JEE MAIN - Mathematics (2016 - 10th April Morning Slot - No. 11)
The mean of 5 observations is 5 and their variance is 124. If three of the observations
are 1, 2 and 6 ; then the mean deviation from the mean of the data is :
2.4
2.8
2.5
2.6
Explanation
Let 5 observations are x1, x2, x3, x4, x5
given, x1 = 1, x2 = 2, x3 = 6
Mean = 5
$$ \therefore $$ Mean$$\left( {\overline x } \right)$$ = $${{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}} \over 5}$$ = 5
$$ \Rightarrow $$ 1 + 2 + 6 + x4 + x5 = 25
$$ \therefore $$ x4 + x5 = 16
$$ \Rightarrow $$ (x4 $$-$$ 5) + (x5 $$-$$ 5) + 10 = 16
$$ \Rightarrow $$ (x4 $$-$$ 5) + (x5 $$-$$ 5) = 6
$$ \therefore $$ Mean deviation about mean,
= $${{\sum {\left| {{x_i} - \overline x } \right|} } \over n}$$
= $${{\left| {1 - 5} \right| + \left| {2 - 5} \right| + \left| {6 - 5} \right| + \left| {{x_4} - 5} \right| + \left| {{x_5} - 5} \right|} \over 5}$$
= $${{4 + 3 + 1 + 6} \over 5}$$
= $${{14} \over 5}$$
= 2.8
given, x1 = 1, x2 = 2, x3 = 6
Mean = 5
$$ \therefore $$ Mean$$\left( {\overline x } \right)$$ = $${{{x_1} + {x_2} + {x_3} + {x_4} + {x_5}} \over 5}$$ = 5
$$ \Rightarrow $$ 1 + 2 + 6 + x4 + x5 = 25
$$ \therefore $$ x4 + x5 = 16
$$ \Rightarrow $$ (x4 $$-$$ 5) + (x5 $$-$$ 5) + 10 = 16
$$ \Rightarrow $$ (x4 $$-$$ 5) + (x5 $$-$$ 5) = 6
$$ \therefore $$ Mean deviation about mean,
= $${{\sum {\left| {{x_i} - \overline x } \right|} } \over n}$$
= $${{\left| {1 - 5} \right| + \left| {2 - 5} \right| + \left| {6 - 5} \right| + \left| {{x_4} - 5} \right| + \left| {{x_5} - 5} \right|} \over 5}$$
= $${{4 + 3 + 1 + 6} \over 5}$$
= $${{14} \over 5}$$
= 2.8
Comments (0)
