Question

Given a binomial random variable with n​ = 100 and p​ = 0.7​, estimate the​ Pr[X ≤ 80​]

Answer 1

Solution:

Given that,

P = 0.7

1 - P = 0.3

n = 100

Here, $X\sim$ BIN ( n , P ) that is , BIN (100 , 0.7)

then,

n*p = 100*0.7 = 70 > 5

n(1- P) = 100*0.3 = 30 > 5

According to normal approximation binomial,

X $\rightarrow$ Normal

Mean = $\mu$ = n*P = 100*0.70 = 70

Standard deviation = $\sigma$ =$\sqrt{}$n*p*(1-p) = $\sqrt{}$ 100*0.70*0.30 = $\sqrt{}$ 21

We using countinuity correction factor

P( X $\leq$ a ) = P(X < a + 0.5)

P(x < 80.5) = P((x - $\mu$ ) / $\sigma$ < (80.5 - 70 ) / $\sqrt{}$ 21)

= P(z < 2.29)

Probability = 0.9890