Use the normal approximation to the binomial to approximate the probability **STATS TUTORS ONLY**

Question

A certain flight arrives on time 80 percent of the time. Suppose 140 flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that 

​(a) exactly 115 flights are on time. P(115)=

​(b) at least 115 flights are on time.

​(c) fewer than 117 flights are on time.

​(d) between 117 and 120​, inclusive are on time.

Answer 1

First, we need to compute the mean and standard deviation for our sample.

For the binomial approximation of the normal distribution, the mean = sample size * population proportion.

Our sample size is 140 and our population proportion is 0.8 (80%).

So the mean = 0.8 * 140 = 112

The standard deviation = square root(sample size * population proportion * (1-population proportion)) =

square root (140 * 0.8 * 0.2) = 4.73

For a binomial variable X, the z statistic computed as (X - mean) / standard deviation

part a) For the normal distribution the probabilities are only computed for intervals, which means the probabilities for individual points are zero. so P(115) = 0. If we didn't use the normal approximation , we could use combinations to compute the probability =

115 25

C (0.8) (0.2) = 0.07

140 115

part b) Here we compute the z-statistic, where X = 115:

Z = (115 - 112) / 4.73 = 0.634

probability at least 115 flights on are on time = P(X > 115)

standardizing X to get Z statistic we get P(Z > 0.634)

since the normal distribution is symmetric P(Z > 0.634) is the same as P(Z < -0.634) = 0.263 (from normal probability tables or using calculator)

part c) Here we compute the z-statistic, where X = 117

Z = (117-112)/4.73 = 1.057

probability fewer than 117 flights on are time = P(X < 117)

standardizing X to get Z statistic we get P (Z < 1.057) = 0.855

part d) Here we need to compute the z-statistics for the lower and upper bounds, X = 117 and 120.

We have already computed the z-statistic for X = 117 as 1.057

The z-statistic for X = 120

Z = (120-112)/4.73 = 1.691

probability between 117 and 120 inclusive, flights are on time P( 117 < X < 120)

= P(X < 120) - P(X < 117)

standardizing x to get z-statistic we get:

P(Z < 1.691) - P(Z < 1.057) = 0.955 - 0.855 = 0.1