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Monica sells pretzels

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Monica sells pretzels

by sanju09 » Fri Nov 04, 2011 5:11 am
Monica sells pretzels in the cafeteria every school day for a week. She sells 14 pretzels on Monday, 12 pretzels on Tuesday, 16 pretzels on Wednesday, and 12 pretzels on Thursday. Then, she calculates the mean, median, and mode of her sales. If she sells 13 pretzels on Friday, then
(A) the mode will increase.
(B) the mean will stay the same.
(C) the median will stay the same.
(D) the median will decrease.
(E) the mean will increase.
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by [email protected] » Fri Nov 04, 2011 5:14 am
The median will decrease.

Intial Mean = 13.5, median - 13, Mode =12

After 13 pretzels on Friday

Mean = 13.4, median = 13, Mode = 12

hence answer is D
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by neelgandham » Fri Nov 04, 2011 5:38 am
Monica sells pretzels in the cafeteria every school day for a week. She sells 14 pretzels on Monday, 12 pretzels on Tuesday, 16 pretzels on Wednesday, and 12 pretzels on Thursday. Then, she calculates the mean, median, and mode of her sales. If she sells 13 pretzels on Friday, then

Sales till Thursday:
Mean : (12+16+12+14)/4 = 13.5
Median(12,12,14,16)=(12+14)/2 = 13

Sales till Friday:
Mean : is less than 13.5 as you are adding a number less then 13.5(13) to the total.
Median(12,12,13,14,16) = 13

So, the answer is Option C IMO
(A) the mode will increase.
(B) the mean will stay the same.
(C) the median will stay the same.
(D) the median will decrease.
(E) the mean will increase.[/quote]
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by user123321 » Fri Nov 04, 2011 6:19 am
should be C.

Median will stay the same.

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by vaibhavgupta » Fri Nov 04, 2011 8:42 am
sanju09 wrote:Monica sells pretzels in the cafeteria every school day for a week. She sells 14 pretzels on Monday, 12 pretzels on Tuesday, 16 pretzels on Wednesday, and 12 pretzels on Thursday. Then, she calculates the mean, median, and mode of her sales. If she sells 13 pretzels on Friday, then
(A) the mode will increase.
(B) the mean will stay the same.
(C) the median will stay the same.
(D) the median will decrease.
(E) the mean will increase.
[spoiler]C, since 12+14/2=13. and the other one will also have a median of 13![/spoiler] :)
If OA is A, IMO B
If OA is B, IMO C
If OA is C, IMO D
If OA is D, IMO E
If OA is E, IMO A

FML!! :/
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by Tani » Fri Nov 04, 2011 9:27 am
AS shown above:
the mode is unchanged
the median is unchanged
the mean decreases

Answer C
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by vaibhavgupta » Fri Nov 04, 2011 9:33 am
Tani Wolff - Kaplan wrote:AS shown above:
the mode is unchanged
the median is unchanged
the mean decreases

Answer C
If i may ask. I am a little confused about mode. could u pls explain! :)
If OA is A, IMO B
If OA is B, IMO C
If OA is C, IMO D
If OA is D, IMO E
If OA is E, IMO A

FML!! :/
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by Tani » Fri Nov 04, 2011 10:24 am
median is the one in the middle when the terms are put in numerical order. When you have an odd number of items, it will be the middle one. When you have an even number it will be the average between the two middle terms.

Mean is the arithmetic average, i.e. the sum of the terms divided by the number of terms

The mode is simply the term that appears most often in the set. A set can have more than one mode: e.g. 2, 3, 3, 4, 5, 6, 7, 7, 8 = here the modes are 3 and 7
Tani Wolff
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