Bunuel
Ten pressure measurements were taken for the liquid flowing through a pipe. The average measurement was 20psi(pounds per square inch). Was the standard deviation of the 10 measurements greater than 2?
(1) Each measurement differed from the mean by less than 10%
(2) The variance of the measurements was 3.75
For the S.D to be greater than \(2\), the variance should be greater than \(4\). SD = \(\sqrt{ variance}\)
For the variance to be greater than 4, the sum of the squares of difference from the mean of all \(10\) measurements should be greater than 40.(1) Each measurement differed from the mean by less than 10%To make it easier to calculate let us consider \(10\%\) rather than less than \(10\%.\)
Hence if each measurement was \(10\%\) of mean, then:
Highest measurement \(= 1.1*20 =22\)
Lowest measurement \(= .90*20=18 \)
Max deviation from mean for the highest measurement \(= 20-22 < 2\) ( Because it's less than \(10\%,\) but we considered \(10\%\))
Max deviation from mean for the lowest measurement\(= 20-18 <2\) ( Because it's less than \(10\%,\) but we considered \(10\%\)))
Hence difference of each measurement from mean even at max would be \(< 2\)
Thus squares of the above differences will at max be \(< 4 \)
Thus the sum of the above squares of all \(10\) data points will at max be \(< 40\)
Thus the variance will at max be less than \(4\)
If the variance is less than \(4\) then S.D has to be less than \(2\).
SUFF.(2) The variance of the measurements was 3.75This option clearly tells that the variance is less than \(4\). Hence S.D has to be less than \(2.\)
SUFF.Ans D
Hope it's clear.