Question

If mean and median of the distribution $a,b,c,d,e,f$ ( where $b>e>f>c>d>a$) are $20$ and $14$ respectively then mean of numbers $a-f,b-c,e,d$ will be

• A. $15.33$
• B. $16$
• C. $10.66$
• D. None

Answer 1

The correct option is D None

Mean is the sum of the data values divided by the total number of data values.

Median is the middle value of the data values after arranging the data values in the ascending or descending order.

If $n$ is odd, then the median is $(\frac{n+1}{2}{)}^{th}$ value.

If $n$ is even, then the median is ${\frac{(\frac{n}{2}{)}^{th}+(\frac{n}{2}+1)}{2}}^{th}$ value.

Given data is $a,b,c,d,e,f$ and $b>e>f>c>d>a$

Therefore, arranging the data in ascending order is $b,e,f,c,d,a$

Given that mean of the data $=20$

$⟹\frac{a+b+c+d+e+f}{6}=20$

$⟹a+b+c+d+e+f=120$ -------(1)

Given that median of the data $=14$

total number of values $=n=6$ (even)

Median = ${\frac{(\frac{n}{2}+\frac{n+1}{2})}{2}}^{th}$ value = ${\frac{(\frac{6}{2}{)}^{th}+(\frac{6}{2}+1)}{2}}^{th}$ value= ${\frac{(3{)}^{rd}+(4)}{2}}^{th}$ value =$\frac{f+c}{2}$

$⟹\frac{f+c}{2}=14$

$⟹f+c=28$ ------(2)

Therefore, mean of the numbers $a-f,b-c,e,d$ is $\frac{a-f+b-c+e+d}{4}$

$⟹mean=\frac{a+b+e+d-(f+c)}{4}$

$⟹mean=\frac{120-(f+c)-(f+c)}{4}$ (from (1))

$⟹mean=\frac{120-2\left(28\right)}{4}$

$⟹mean=\frac{64}{4}=16$