Question

Consider a binomial experiment with 16 trials and probability 0.60 of success on a single trial.

(a) Use the binomial distribution to find the probability of exactly 10 successes.

(b) Use the normal distribution to approximate the probability of exactly 10 successes.

(c) Compare the results of parts (a) and (b).

Answer 1

Solution

Given that ,

p = 0.60

1 - p = 0.40

n = 16

x = 10

a)

Using binomial probability formula ,

P(X = x) = ((n! / x! (n - x)!) * p^x * (1 - p)^{n - x}

P(X = 10) = ((16! / 10! (16-10)!) * 0.60¹⁰ * (0.40)^16-10

=  ((16! / 10! (6)!) * 0.60¹⁰ * (0.40)⁶

= 0.1983

Probability = 0.1983

b)

Here, BIN ( n , P ) that is , BIN (16 , 0.60)

then,

n*p = 16 * 0.60 = 9.6

n(1- P) = 16 * 0.40 = 6.4

According to normal approximation binomial,

X Normal

Mean = = n*P = 9.6

Standard deviation = =n*p*(1-p) = 16 * 0.60 * 0.40= 3.84

We using continuity correction factor

P(X = a) = P( a - 0.5 < X < a + 0.5)

P(9.5 < x < 10.5) = P((9.5 - 9.6)/ 3.84) < (x - ) /  < (10.5 - 9.6) / 3.84) )

= P(-0.05 < z < 0.46)

= P(z < 0.46) - P(z < -0.05)

= 0.6772 - 0.4801

= 0.1971

Probability = 0.1971

c)

The results of Part a) is more accurate than part b).