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In a distribution of 8,500 different measurements of the var
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11 Dec 2017, 10:08
11 Dec 2017, 10:08
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Question Stats:
75% (01:20) correct
24% (02:15) wrong based on 37 sessions
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In a distribution of 8,500 different measurements of the variable x, 26.5 is the 56 th percentile and 37.1 is the 78 th percentile. Which of the following is closest to the number of measurements of x that are in the distribution such that 26.5 ≤ x ≤ 37.1 ?
A. 1,850
B. 2,200
C. 3,500
D. 4,750
E. 6,650
A. 1,850
B. 2,200
C. 3,500
D. 4,750
E. 6,650
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Re: In a distribution of 8,500 different measurements of the var
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30 Jan 2018, 17:33
30 Jan 2018, 17:33
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Expert Reply
Notice that the problem doesn't mention the word "normal." So this distribution of measurements hasn't necessarily got anything to do with standard deviation.
Anyway, the top 1%ile make up 1/100th of the total, as will the bottom 1%ile. Don't be fooled by those sloppy measurements. The actual measurements aren't important in this problem. What's important is their percentiles. The first measurement is at the 56th %ile and the second is at the 78th %ile, which means that they are 22% apart from each other. So if each percentile equals 85 (since 8500/100 = 85), then there must be 22x85 measurements between them.
These numbers look a little sloppy but an easy way to do the math in your head is to find 20x85 and then find 2x85 and then add them. What's 20x85? If you multiply 85 by 2 you get 170, then multiply by 10 by adding a zero and we get 1700. Just remember that for a second. Then we know that 2x85 must be one tenth of 1700, which is 170, so if we then add 1700 and 170 we get 1870. Pretty close to A. And since the problem said "closest" then that's good enough.
Anyway, the top 1%ile make up 1/100th of the total, as will the bottom 1%ile. Don't be fooled by those sloppy measurements. The actual measurements aren't important in this problem. What's important is their percentiles. The first measurement is at the 56th %ile and the second is at the 78th %ile, which means that they are 22% apart from each other. So if each percentile equals 85 (since 8500/100 = 85), then there must be 22x85 measurements between them.
These numbers look a little sloppy but an easy way to do the math in your head is to find 20x85 and then find 2x85 and then add them. What's 20x85? If you multiply 85 by 2 you get 170, then multiply by 10 by adding a zero and we get 1700. Just remember that for a second. Then we know that 2x85 must be one tenth of 1700, which is 170, so if we then add 1700 and 170 we get 1870. Pretty close to A. And since the problem said "closest" then that's good enough.
Re: In a distribution of 8,500 different measurements of the var
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29 Sep 2019, 13:37
29 Sep 2019, 13:37
2
The first measurement is at the 56th %ile and the second is at the 78th %ile, which means that they are 22% apart from each other. So, we can readily do this as, 8500*.22 = 1870, closer to 1850
