Let a random variable X have a binomial distribution with mean 8 and variance 4.

Question

Let a random variable X have a binomial distribution with mean 8 and variance 4. If $$P\left( {X \le 2} \right) = {k \over {{2^{16}}}}$$, then k
is equal to :

Accepted Answer

Let number of trials be n and probability of success = p, probability of failure = q

Given np = 8, npq = 4

$$ \Rightarrow $$ q = $${1 \over 2}$$, p = $${1 \over 2}$$, n = 16 (as p + q = 1)

p$$\left( {x \le 2} \right)$$ = $${{{}^{16}{C_0} + {}^{16}{C_1} + {}^{16}{C_2}} \over {{2^{16}}}} = {{1 + 16 + 120} \over {{2^{16}}}} = {{137} \over {{2^{16}}}}$$

Solutions · 1