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14 May 2021, 07:45
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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:
35%
(medium)
Question Stats:
71% (01:29) correct
29%
(01:31)
wrong
based on 55
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P = {22, 24, 26, 28, 30, 32}
Q = {2, 24, 25, 26, 27, 28}
R = {4, 7, 10, 13, 16, 19}
If p is the standard deviation of set P, q is the standard deviation of set Q, and r is the standard deviation of set R, which of the following must be true?
A. p<q<r
B. p<r<q
C. q<r<p
D. q<p<r
E. r<q<p
Q = {2, 24, 25, 26, 27, 28}
R = {4, 7, 10, 13, 16, 19}
If p is the standard deviation of set P, q is the standard deviation of set Q, and r is the standard deviation of set R, which of the following must be true?
A. p<q<r
B. p<r<q
C. q<r<p
D. q<p<r
E. r<q<p
16 May 2021, 10:11
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Standard deviation increases if the sum of the difference, between each number of the set and the mean average, increases.
At a glance we can see that the mean for set P and set R are 27 and 11.5, The difference between each number and the mean average in set P is smaller than set R, thus we know:
standard deviation of set P < standard deviation of set R .....(I)
The problem is made complex with set Q, as you see there is an outlier in set Q, this, in turn, will drive the mean average lower. Upon calculation, the mean average is 22. Based on this, the sum of the difference between each number of the set Q and the mean average is larger than the set R.
28-22 = 6 19-11.5 = 7.5
27-22 = 5 16-11.5 = 4.5
26-22 = 4 13-11.5 = 1.5
25-22 = 3 11.5-10 = 1.5
24-22 = 2 11.5-7 = 4.5
22-2 = 20 11.5-4 = 7.5
sum difference for set Q: 40
sum difference for set R: 27
Therefore, standard deviation of set R < standard deviation of set Q .....(II)
Combining the (I) and (II), we get that
std. dev. P < std. dev. R < std. dev. Q
And therefore the answer is (B)
At a glance we can see that the mean for set P and set R are 27 and 11.5, The difference between each number and the mean average in set P is smaller than set R, thus we know:
standard deviation of set P < standard deviation of set R .....(I)
The problem is made complex with set Q, as you see there is an outlier in set Q, this, in turn, will drive the mean average lower. Upon calculation, the mean average is 22. Based on this, the sum of the difference between each number of the set Q and the mean average is larger than the set R.
28-22 = 6 19-11.5 = 7.5
27-22 = 5 16-11.5 = 4.5
26-22 = 4 13-11.5 = 1.5
25-22 = 3 11.5-10 = 1.5
24-22 = 2 11.5-7 = 4.5
22-2 = 20 11.5-4 = 7.5
sum difference for set Q: 40
sum difference for set R: 27
Therefore, standard deviation of set R < standard deviation of set Q .....(II)
Combining the (I) and (II), we get that
std. dev. P < std. dev. R < std. dev. Q
And therefore the answer is (B)
16 May 2021, 23:26
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P = {22, 24, 26, 28, 30, 32} Mean P = \(\frac{22 + 32 }{ 2} = 27\) [Consecutive even numbers]
Q = {2, 24, 25, 26, 27, 28} Mean Q = \(\frac{2 + 24 + 25 + 26 + 27 + 28 }{ 6} = 22\)
R = {4, 7, 10, 13, 16, 19} Mean R = \(\frac{4 + 7 + 10 + 13 + 16 + 19 }{ 6} = 11.5\)
S.D. = Take center distance from mean for all sets
S.D. of P = p = {27 - 22, 27 - 24, 27 - 26, 28 - 27, 30 - 27, 32 - 27} = {5, 3, 1, 1, 3, 5}
S.D. of Q = q = {22 - 2, 24 - 22, 25 - 22, 26 - 22, 27 - 22, 28 - 22} = {20, 2, 3, 4, 5, 6}
S.D. of R = r = {11.5 - 4 , 11.5 - 7, 11.5 - 10, 13 - 11.5, 16 - 11.5, 19 - 11.5} = {7.5, 4.5, 1.5, 1.5, 4.5, 7.5}
S.D. of Q(q) >S.D. of R(r)>S.D. of P(p) => q > r > p
=> p < r < q
Answer B
Q = {2, 24, 25, 26, 27, 28} Mean Q = \(\frac{2 + 24 + 25 + 26 + 27 + 28 }{ 6} = 22\)
R = {4, 7, 10, 13, 16, 19} Mean R = \(\frac{4 + 7 + 10 + 13 + 16 + 19 }{ 6} = 11.5\)
S.D. = Take center distance from mean for all sets
S.D. of P = p = {27 - 22, 27 - 24, 27 - 26, 28 - 27, 30 - 27, 32 - 27} = {5, 3, 1, 1, 3, 5}
S.D. of Q = q = {22 - 2, 24 - 22, 25 - 22, 26 - 22, 27 - 22, 28 - 22} = {20, 2, 3, 4, 5, 6}
S.D. of R = r = {11.5 - 4 , 11.5 - 7, 11.5 - 10, 13 - 11.5, 16 - 11.5, 19 - 11.5} = {7.5, 4.5, 1.5, 1.5, 4.5, 7.5}
S.D. of Q(q) >S.D. of R(r)>S.D. of P(p) => q > r > p
=> p < r < q
Answer B
