# Statistic : Test

https://gmatclub.com/forum/statistics-made-easy-all-in-one-topic-203966.html

**The average age of a group of n people is 15 yrs. One more person aged 39 joins the group and the new average is 17 yrs. What is the value of n?**(A) 9

(B) 10

( C)11

(D) 12

(E) 13**When a person aged 39 is added to a group of n people, the average age increases by 2. When a person aged 15 is added instead, the average age decreases by 1. What is the value of n?**(A) 7

(B) 8

© 9

(D) 10

(E) 11**A, B and C have received their Math midterm scores today. They find that the arithmetic mean of the three scores is 78. What is the median of the three scores?**

1. A scored a 73 on her exam.

2. C scored a 78 on her exam.

(A) Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient

(B) Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient

(C) Both statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient

(D) EACH statement ALONE is sufficient

(E) Statements (1) and (2) TOGETHER are NOT sufficient**Five logs of wood have an average length of 100 cm and a median length of 116 cm. What is the maximum possible length, in cm, of the shortest piece of wood?**(A) 50

(B) 76

© 84

(D) 96

(E) 100**For the past n days the average daily production at a company was 60 units. If today’s production of 100 units raises the average to 65 units per day, what is the value of n?**(A) 30

(B) 18

© 10

(D) 9

(E) 7**When Anna makes a contribution to a charity fund at school, the average contribution size increases by 50%, reaching $75 per person. If there were 5 other contributions made before Anna’s, what is the size of her donation?**(A) $100

(B) $150

© $200

(D) $250

(E) $450**A set of numbers has an average of 50. If the largest element is 4 greater than 3 times the smallest element, which of the following values cannot be in the set?**(A) 85

(B) 90

© 123

(D) 150

(E) 155**If x is the sum of the even integers from 200 to 600 inclusive, and y is the number of even integers from 200 to 600 inclusive, what is the value of x + y?**(A) 200*400

(B) 201*400

© 200*402

(D) 201*401

(E) 400*401**The sum of n consecutive positive integers is 45. What is the value of n?**Statement I: n is even

Statement II: n < 9**Which of the following cannot be the range of a set consisting of 5 odd multiples of 9?**(A) 72

(B) 144

© 288

(D) 324

(E) 436**If the arithmetic mean of n consecutive odd integers is 20, what is the greatest of the integers?**(1) The range of the n integers is 18.

(2) The least of the n integers is 11.**A certain list of 300 test scores has an arithmetic mean of 75 and a standard deviation of d, where d is positive. Which of the following two test scores, when added to the list, must result in a list of 302 test scores with a standard deviation less than d?**(A) 75 and 80

(B) 80 and 85

© 70 and 75

(D) 75 and 75

(E) 70 and 80**If 100 is included in each of sets A, B and C (given A= {30, 50, 70, 90, 110}, B = {-20, -10, 0, 10, 20} and C= {30, 35, 40, 45, 50}), which of the following represents the correct ordering (largest to smallest) of the sets in terms of the absolute increase in their standard deviation?**(A) A, C, B

(B) A, B, C

© C, A, B

(D) B, A, C

(E) B, C, A**During an experiment, some water was removed from each of the 8 water tanks. If the standard deviation of the volumes of water in the tanks at the beginning of the experiment was 20 gallons, what was the standard deviation of the volumes of water in the tanks at the end of the experiment?**Statement 1: For each tank, 40% of the volume of water that was in the tank at the beginning of the experiment was removed during the experiment.

