How do you solve this?

Question

A certain flight arrives on time

88

percent of the time. Suppose

124

flights are randomly selected. Use the normal approximation to the binomial to approximate the probability that

​(a) exactly

120

flights are on time.

​(b) at least

120

flights are on time.

​(c) fewer than

98

flights are on time.

​(d) between

98

and

120

inclusive are on time.

(a)

​P(1120​)equals=nothing

​(Round to four decimal places as​ needed.)

​(b)

​P(Xgreater than or equals≥120​)equals=nothing

​(Round to four decimal places as​ needed.)

​(c)

​P(Xless than<98​)equals=nothing

​(Round to four decimal places as​ needed.)

​(d)

​P(98less than or equals≤Xless than or equals≤1120​)equals=nothing

​(Round to four decimal places as​ needed.)

Answer 1

To use the normal approximation you will need to compute the mean (sample size * proportion "success") and standard deviation (square root of (sample size * proportion "success" * (1 - proportion "success")).

Let X be binomial variable = number of flights on time.

a) You can only compute probabilities under the continuous normal curve for intervals, so the probability of a single point (120) is zero.

b) You want to compute P(X > 120). Standardizing to z-score you get: P(Z > (120 - mean)/standard deviation), which you can compute from normal distribution tables or by using a calculator.

c) Here you want to compute P(X < 98). You standardize to z-score and compute probability as in b).

d) Here you want to compute P(98 < X 120). Just compute 1 - the sum of probabilities from b) and c).