Statistics: Mean Median Mode
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1. A set of four numbers, a, b, c, and d has mean x, median y, and mode z. Which of the following is possible?

I. x = y = z
II. y = a
III. x > 4y

(A) none
(B) I and II, only
(C) I and III, only
(D) II and III, only
(E) I, II, and III

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The mean, median, and mode can all be equal (for example: 1, 2, 2, 3) so (I) is true. The median can be one of the numbers and a, b, c, d aren't necessarily listed in order so (II) is true. The mean can be much bigger than the median (for example: 1, 2, 2, 1000 has mean 251.25 and median 2) so (III) is also true and the answer is (E).

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2. Four positive numbers a, b, c, and d have mean x and median y. If a < b < c < d then which of the following is possible?

I. x < y/2
II. y = b
III. x > c

(A) I, only
(B) II, only
(C) III, only
(D) I and III, only
(E) II and III, only

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The numbers are in order so the median is (b+c)/2. The mean is (a+b+c+d)/4. You can write the mean (x) in terms of the median (y) like so: x = (a+d)/4 + (b+c)/4 = (a+d)/4 + y/2. So (I) is not true. Since b and c are different, (II) is not true. You can have the average bigger than the third number (for example: 1, 2, 3, 1000) so (III) is true and the answer is (C).

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3. A set consists of an odd number of integers with mean x, median y and mode z. If z is a single mode (there are no other numbers with as many multiple occurrences as z), then which of the following is always true?

I. x is an integer
II. y is an integer
III. z is an integer

(A) II, only
(B) III, only
(C) II and III, only
(D) I and III, only
(E) I, II, and III

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The mean is an integer only if the sum happens to be divisible by 5 so (I) is not always true. If there are an odd number of numbers, the median is one of the numbers so (II) is true. The mode is always one of the numbers, so (III) is true. The answer is (C).

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4. Three integers a, b, and c have a median that is two more than the mean. If a < b < c and a = 2 and c = 12, then what is the median?

(A) 3
(B) 4
(C) 16/3
(D) 8
(E) 10

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The median (b) is two more than the mean ([a+b+c]/3). They give you a and c so b must equal (14+b)/3 + 2. Solve for b to get b=10. The answer is (E).

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5. Four numbers a, b, c, and d have a median that is equal to the average. If a < b < c < d, then which statement is true?

(A) (b–a) = (d–c)
(B) (c+d) = 2 · (a+b)
(C) d = 4a
(D) (a+c) = (b+d)
(E) none of these

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The median is (b+c)/2 and the average (or mean) is (a+b+c+d)/4. Setting these equal leads to: (b+c) = (a+d) which is the same as (b–a)=(d–c). So the answer is (A).

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6. A set of N consecutive integers has average A and median M. What is M – A?

(A) 0
(B) 0.25
(C) 0.5
(D) 1
(E) cannot be determined

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For consecutive integers, the median and the average are always the same (for example: 1, 2, 3, 4, 5, 6 has median and average 3.5). So the answer is 0 or (A).

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7. The first member of a set of N integers (N > 2) is 1. Each member of the set after the first is equal to twice the previous member. Which of the following must be true of the set?

I. The mean is greater than the median.
II. The median is an integer.
III. The mean is an integer.

(A) I, only
(B) II, only
(C) III, only
(D) I and II, only
(E) I, II, and III

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If the numbers to the right of the median get rapidly bigger, the mean gets "pulled" to the right. For example: 1, 2, 4, 8, 16 has a much bigger mean than 1, 2, 4, 6, 7. So (I) is true. All the numbers in the set except for the first are even, so the median will either be one of the numbers or the average of two even numbers. Either way, the median will be an integer so (II) is true. The very first mean (for 1, 2, 4) is not an integer so (III) is not true and the answer is (D).

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8. Five consecutive positive integers are represented by v, w, x, y, and z in order from smallest to largest. If you replace v with double its value which of the following will be true about the new set of five integers?

I. The mean will be larger than x.
II. The median will be equal to x.
III. The median will be equal to y.

(A) I, only
(B) II, only
(C) III, only
(D) I and II, only
(E) I and III, only

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Obviously, making a positive number bigger will increase the average so (I) is true. If doubling the first number makes it bigger than the middle number (original set: 10, 11, 12, 13, 14), then there will be a new middle number so the new set might have a different median. Thus, (II) is not true. If doubling the first number doesn't make it bigger than the middle number (original set: 1, 2, 3, 4, 5), then the median will stay the same, so (III) isn't true either. The answer is (A).

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9. A set contains four numbers. The numbers are represented by x, 4x, 5x, and 5x–2. The set contains three different numbers and one pair of equal numbers. Which of the following is possible?

I. The mode is x.
II. The mode is 4x.
III. The mode is 5x.

(A) III, only
(B) I and II, only
(C) I and III, only
(D) II and III, only
(E) I, II, and III

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Of course, x, 4x, 5x all have to be different. However, you can make x equal to 5x–22 so the mode could be x. You can also make 4x equal to 5x–2 so the mode could be 4x. But you can't make 5x equal to 5x–2 so the mode can't be 5x. The answer is (B).

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10. Two of the numbers in a set of 9 positive numbers are changed. As a result of these changes, the median of the set decreases and the mean of the set increases. Which of the following changes could have this result?

(A) double the highest number and halve the second highest
(B) double the highest number and double the lowest number
(C) double the highest number of double the second highest
(D) double the highest number and halve the lowest number
(E) none of these

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Choice A: If you double the highest and halve the second highest, you gain more than you lose so the mean will increase. If half of the second highest happens to be below the original median, then the median will shift down a notch. So choice (A) works. Choice B (doubling the highest and the lowest) would never decrease the median. Choice C (doubling the highest and second highest) wouldn't change the median. Choice D (doubling the highest and halving the lowest) wouldn't change the median either.

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