Statistics

JEE Main 2024 (Online) 27th January Morning ShiftMCQEasy📈 Topic in 45% papers+4−1

Ques. 1JEE Main 2024, 27 January (Morning Shift)

Let a1, a2, a10 be 10 observations such that k=110 ak=50 and k<j ak aj=1100. Then the standard deviation of a1, a2, , a10 is equal to :

Answer

(4) 5

Solution

( ak )2= ak2+2 k<jak aj

⇒ 2500= ak2+2200 ⇒ ak2=300

∵ 2= ak2n- ( akn )2

= 30010- ( 5010 )2=30-25=5

∴ = 5

Answer: (4)

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Ques. 2JEE Main 2024, 27 January (Evening Shift)

The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking it was found that an observation was read as 10 in place of 12 . If and 2 denote the mean and variance of the correct observations respectively, then 15 ( + 2+ 2 ) is equal to .

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