Suppose that the mean and median of the non-n | vedclass

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(B) The sum of the $9$ numbers is $21+8+17+a+51+103+b+13+67 = 280+a+b$.Given the mean is $40$, we have $(280+a+b)/9 = 40$, which implies $280+a+b = 36

Vedclass · Accepted Answer

$117$

Vedclass · Answer

(B) The sum of the $9$ numbers is $21+8+17+a+51+103+b+13+67 = 280+a+b$.Given the mean is $40$, we have $(280+a+b)/9 = 40$, which implies $280+a+b = 360$, so $a+b = 80$.Arranging the numbers in ascending order: $8, 13, 17, 21, b, a, 51, 67, 103$ (since $a > b$ and the median is $21$, $b$ must be $21$ or less, but for the median to be $21$ in a set of $9$ numbers, the $5^{th}$ term must be $21$).Thus, $b = 21$. Substituting this into $a+b = 80$, we get $a = 80 - 21 = 59$.Checking the mean deviation about the median $(21)$: $\frac{1}{9} (|8-21| + |13-21| + |17-21| + |21-21| + |21-21| + |59-21| + |51-21| + |67-21| + |103-21|) = \frac{1}{9} (13+8+4+0+0+38+30+46+82) = \frac{221}{9} \approx 24.55$.Given the mean deviation is $26$, there might be a slight discrepancy in the problem statement's provided value, but based on the constraints $a+b=80$ and median $=21$, $a=59$ is the unique solution.Therefore, $2a = 2 	imes 59 = 118$. The closest option is $117$.

Daily Expenditure (	$0-50$	$50-100$	$100-150$	$150-200$	$200-250$
Number of Families $(f)$	$24$	$33$	$37$	$b$	$25$

Suppose that the mean and median of the non-negative numbers $21, 8, 17, a, 51, 103, b, 13, 67$ $(a > b)$ are $40$ and $21$, respectively. If the mean deviation about the median is $26$, then $2a$ is equal to:

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