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28 Jul 2020, 03:34
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
48% (02:13) correct
52%
(02:20)
wrong
based on 201
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58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88
The number of students in each of 10 different classes is a different number from the above list. What is the standard deviation of the number of students in the 10 classes?
(1) The median number of students is equal to the mean number of students in the 10 classes.
(2) The number of students in any class is more than 63 and the average number of students in the 10 classes is same as the average of the above list.
Are You Up For the Challenge: 700 Level Questions
The number of students in each of 10 different classes is a different number from the above list. What is the standard deviation of the number of students in the 10 classes?
(1) The median number of students is equal to the mean number of students in the 10 classes.
(2) The number of students in any class is more than 63 and the average number of students in the 10 classes is same as the average of the above list.
Are You Up For the Challenge: 700 Level Questions
yashikaaggarwal
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28 Jul 2020, 03:50
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(1) The median number of students is equal to the mean number of students in the 10 classes.
The mean and median satisfying the above constraint can be 67,69,71,73,75,77,79 (since the numbers are at even consecutive difference)
There are 7 set possible, hence the standard deviation will vary as well.
(Insufficient)
(2) The number of students in any class is more than 63 and the average number of students in the 10 classes is same as the average of the above list.
The mean of the list 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88 = middle Two numbers/2
=> 72+74/2 = 73
Plus the number of students are more than 63
So the given possible set no. = 4
Also the mean of all number = mean of 10 class students number
The only set possible having 73 as mean ranges from 64, 66, 68, 70, 72, 74, 76, 78, 80, 82.
Sum of (x-mean)^2 = 81+49+25+9+1+1+9+25+49+81 = 330/10 = 33
SD = √33
(Sufficient)
IMO B
Posted from my mobile device
The mean and median satisfying the above constraint can be 67,69,71,73,75,77,79 (since the numbers are at even consecutive difference)
There are 7 set possible, hence the standard deviation will vary as well.
(Insufficient)
(2) The number of students in any class is more than 63 and the average number of students in the 10 classes is same as the average of the above list.
The mean of the list 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88 = middle Two numbers/2
=> 72+74/2 = 73
Plus the number of students are more than 63
So the given possible set no. = 4
Also the mean of all number = mean of 10 class students number
The only set possible having 73 as mean ranges from 64, 66, 68, 70, 72, 74, 76, 78, 80, 82.
Sum of (x-mean)^2 = 81+49+25+9+1+1+9+25+49+81 = 330/10 = 33
SD = √33
(Sufficient)
IMO B
Posted from my mobile device
28 Jul 2020, 10:11
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Bunuel
We have (88 - 58)/2 + 1 = 16 values of consecutive even integers. To find the standard deviation likely have to confirm all 10 values.
Statement 1:
Let's take consecutive evens from 60 through 78 (inclusive) for example. It is an arithmetic sequence so it satisfies the statement, however, if we replace 60 and 78 with 58 and 80 we can still have median = mean but the std is bigger. Hence insufficient.
(An important note is that an arith sequence means (mean = median) but vice-versa is not always true!)
Statement 2:
The average value of the list is the average of 58 and 88 which is 73. Then we need to select 10 values from above with an average of 73, under the condition that all values are greater than 63. The smallest list we are allowed to pick is 64 through 82 inclusive, and it already has an average of (64+82)/2 = 73. Then we can only pick this one list, sufficient.
Ans: B
