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In a distribution of 8500 parameters, 26.7 has $56^{\text {th $ perc
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18 Mar 2025, 01:07
18 Mar 2025, 01:07
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In a distribution of 8500 parameters, 26.7 has $\(56^{\text {th \)$ percentile and 37.1 has $\(78^{\text {th \)$ percentile. If $A$ is the set of parameters having values between 26.7 and 37.1 , what is the number of terms in set $A$ ?
(A) 1869
(B) 2475
(C) 3657
(D) 4741
(E) 6629
(A) 1869
(B) 2475
(C) 3657
(D) 4741
(E) 6629
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Re: In a distribution of 8500 parameters, 26.7 has $56^{\text {th $ perc
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22 Mar 2025, 04:15
22 Mar 2025, 04:15
Expert Reply
OFFICIAL EXPLANATION
We know that in a distribution of 8500 parameters, 26.7 has $\(56^{\text {th \)$ percentile which implies $\(56 \%\)$ of $\(8500=0.56 \times 8500=4760\)$ parameters are less than 26.7
Also as 37.1 has $\(78^{\text {th \)$ percentile which means there are $\(78 \%$ of $8500=0.78 \times 8500=6630\)$ parameters less than 37.1
Note:-6630 values less than 37.1 include 26.7 which is not included in 4760
Finally the values exactly between $\(26.7 \& 37.1\)$ will be $\(6630-4760-1=1869\)$ ( 1 is subtracted as 26.7 was included in the numbers).
Hence the answer is (A).
We know that in a distribution of 8500 parameters, 26.7 has $\(56^{\text {th \)$ percentile which implies $\(56 \%\)$ of $\(8500=0.56 \times 8500=4760\)$ parameters are less than 26.7
Also as 37.1 has $\(78^{\text {th \)$ percentile which means there are $\(78 \%$ of $8500=0.78 \times 8500=6630\)$ parameters less than 37.1
Note:-6630 values less than 37.1 include 26.7 which is not included in 4760
Finally the values exactly between $\(26.7 \& 37.1\)$ will be $\(6630-4760-1=1869\)$ ( 1 is subtracted as 26.7 was included in the numbers).
Hence the answer is (A).
gmatclubot
Re: In a distribution of 8500 parameters, 26.7 has $56^{\text {th $ perc [#permalink]
22 Mar 2025, 04:15
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