Question

Consider the probability that more than 92 out of 157 flights will be on-time. Assume the probability that a given flight will be on-time is 62%.

Approximate the probability using the normal distribution. Round your answer to four decimal places.

Answer 1

Solution:

Given that,

p = 62% = 0.62

1 - p = 1 - 0.62 = 0.38

n = 157

Here, BIN ( n , p ) that is , BIN (157 , 0.62)

According to normal approximation binomial,

X Normal

Mean = = n*p = 157 * 0.62 = 97.34

Standard deviation = =n*p*(1-p) = 157*0.62*0.38 = 36.9892 = 6.0818747

Find P(X > 92)

= P(X > 92.5) ...continuity correction

=  P((x - ) / < (92.5 - 97.34) / 6.0818747)

= P(Z > -0.796)

= 1 - P(Z < -0.796)

= 1 - 0.2130

= 0.7870

Probability that more than 92 out of 157 flights will be on-time is 0.7870