JAWAHARLAL NEHRU TECHNOLOGICAL UNIVERSITY HYDERABAD B.Tech I Year MATHEMATICS-III Previous Question Paper (November/December - 2020 - R16)

A binomial distribution represents the number of successes in n independent Bernoulli trials, each with a probability of success p and failure q = 1 - p. The probability mass function is given by P(X = x) = ⁿC_x p^x q^n-x for x = 0, 1, 2, ..., n.

The mean (expected value) E[X] is defined as:

Since the term for x = 0 is zero, we start the summation from x = 1:

Using the identity x / x! = 1 / (x-1)!, we simplify:

Factor out np from the summation:

Let k = x - 1 and m = n - 1. As x goes from 1 to n, k goes from 0 to m:

The summation term represents the binomial expansion of (p + q)^m. Since p + q = 1:

Therefore, the mean of the binomial distribution is np.

The given probability mass function is f(x) = (1/8) * 3Cx for x = 0, 1, 2, 3. This corresponds to a Binomial distribution B(n, p) with n = 3 and p = 0.5, since 3C0=1, 3C1=3, 3C2=3, 3C3=1, and the sum is (1+3+3+1)/8 = 1.

The moment generating function (MGF) of a Binomial distribution is given by M_X(t) = (q + pe^t)^n. Substituting p = 0.5, q = 0.5, and n = 3:

To find the mean and variance, we can use the properties of the Binomial distribution or the derivatives of the MGF:

1. Mean (E[X]): The mean is given by np. E[X] = 3 * 0.5 = 1.5.
2. Variance (Var(X)): The variance is given by npq. Var(X) = 3 * 0.5 * 0.5 = 0.75.

Alternatively, using the MGF: M'_X(t) = (1/8) * 3(1 + e^t)^2 * e^t. At t=0, E[X] = M'_X(0) = (3/8)(2^2) = 1.5. M''_X(t) = (3/8) * [2(1 + e^t) * e^t * e^t + (1 + e^t)^2 * e^t]. At t=0, E[X^2] = M''_X(0) = (3/8) * [2(2) + 4] = (3/8) * 8 = 3. Variance = E[X^2] - (E[X])^2 = 3 - (1.5)^2 = 3 - 2.25 = 0.75.