JEE Main 2019MathematicsStatisticsMediumMCQ

JEE Main 2019 — Statistics Question with Solution

JEE Main 2019 (10 Jan Shift 2)

Question

If the mean and standard deviation of 5 observations x1, x2, x3, x4, x5 are 10 and 3, respectively, then the variance of 6 observations x1, x2, …, x5 and -50 is equal to

Choose an option

Show full solutionCorrect option: B
Correct answer
B507.5

Step-by-step explanation

We have,

Mean =x1+x2+x3+x4+x55=10

⇒x1+x2+x3+x4+x5=50   ...i 

Variance

=S.D2=Σxi25-x¯2=9 

⇒x12+x22+x32+x42+x525-100=9

⇒x12+x22+x32+x42+x52=545   ...ii 

Now,

x¯new=x1+x2+x3+x4+x5-506=0 

New variance

=∑i=16xi26-x¯new2 

=x12+x22+x32+x42+x52+25006-0

=545+25006=30456=507.5

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Statistics chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2019, 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.