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13.Statistics
hard
The mean of five observations is $5$ and their variance is $9.20$. If three of the given five observations are $1, 3$ and $8$, then a ratio of other two observations is
A
$10 : 3$
B
$4 : 9$
C
$5 : 8$
D
$6 : 7$
(JEE MAIN-2019)
Solution
$\mu = \frac{{1 + 3 + 8 + x + y}}{5}$
$25 = 12 + x + y \Rightarrow x + y = 13\,\,\,\,\,\,\,\,……..\left( 1 \right)$
${\sigma ^2} = \frac{{\sum {{{\left( {{x_i} – \mu } \right)}^2}} }}{N}$
$9.2 = \frac{{1 + 9 + 64 + {x^2} + {y^2}}}{5} – 25$
$34.2 \times 5 = 74 + {x^2} + {y^2}$
$171 = 74 + {x^2} + {y^2}$
$97 = {x^2} + {y^2}……….\left( 2 \right)$
${\left( {x + y} \right)^2} = {x^2} + {y^2} + 2xy$
$169 – 97 = 2xy \Rightarrow xy = 36$
$T = 4,9$
So, ratio is $\frac{4}{9}$ or $\frac{9}{4}$
Standard 11
Mathematics
