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JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Probability

Let X be a random variable with distribution.

x	-2	-1	3	4	6
P(X=x)	$\frac{1}{5}$	a	$\frac{1}{3}$	$\frac{1}{5}$	b

If the mean of X is 2.3 and variance of X is

σ^{2}

, then

100 σ^{2}

is equal to:

Answer: 781

Solution

From

\sum P = 1

:

\frac{1}{5} + a + \frac{1}{3} + \frac{1}{5} + b = 1

, so

a + b = \frac{4}{15}

. The mean gives

- \frac{2}{5} - a + 1 + \frac{4}{5} + 6 b = 2.3

, i.e.

- a + 6 b = 0.9

. Solving the two equations yields

a = \frac{1}{10}

and

b = \frac{1}{6}

. Then

E (X^{2}) = \frac{4}{5} + a + 3 + \frac{16}{5} + 36 b = 7 + 0.1 + 6 = 13.1,

so

σ^{2} = E (X^{2}) - (E (X))^{2} = 13.1 - (2.3)^{2} = 7.81

, and

100 σ^{2} = 781

.

JEE Main · 2021 · 01 Sep 2021 · Shift 2 · Numerical answer

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