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Maths Question 9 – JEE-MAIN 2025

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Let the Mean and Variance of five observations x1=1,x2=3,x3=a,x4=7 and x5=b,a>b, be 5 and 10 respectively. Then the Variance of the observations xn+n,n=1,2,.....,5 is

Recall the definitions of mean and variance for a set of observations to set up initial equations.

Step 1: Determine the unknown observations a and b✦ Active

The mean of the five observations x1=1,x2=3,x3=a,x4=7,x5=b is given as 5. Using the mean formula:

1+3+a+7+b5=5⟹11+a+b=25⟹a+b=14(1)

The variance is given as 10. Using the variance formula σ2=∑xi2N−(x¯)2:

12+32+a2+72+b25−52=10⟹1+9+a2+49+b25−25=10 59+a2+b25=35⟹59+a2+b2=175⟹a2+b2=116(2)

Substitute b=14−a from (1) into (2) and solve the quadratic equation for a:

a2+(14−a)2=116⟹a2+196−28a+a2=116 2a2−28a+80=0⟹a2−14a+40=0 (a−4)(a−10)=0

This gives a=4 or a=10. If a=4, b=10. If a=10, b=4. Given the condition a>b, we choose a=10 and b=4. The original observations are x1=1,x2=3,x3=10,x4=7,x5=4.

Step 2: Calculate the new observations yn=xn+n○ Expand

The new observations are formed by adding n to each xn:

y1=x1+1=1+1=2 y2=x2+2=3+2=5 y3=x3+3=10+3=13 y4=x4+4=7+4=11 y5=x5+5=4+5=9

The new set of observations is {2,5,13,11,9}.

Step 3: Calculate the Variance of the new observations○ Expand

First, find the mean of the new observations, y¯:

y¯=2+5+13+11+95=405=8

Now, calculate the variance σy2 using the formula σy2=∑yi2N−(y¯)2:

∑yi2=22+52+132+112+92=4+25+169+121+81=400 σy2=4005−(8)2=80−64=16

The variance of the new observations is 16.

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