Statement 2: The average volume of water in the tanks at the end of the experiment was 80 gallons**M is a collection of four odd integers. The range of set M is 4. How many distinct values can standard deviation of M take?**(A) 3

(B) 4

© 5

(D) 6

(E) 7**For the set {2, 2, 3, 3, 4, 4, 5, 5, x}, which of the following values of x will most increase the standard deviation?**(A) 1

(B) 2

© 3

(D) 4

(E) 5**An automated manufacturing unit employs N experts such that the range of their monthly salaries is $10,000. Their average monthly salary is $7000 above the lowest salary while the median monthly salary is only $5000 above the lowest salary. What is the minimum value of N?**(A)10

(B)12

©14

(D)15

(E)20**Given that set S has four odd integers and their range is 4, how many distinct values can the standard deviation of S take?**(A) 3

(B) 4

© 5

(D) 6

(E) 7**Which of the following distribution of numbers has the greatest standard deviation?**(A) {-3, 1, 2}

(B) {-2, -1, 1, 2}

© {3, 5, 7}

(D) {-1, 2, 3, 4}

(E) {0, 2, 4}**Which of the following data sets has the third largest standard deviation?**(A) {1, 2, 3, 4, 5}

(B) {2, 3, 3, 3, 4}

© {2, 2, 2, 4, 5}

(D) {0, 2, 3, 4, 6}

(E) {-1, 1, 3, 5, 7}**7.51; 8.22; 7.86; 8.36**

8.09; 7.83; 8.30; 8.01

7.73; 8.25; 7.96; 8.53

A vending machine is designed to dispense 8 ounces of coffee into a cup. After a test that recorded the number of ounces of coffee in each of 1000 cups dispensed by the vending machine, the 12 listed amounts, in ounces, were selected from the data above. If the 1000 recorded amounts have a mean of 8.1 ounces and a standard deviation of 0.3 ounces, how many of the 12 listed amounts are within 1.5 standard deviations of the mean?

A)Four

B) Six

C) Nine

D) Ten

E) Eleven

_________________________________________________________________

**C . Solution:**First tell me, if the age of the additional person were 15 yrs, what would have happened to the average? The average would have remained the same since this new person’s age would have been the same as the age that represents the group. But his age is 39–15 = 24 more than the average. We know that we need to evenly split the extra among all the people to get the new average. When 24 is split evenly among all the people (including the new guy), everyone gets 2 extra (since average age increased from 15 to 17). There must be 24/2 = 12 people now (including the new guy) i.e. n must be 11 (without including the new guy).**A. Solution:**What is the first thing you can say about the initial average? It must have been between 39 and 15. When a person aged 39 is added to the group, the average increases and when a person aged 15 is added, the average decreases.

Let’s look at the second case first. When the person aged 15 is added to the group, the average becomes (initial average — 1). If instead, the person aged 39 were added to the group, there would be 39–15 = 24 extra which would make the average = (initial average + 2). This difference of 24 creates a difference of 3 in the average. This means there must have been 24/3 = 8 people (after adding the extra person). The value of n must be 8–1 = 7.**B. Solution:**Recall from the arithmetic mean post that the sum of deviations of all scores from the mean is 0.

i.e. if one score is less than mean, there has to be one score that is more than the mean.

e.g. If mean is 78, one of the following must be true:

All scores are equal to 78.

At least one score is less than 78 and at least one is greater than 78.

For example, if one score is 70 i.e. 8 less than 78, another score has to make up this deficit of 8. Therefore, there could be a score that is 86 (8 more than 78) or there could be two scores of 82 each etc.**Statement 1: A scored 73 on her exam.**

For the mean to be 78, there must be at least one score higher than 78. But what exactly are the other two scores? We have no idea! Various cases are possible:

73, 78, 83 or 73, 74, 87 or 70, 73, 91 etc.

In each case, the median will be different. Hence this statement alone is not sufficient.**Statement 2: C scored 78 on her exam.**Now we know that one score is 78. Either the other two will also be 78 or one will be less than 78 and the other will be greater than 78. In either case, 78 will be the middle number and hence will be the median. This statement alone is sufficient.**B.**First thing that comes to mind — median is the 3rd term out of 5 so the lengths arranged in increasing order must look like this:

___ ___ 116 ___ ___The mean is given and we need to maximize the smallest number. Basically, the smallest number should be as close to the mean as possible. This means the greatest number should be as close to the mean as possible too (if the shortfall deviation is small, the excess deviation should by equally small).

