Let the mean of 6 observations 1, 2, 4, 5, and be 5 and their variance be 10. Then their mean deviation about the mean is equal to [2023]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
Let the mean of 6 observations 1, 2, 4, 5, $x$ and $y$ be 5 and their variance be 10. Then their mean deviation about the mean is equal to [2023]

1 $\frac{7}{3}$

2 $\frac{8}{3}$

3 $\frac{10}{3}$

4 $3$

Ans.
(2)

$Mean (\bar{x}) = \frac{1 + 2 + 4 + 5 + x + y}{6} = 5$

$\Rightarrow x + y = 18 (i)$

$Variance = 10$

$\Rightarrow \frac{1^{2} + 2^{2} + 4^{2} + 5^{2} + x^{2} + y^{2}}{6} - 25 = 10$

$\Rightarrow x^{2} + y^{2} = 164 (ii)$

$Solving (i) and (ii), we get:$

$x = 10, y = 8$

$Now, M . D (\bar{x}) = \frac{\sum | x_{i} - \bar{x} |}{6} = \frac{4 + 3 + 1 + 0 + 5 + 3}{6} = \frac{16}{6} = \frac{8}{3}$

Want Us to Email you About Special Offers & Updates?