The mean and variance of the data 4, 5, 6, 6, 7, 8, x, y, where x
JEE · Math · previous-year question
- A.162
- B.320correct
- C.674
- D.420
Answer
B. 320
Explanation
Mean $$ = {{4 + 5 + 6 + 6 + 7 + 8 + x + y} \over 8} = 6$$ $$\therefore$$ $$x + y = 12$$ ..... (i) And variance $$ = {{{2^2} + {1^2} + {0^2} + {0^2} + {1^2} + {2^2} + {{(x - 6)}^2} + {{(y - 6)}^2}} \over 8}$$ $$ = {9 \over 4}$$ $$\therefore$$ $${(x - 6)^2} + {(y - 6)^2} = 8$$ ..... (ii) From (i) and (ii) x = 4 and y = 8 $$\therefore$$ $${x^4} + {y^2} = 320$$
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