If this doesn’t make sense, think of a set with mean 20:19, 20, 21 (smallest number is very close to mean; greatest number is very close to the mean too)

1, 20, 39 (smallest number is far away from the mean, greatest number is far away too)

Using the same logic, let’s make the greater numbers as small as possible (so the smallest number can be as large as possible). The two greatest numbers should both be at least 116 (since 116 is the median). Now the lengths arranged look like this:___ ___ 116 116 116

Since the mean is 100 and each of the 3 large numbers are already 16 more than 100 i.e. total 16*3 = 48 more than the mean (excess deviation is 48), the deviations of the two small numbers should be a total of 48 less than the mean. To make the smallest number as great as possible, each of the small numbers should be 48/2 = 24 less than the mean i.e. they both should be 76.**E. Solution:**If today’s production were also 60 units, what would have happened to the average? Obviously, it would have stayed the same! But today’s production is 40 units extra and hence it raised the average. It raised the average by 5 units which means that each one of the n observations and today’s observation got an extra 5. Since 40 got distributed and each was given 5, there must have been a total of 40/5 = 8 observations including today’s. Therefore, the value of n must have been 8–1 = 7.**C. Solution:**After Anna’s contribution, the average size increases by 50% and reaches $75. What must have been the average size of contribution before Anna’s donation? It must have been $50 since a 50% increase would lead us to $75. So, $50 was the average size of 5 donations before Anna made her donation. Had Anna donated $50 as well, the average would have stayed the same i.e. $50. But the average increased to $75 which means that Anna donated an extra $25 for each of the 6 observations (including her) in addition to the $50 she would have donated to keep the average same.

Hence, the amount Anna donated = 50 + 6*25 = $200**E. Solution:**This question might look a little ominous but it isn’t very tough, really! The set has an average of 50 so that already tells us that we can represent each element of the set by 50. If there is an element which is a little less than 50, there will be another element which is a little more than 50.

The largest element is 4 greater than 3 times the smallest element so

L = 4 + 3S.

The smallest element must be less than 50 and the largest must be greater than 50. Say, if the smallest element is 20, the largest will be 4 + 3*20 = 64.

Is there any limit imposed on the largest value of the largest element? Yes, because there is a limit on the largest value of the smallest element. The smallest element must be less than 50. The smallest member of the set can be 49.9999… The limiting value of the smallest number is 50. As long as the smallest number is a tiny bit less than 50, you can have the greatest number a tiny bit less than 4 + 3*50 = 154. The number 154 and all numbers greater than 154 cannot be a part of the set.**D.**There are various ways of getting the answer here. We will use the concepts we learned last week.The given sequence is 200, 202, 204, … 600It is an arithmetic progression. What is the total number of terms here?**Using Logic**In every 100 consecutive integers, there are 50 odd integers and 50 even integers. So we will get 50 even integers from each of 200–299, 300–399, 400–499 and 500–599 i.e. a total of 50*4 = 200 even integers. Also, since the sequence includes 600, number of even integers = 200 + 1 = 201**Recall that in our****arithmetic progressions post****,**we saw that the last term of a sequence which has n terms will be first term + (n — 1)* common difference.

600=200+(n–1)∗2 ;n=201

Hence y=201 (because y is the number of even integers from 200 to 600)

Let’s go on now. What is the average of the sequence? Since it is an arithmetic progression with odd number of integers, the average must be the middle number i.e. 400.

Notice that since this arithmetic progressions looks like this:

(n — m), … (n — 6), (n — 4), ( n — 2), n, (n + 2), (n + 4), (n + 6), … (n + m)

We can find the middle number i.e. the average by just averaging the first and the last terms.

(n–m)+(n+m)/2=2n/2=n

Average=(200+600)/2=400

Sum of all terms in the sequence = x = Arithmetic Mean * Number of terms = 400*201

x+y=400∗201+201=401∗201**E. Solution:**First I will give the solution of this question and then discuss the logic used to solve it.

