Let the mean and standard deviation of marks of class A of 100 students be respectively 40 and \(\alpha(>0)\), and the mean and standard deviation of marks of class B of n students be respectively 55 and \(30-\alpha\). If the mean and variance of the marks of the combined class of \(100+\mathrm{n}\) students are respectively 50 and 350, then the sum of variances of classes A and B is:

JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Statistics

Let the mean and standard deviation of marks of class A of 100 students be respectively 40 and

α (> 0)

, and the mean and standard deviation of marks of class B of n students be respectively 55 and

30 - α

. If the mean and variance of the marks of the combined class of

100 + n

students are respectively 50 and 350, then the sum of variances of classes A and B is:

Answer: (A), (D)

The combined mean

\frac{100 \cdot 40 + 55 n}{100 + n} = 50

gives

n = 200.

With class deviations

40 - 50 = - 10

and

55 - 50 = 5

, the combined variance

\frac{100 ( α ^{2} + 100 ) + 200 (( 30 - α ) ^{2} + 25 )}{300} = 350

reduces to

α^{2} - 40 α + 300 = 0

, so

α = 10

α = 30.

For

α = 10

the two variances are

100

and

400

, summing to

500

; for

α = 30

they are

900

and

0

, summing to

900.

The sum of variances is

500

900.

JEE Main · 2023 · 31 Jan 2023 · Shift 2 · One or more correct options