A Travel Weekly International Air Transport Association survey asked business travelers about the purpose for their most recent business trip. 19% responded that it was for an internal company visit. Suppose 950 business travelers are randomly selected.

a. What is the probability that more than 25% of the business travelers say that the reason for their most recent business trip was an internal company visit?

b. What is the probability that between 15% and 20% of the business travelers say that the reason for their most recent business trip was an internal company visit?

c. What is the probability that between 133 and 171 of the business travelers say that the reason for their most recent business trip was an internal company visit?

(a) P (p' > 0.25) = 0.

(b) P (0.15 < p' < 0.20) = 0.7784.

(c) P (133 < X < 171) = 0.2205

Step-by-step explanation:

Let p = proportion of business trip that were for  internal company visit.

It is provided that 19% responded that it was for an internal company visit, i.e.

p = 0.19.

A sample of n = 950 business travelers are randomly selected.

According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

The standard deviation of this sampling distribution of sample proportion is:

Thus, the distribution of is N (μ = 0.19, σ = 0.013).

(a)

Compute the probability that more than 25% of the business travelers say that the reason for their most recent business trip was an internal company visit as follows:

P (p' > 0.25) = P (Z > 4.62) = 0

*Use a z-table for the probability.

Thus, the probability that more than 25% of the business travelers say that the reason for their most recent business trip was an internal company visit is 0.

(b)

Compute the probability that between 15% and 20% of the business travelers say that the reason for their most recent business trip was an internal company visit as follows:

P (0.15 < p' < 0.20) = P (-3.08 < Z < 0.77)

                               = P (Z < 0.77) - P (Z < -3.08)

                               = 0.7794 - 0.0010

                               = 0.7784

*Use a z-table for the probability.

Thus, the probability that between 15% and 20% of the business travelers say that the reason for their most recent business trip was an internal company visit is 0.7784.

(c)

Compute the proportion of X = 133  and X = 171 as follows:

p' = 133/950 = 0.14

p' = 171/950 = 0.18

Compute the value of P (0.14 < p' < 0.20) as follows:

P (0.14 < p' < 0.18) = P (-3.85< Z < -0.77)

                               = P (Z < -0.77) - P (Z < -3.85)

                               = 0.2206- 0.0001

                               = 0.2205

*Use a z-table for the probability.

Thus, the probability that between 133 and 171 of the business travelers say that the reason for their most recent business trip was an internal company visit is 0.2205.

no this is a nonlinear negative

answer: all of the odd numbers were opened

the first person was the one that was opening all of the lockers. since one is odd, this causes all of the odd number lockers to be opened in the end.