In how many ways can you write n consecutive integers such that their sum is 45? Let’s see whether we can get such numbers for some values of n.

n = 1 -> Numbers: 45

n = 2 -> Numbers: 22 + 23 = 45

n = 3 -> Numbers: 14 + 15 + 16 = 45

n = 4 -> No such numbers

n = 5 -> Numbers: 7 + 8 + 9 + 10 + 11 = 45

n = 6 -> Numbers: 5 + 6 + 7 + 8 + 9 + 10 = 45

Let’s stop right here.

Statement I: n must be even.

n could be 2 or 6. Statement I alone is not sufficient.

Statement II: n < 9

n can take many values less than 9 hence statement 2 alone is not sufficient.

Both statements together: Since n can take values 2 or 6 which are even and less than 9, both statements together are not sufficient.**E.**There are infinite possibilities regarding the multiples of 9 that can be included in the set. The set could be any one of the following (or any one of the other infinite possibilities):

S = {9, 27, 45, 63, 81} or

S = {9, 63, 81, 99, 153} or

S = {99, 135, 153, 243, 1071}

The range in each case will be different. The question asks us for the option that ‘cannot’ be the range. Let’s figure out the constraints on the range.

A set consisting of only odd multiples of 9 will have a range that is an even number (Odd Number — Odd Number = Even number)

Also, the range will be a multiple of 9 since both, the smallest and the greatest numbers, will be multiples of 9. So their difference will also be a multiple of 9.

Only one option will not satisfy these constraints. Do you remember the divisibility rule of 9? The sum of the digits of the number should be divisible by 9 for the number to be divisible by 9. The sum of the digits of 436 is 4 + 3 + 6 = 13 which is not divisible by 9. Hence 436 cannot be divisible by 9 and therefore, cannot be the range of the set.**D. Solution:**We have discussed mean in case of arithmetic progressions in the previous posts. If mean of consecutive odd integers is 20, what do you think the integers will look like?

19, 21 or

17, 19, 21, 23 or

15, 17, 19, 21, 23, 25 or

13, 15, 17, 19, 21, 23, 25, 27 or

11, 13, 15, 17, 19, 21, 23, 25, 27, 29

etc.

Does it make sense that the required numbers will represent one such sequence? The numbers in the sequence will be equally distributed around 20. Every time you add a number to the left, you need to add one to the right to keep the mean 20. The smallest sequence will have 2 numbers 19 and 21, the largest will have infinite numbers. Did you notice that each one of these sequences has a unique “range,” a unique “least number” and a unique “greatest number?” So if you are given any one statistic of the sequence, you will know the entire desired sequence.**Statement 1:**Only one possible sequence: 11, 13, 15, 17, 19, 21, 23, 25, 27, 29 will have the range 18. The greatest number here is 29. This statement alone is sufficient.**Statement 2:**Only one possible sequence: 11, 13, 15, 17, 19, 21, 23, 25, 27, 29 will have 11 as the least number. The greatest number here is 29. This statement alone is sufficient too.**D. Solution:**As discussed above, the standard deviation of a set measures the deviation from the mean. A low standard deviation indicates that the data points are very close to the mean whereas a high standard deviation indicates that the data points are spread far apart from the mean.

When we add numbers that are far from the mean, we are stretching the set and hence, increasing the SD. When we add numbers which are close to the mean, we are shrinking the set and hence, decreasing the SD.

Therefore, adding two numbers which are closest to the mean will shrink the set the most, thus decreasing SD by the greatest amount.

Numbers closest to the mean are 75 and 75 (they are equal to the mean) and thus adding them will decrease SD the most.**E. Solution:**The question looks a little convoluted but actually you don’t have to calculate anything. SD measures the deviation of the elements from the mean. If a new element is added which is far away from the mean, it will add much more to the deviations than if it were added close to the mean.

The means of A, B and C are 70, 0 and 40, respectively.

