Question

Suppose there are 54% female students on CMU campus. A random sample of 100 students was obtained. What is the probability there will be equal to or more than 58 female students?

Answer 1

Solution:

Given that,

P = 0.54

1 - P = 0.46

n = 100

Here, BIN ( n , P ) that is , BIN (100 , 0.54)

then,

n*p = 100 * 0.54 = 54 > 5

n(1- P) = 100 * 0.46 = 46 > 5

According to normal approximation binomial,

X Normal

Mean = = n*P = 54

Standard deviation = =n*p*(1-p) = 100 * 0.54 * 0.46 = 24.84

We using continuity correction factor

P(X a ) = P(X > a - 0.5)

P(x > 57.5) = 1 - P(x < 57.5)

= 1 - P((x - ) / < (57.5 - 54) / 24.84)

= 1 - P(z < 0.70)

= 1 - 0.7580   

= 0.2420

Probability = 0.2420