JEE MAIN - Mathematics (2019 - 12th January Evening Slot - No. 11)
The mean and the variance of five observations are 4 and 5.20, respectively. If three of the observations are
3, 4 and 4 ; then the absolute value of the difference of the other two observations, is :
1
7
3
5
Explanation
mean $$\overline x $$ = 4, $$\sigma $$2 = 5.2, n = 5, . x1 = 3 x2 = 4 = x3
$$\sum {{x_i}} = 20$$
x4 + x5 = 9 . . . . . . (i)
$${{\sum {x_i^2} } \over x} - {\left( {\overline x } \right)^2} = \sigma \Rightarrow \sum {x_i^2} = 106$$
$$x_4^2 + x_5^2 = 65$$ . . . . . .(ii)
Using (i) and (ii) (x4 $$-$$ x5)2 = 49
$$\left| {{x_4} - {x_5}} \right| = 7$$
$$\sum {{x_i}} = 20$$
x4 + x5 = 9 . . . . . . (i)
$${{\sum {x_i^2} } \over x} - {\left( {\overline x } \right)^2} = \sigma \Rightarrow \sum {x_i^2} = 106$$
$$x_4^2 + x_5^2 = 65$$ . . . . . .(ii)
Using (i) and (ii) (x4 $$-$$ x5)2 = 49
$$\left| {{x_4} - {x_5}} \right| = 7$$
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