100 is farthest from 0 so it will change the SD of set B the most (in terms of absolute increase). It is closest to 70 so it will change the SD of set A the least. Hence the correct ordering is B, C, A.**A.**We have 8 water tanks. This implies that we have 8 elements in the set (volume of water in each of the 8 tanks). SD of the volume of water in the tanks is 20 gallons. We need to find the new SD i.e. the SD after water was removed from the tanks.**Statement 1:**For each tank, 40% of the volume of water that was in the tank at the beginning of the experiment was removed during the experiment.Initial SD is 20. When 40% of the water is removed from each tank, the leftover water is 60% of the initial volume of water i.e. 0.6*initial volume of water. This means that each element of the initial set was multiplied by 0.6 to obtain the new set. The SD will change. It will become 0.6*previous SD i.e. 0.6*20 = 12 (think of the formula of SD we discussed in the first SD post). This statement alone is sufficient.**Statement 2:**The average volume of water in the tanks at the end of the experiment was 80 gallons.The average volume doesn’t give us the SD of the new set. Hence, this statement alone is not sufficient.**B. Solution:**Since the range of M is 4, it means the greatest difference between any two elements is 4. One way of doing this will be M = {1, x, y, 5} (obviously, there are innumerable ways of writing M)

Here, x and y can take one of 3 different values: 1, 3 and 5 (x and y cannot be less than 1 or greater than 5 because the range of the set is 4).

Both x and y could be same. This can be done in 3 ways. Or x and y could be different. This can be done in 3C2 = 3 ways. Total x and y can take values in 3 + 3 = 6 ways.

For clarification, let me enumerate the 6 ways in which you can get the desired set:

{1, 1, 1, 5}, {1, 3, 3, 5}, {1, 5, 5, 5}, {1, 1, 3, 5}, {1, 1, 5, 5}, {1, 3, 5, 5}

Note here that standard deviations of {1, 1, 1, 5} and {1, 5, 5, 5} are same. Why? Because SD measures deviation from mean. It has nothing to do with the actual value of mean and actual value of numbers

Mean of {1, 1, 1, 5} is 2. Three of the numbers are distance 1 away from mean and one number is distance 3 away from mean. Mean of {1, 5, 5, 5} is 4. Three of the numbers are distance 1 away from mean and one number is distance 3 away from mean. Sum of the squared deviations will be the same in both the cases and the number of elements is also the same in both the cases. Therefore, both these sets will have the same SD.

Similarly, {1, 1, 3, 5} and {1, 3, 5, 5} will have the same SD.

From the leftover sets, {1, 3, 3, 5} will have a distinct SD and {1, 1, 5, 5} will have a distinct SD.

In all, there are 4 different values that SD can take in such a case.**A.**you can spend a lot of time calculating every last detail of this question, what it actually comes down to is “which of these numbers is furthest from 3.5”.- there are at least 2 people — say one with salary 0 and the other with 10,000. No salary will lie outside this range.Median is $5000 — i.e. when all salaries are listed in increasing order, the middle salary (or average of middle two) is $5000. With 2 people, one at 0 and the other at 10,000, the median will be the average of the two i.e. (0 + 10,000)/2 = $5000. Since there are at least 10 people, there is probably someone earning $5000. Let’s put in 5000 there for reference.

0 … 5000 … 10,000

Arithmetic mean of all the salaries is $7000. Now, mean of 0, 5000 and 10,000 is $5000, not $7000 so this means that we need to add some more people. We need to add them more toward 10,000 than toward 0 to get a higher mean. So we will try to get a mean of $7000.

Let’s use deviations from the mean method to find where we need to add more people.

0 is 7000 less than 7000 and 5000 is 2000 less than 7000 which means we have a total of $9000 less than 7000. On the other hand, 10,000 is 3000 more than 7000. The deviations on the two sides of mean do not balance out. To balance, we need to add two more people at a salary of $10,000 so that the total deviation on the right of 7000 is also $9000. Note that since we need the minimum number of experts, we should add new people at 10,000 so that they quickly make up the deficit in the deviation. If we add them at 8000 or 9000 etc, we will need to add more people to make up the deficit at the right.

