2.5 SD more away then we can find the new value = 12.

I don't quite understand your question... Again the question is from GmarPrep and there is no word "away" in it (at least in the version I have).

Value is more than 2.5SD from the mean means that the distance between the mean and the value must be more than 2.5SD=7.5. So the value either < 12.5 or > 27.5.

Answer: A

Bunuel,
First of all thanks for all the wonderful material that you compile and post here on this forum. I have been following lot of your math related posts for past few days. Your posts are great help in the gmat prep.

I was wondering if you have solutions for these PS SD questions? - specially to Q#8 & 9?

Q#8: ps-questions-about-standard-deviation-85897-20.html#p664302

Q#9: E is a collection of four ODD integers and the greatest difference between any two integers in E is 4. The standard deviation of E must be one of how many numbers?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7

Let the smallest odd integer be 1, thus the largest one will be 5. We can have following 6 types of sets:

1. {1, 1, 1, 5} --> mean=2 --> |mean-x|=(1, 1, 1, 3);
2. {1, 1, 3, 5} --> mean=2.5 --> |mean-x|=(1.5, 1.5, 0.5, 2.5);
3. {1, 1, 5, 5} --> mean=3 --> |mean-x|=(2, 2, 2, 2);
4. {1, 3, 3, 5} --> mean=3 --> |mean-x|=(2, 0, 0, 2);
5. {1, 3, 5, 5} --> mean=3.5 --> |mean-x|=(2.5, 0.5, 1.5, 1.5);
6. {1, 5, 5, 5} --> mean=4 --> |mean-x|=(3, 1, 1, 1).

CALCULATING STANDARD DEVIATION OF A SET {x1, x2, ... xn}:
1. Find the mean, \(m\), of the values.
2. For each value \(x_i\) calculate its deviation (\(m-x_i\)) from the mean.
3. Calculate the squares of these deviations.
4. Find the mean of the squared deviations. This quantity is the variance.
5. Take the square root of the variance. The quantity is th SD.

Expressed by formula: \(standard \ deviation= \sqrt{variance} = \sqrt{\frac{\sum(m-x_i)^2}{N}}\).

You can see that deviation from the mean for 2 pairs of the set is the same, which means that SD of 1 and 6 will be the same and SD of 2 and 5 also will be the same. So SD of such set can take only 4 values.

Answer: B.

Solutions and OA's for other questions are on previous pages.

Hope it's clear.

how did you solve this x - 2sd = 58 and x + 3sd = 98 ?

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)

@ Bunuel

please explain the logical understanding/ difference between questions 10 and 11.
my doubt for q-11 is m getting

value=13.5+ 3= 16.5 and 10.5 so the value should be between 10.5 and 16.5,

however the options and the corresponding OA just confuses me.

pls explain

10. If a certain sample of data has a mean of 20.0 and a standard deviation of 3.0, which of the following values is more than 2.5 standard deviations from the mean?
(A) 12.0
(B) 13.5
(C) 17.0
(D) 23.5
(E) 26.5

Value is more than 2.5SD from the mean means that the distance between the mean and the value must be more than 2.5*SD=7.5. So the value must be either less than 20-7.5=12.5 or more than 20+7.5=27.5.

Answer: A.

11. Arithmetic mean and standard deviation of a certain normal distribution are 13.5 and 1.5. What value is exactly 2 standard deviations less than the mean?
(A) 10.5
(B) 11
(C) 11.5
(D) 12
(E) 12.5

The value which is exactly two SD less than the mean is: mean-2*SD=13.5-2*1.5=10.5.

Answer: A.

Hope it's clear.

:@Ritesh Gupta: Let me try to explain this to you.

consider a straight line that represents the mean (m). and lets consider S.D d as one unit. so you have m+d above the line m and m-d below the line m. Now given that 68% lies within one standard deviation from mean, that means 68% = m+d and m-d, which means 34% each. Next, remaining 32% (100-68) is above m+d and below m-d, again equally distributed, hence, 16% each.

Thus, if you graphically visualize, the question is asking you, (m+d) + (m-d) + below (m-d) = 34 + 34 + 16 = 84.

I hope its clear!