The mean and variance of 10 observations are | vedclass

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(A) Let the two missing observations be $a$ and $b$. The sum of the $10$ observations is $10 	imes 9 = 90$. Sum of given $8$ observations $= 2+3+5+10

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) Let the two missing observations be $a$ and $b$. The sum of the $10$ observations is $10 	imes 9 = 90$. Sum of given $8$ observations $= 2+3+5+10+11+13+15+21 = 80$. So, $a+b = 90 - 80 = 10$. Given variance $\sigma^2 = 34.2$. The formula for variance is $\frac{\Sigma x_i^2}{n} - (\bar{x})^2 = 34.2$. $\frac{2^2+3^2+5^2+10^2+11^2+13^2+15^2+21^2+a^2+b^2}{10} - 9^2 = 34.2$. $\frac{4+9+25+100+121+169+225+441+a^2+b^2}{10} - 81 = 34.2$. $1094 + a^2 + b^2 = 1152 \Rightarrow a^2 + b^2 = 58$. Since $a+b=10$ and $a^2+b^2=58$, we solve to get $a=3$ and $b=7$. The $10$ observations are $2, 3, 3, 5, 7, 10, 11, 13, 15, 21$. The median is the average of the $5^{th}$ and $6^{th}$ terms: $\frac{7+10}{2} = 8.5$. Mean deviation about median $= \frac{\Sigma |x_i - 8.5|}{10} = \frac{|2-8.5| + |3-8.5| + |3-8.5| + |5-8.5| + |7-8.5| + |10-8.5| + |11-8.5| + |13-8.5| + |15-8.5| + |21-8.5|}{10}$. $= \frac{6.5 + 5.5 + 5.5 + 3.5 + 1.5 + 1.5 + 2.5 + 4.5 + 6.5 + 12.5}{10} = \frac{50}{10} = 5$.

Class Interval	$0-4$	$4-8$	$8-12$	$12-16$
Frequency	$4$	$3$	$2$	$1$

Marks obtained	$0-20$	$20-40$	$40-60$	$60-80$	$80-100$
Number of students	$10$	$8$	$12$	$9$	$11$

The mean and variance of $10$ observations are $9$ and $34.2$, respectively. If $8$ of these observations are $2, 3, 5, 10, 11, 13, 15, 21$, then the mean deviation about the median of all the $10$ observations is

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