13.Statistics
hard

सात प्रेक्षणों के माध्य तथा प्रसरण क्रमशः $8$ तथा $16$ है। यदि इनमें से $5$ प्रेक्षण $2,4,10,12,14$ है, तो शेष दो प्रेक्षणों का गुणनफल है 

A

$40$

B

$45$

C

$49$

D

$48$

(JEE MAIN-2019)

Solution

Let $7$ observation be ${x_1},{x_2},{x_3},{x_4},{x_5},{x_6},{x_7}$

$\bar x = 8 \Rightarrow \sum\limits_{i = 1}^7 {{x_i}}  = 56\,\,\,\,\,\,…….\left( 1 \right)$

Also ${\sigma ^2} = 16$

$ \Rightarrow 16 = \frac{1}{7}\left( {\sum\limits_{i = 1}^7 {x_i^2} } \right) – {\left( {\bar x} \right)^2}$

$ \Rightarrow 16 = \frac{1}{7}\left( {\sum\limits_{i = 1}^7 {x_i^2} } \right) – 64$

$ \Rightarrow \left( {\sum\limits_{i = 1}^7 {x_i^2} } \right) = 560\,\,\,\,\,\,\,\,\,…….\left( 2 \right)$

Now, ${x_1} = 2,{x_2} = 4,{x_3} = 10,{x_4} = 12,{x_5} = 14$

$ \Rightarrow {x_6} + {x_7} = 14$ (from $(1)$) and $x_6^2 + x_7^2 = 100$ (from$(2)$)

$\therefore x_6^2 + x_7^2 = {\left( {{x_6} + {x_7}} \right)^2} – 2{x_6}{x_7} \Rightarrow {x_6}{x_7} = 48$

Standard 11
Mathematics

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