JEE Main 2022MathematicsStatisticsEasyMCQ

JEE Main 2022 — Statistics Question with Solution

JEE Main 2022 (29 Jun Shift 1)

Question

Let the mean and the variance of 5 observations x1,x2,x3,x4,x5 be 245 and 19425 respectively. If the mean and variance of the first 4 observation are 72 and a respectively, then 4a+x5 is equal to

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Show full solutionCorrect option: B
Correct answer
B15

Step-by-step explanation

Mean x¯=∑xi5=245⇒∑xi=24   ...i

Variance σ2=∑xi25-2452=19425

⇒∑xi2=154   ...ii

Also given meanx1+x2+x3+x44=72⇒x1+x2+x3+x4=14

⇒x5=10 (from i)

and variance σ2=x12+x22+x32+x424-494=a

⇒x12+x22+x32+x42=4a+49

⇒x52=154-4a-49 (from ii)

⇒100=105-4a⇒4a=5

Hence 4a+x5=15

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About this question

This is a previous-year question from JEE Main 2022, covering the Statistics chapter of Mathematics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.