0 … 5000 … 10000, 10000, 10000

Now the mean is 7000 but note that the median has gone awry. It is 10,000 now instead of the 5000 that is required. So we will need to add more people at 5000 to bring the median back to 5000. But that will disturb our mean again! So when we add some people at 5000, we will need to add some at 10,000 too to keep the mean at 7000.

5000 is 2000 less than 7000 and 10,000 is 3000 more than 7000. We don’t want to disturb the total deviation from 7000. So every time we add 3 people at 5000 (which will be a total deviation of 6000 less than 7000), we will need to add 2 people at 10,000 (which will be a total deviation of 6000 more than 7000), to keep the mean at 7000 — this is the most important step. Ensure that you have understood this before moving ahead.

When we add 3 people at 5000 and 2 at 10,000, we are in effect adding an extra person at 5000 and hence it moves our median a bit to the left.

0 … 5000, 5000, 5000, 5000 … 10000, 10000, 10000, 10000, 10000

The median is not $5000 yet. Let’s try one more set of addition.

0 … 5000, 5000, 5000, 5000, 5000, 5000, 5000 … 10000, 10000, 10000, 10000, 10000, 10000, 10000

The median now is $5000 and we have maintained the mean at $7000.

This gives us a total of 15 people. - เฉลยอยู่ข้อ 15
**A.**For answer choice A, the mean = 0 and the deviations are 3, 1, 2

For answer choice B, the mean = 0 and the deviations are 2, 1, 1, 2

For answer choice D, the mean = 2 and the deviations are 3, 0, 1, 2

Comparing answer choices A and D, we see that they both have the same deviations, but D has more elements. This means its denominator will be greater, and therefore, the SD of answer D is smaller than the SD of answer A. This leaves us with options A and B:

For answer choice A, the mean = 0 and the deviations are 3, 1, 2

For answer choice B, the mean = 0 and the deviations are 2, 1, 1, 2

Now notice that although two deviations of answers A and B are the same, answer choice A has a higher deviation of 3 but fewer elements than answer choice B. This means the SD of A will be higher than the SD of B, so the SD of A will be the highest.- Deviations of answer choice A: 2, 1, 0, 1, 2

Deviations of answer choice B: 1, 0, 0, 0, 1 (lowest SD)

Deviations of answer choice C: 1, 1, 1, 1, 2

Deviations of answer choice D: 3, 1, 0, 1, 3

Deviations of answer choice E: 4, 2, 0, 2, 4 (highest SD)

Obviously, option B has the lowest SD (the deviations are the smallest) and option E has the highest SD (the deviations are the greatest). This means we can automatically rule these answers out, as they cannot have the third largest SD.

Deviations of answer choice A: 2, 1, 0, 1, 2

Deviations of answer choice C: 1, 1, 1, 1, 2

Deviations of answer choice D: 3, 1, 0, 1, 3

Out of these options, answer choice D has a higher SD than answer choice A, since it has higher deviations of two 3s (whereas A has deviations of two 2s). Also, C is more tightly packed than A, with four deviations of 1. If you are not sure why, consider this:

The square of deviations for C will be 1 + 1+ 1 + 1 + 4 = 8

The square of deviations for A will be 4 + 1 + 0 + 1 + 4 = 10

So, A will have a higher SD than C but a lower SD than D. Arranging from lowest to highest SD’s, we get: B, C, A, D, E. Answer choice A has the third highest SD, - Okay, so the standard deviation is 0.3 ounces. We want the values that are within 1.5 standard deviations of the mean. 1.5 standard deviations would be (1.5)(0.3) = 0.45 ounces, so we want all of the values that are within 0.45 ounces of the mean. If the mean is 8.1 ounces, this means that we want everything that falls between a lower bound of (8.1–0.45) and an upper bound of (8.1 + 4.5). Put another way, we want the number of values that fall between 8.1–0.45 = 7.65 and 8.1 + 0.45 = 8.55.

Looking at our 12 values, we can see that only one value, 7.51, falls outside of this range. If we have 12 total values and only 1 falls outside the range, then the other 11 are clearly within the range