SOLUTION: 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 i

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Question 1169840:
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 20% 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 18% and 20% of the business travelers say that the reason for their most recent business trip was an internal company visit?

Answer by VFBundy(438) About Me  (Show Source):
You can put this solution on YOUR website!
n = 950
p = 0.19
q = 0.81

Mean = np = (950)(0.19) = 180.5
SD = sqrt%28npq%29 = sqrt%28950%2A%280.19%29%2A%280.81%29%29 = 12.0915

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a. What is the probability that more than 20% of the business travelers say that the reason for their most recent business trip was an internal company visit?

0.20(950) = 190

The question says "more than 20%", whereas 190 is exactly 20%. So, we are computing the probability that 191 or more business travelers say their most recent business trip was an internal company visit.

P(x ≥ 191) ----> P(x > 191 - 0.5) ----> P(x > 190.5)

z = %28190.5+-+180.5%29%2F12.0915 = 10%2F12.0915 = 0.83

When you check the z-table, you will see that a z-score of 0.83 equals 0.7967. This is the probability that FEWER than 191 business travelers say their most recent business trip was an internal company visit. We want the probability that 191 or MORE say this. Therefore, the answer is 0.2033, since 1 - 0.7967 = 0.2033.

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b. What is the probability that between 18% and 20% of the business travelers say that the reason for their most recent business trip was an internal company visit?

0.18(950) = 171
0.20(950) = 190

P(x ≥ 171) ----> P(x > 171 - 0.5) ----> P(x > 170.5)
P(x ≤ 190) ----> P(x < 190 + 0.5) ----> P(x < 190.5)
P(170.5 < x < 190.5)

z1 = %28170.5+-+180.5%29%2F12.0915 = %28-10%29%2F12.0915 = -0.83

z2 = %28190.5+-+180.5%29%2F12.0915 = 10%2F12.0915 = 0.83

When you check the z-table, you will see that a z-score (z1) of -0.83 equals 0.2033. This is the probability that FEWER than 171 travelers say that the reason for their most recent business trip was an internal company visit.

When you check the z-table, you will see that a z-score (z2) of 0.83 equals 0.7967. This is the probability that 190 or FEWER travelers say that the reason for their most recent business trip was an internal company visit.

Therefore, the probability that between 18% and 20% of the business travelers say that the reason for their most recent business trip was an internal company visit is 0.5934, since 0.7967 - 0.2033 equals 0.5934.