SOLUTION: The speed limit in the state of Ohio is 55 miles per hour. On a certain stretch of interstate highway the mean speed of traffic (as checked by radar) is 58 miles per hour with a s

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Question 451595: The speed limit in the state of Ohio is 55 miles per hour. On a certain stretch of interstate highway the mean speed of traffic (as checked by radar) is 58 miles per hour with a standard deviation of 4 miles per hour. The speeds of vehicles on this section of highway are normally distributed.
a) What percentage of vehicles are exceeding the speed limit?
b) Above what speed are the top 5% of the speeders travelling?
A cruise ship has 74% chance of having leftover rooms the week before the cruise. There are 25 cruise ships setting sail this month.
a) Is it possible to approximate this binomial distribution using a normal approximation?
b) What is the mean and the standard deviation for this approximation?
c) What is the probability that exactly 20 cruise ships will have leftover rooms using the normal approximation?
d) How different is the normal approximation from the binomial distribution value? Calculate the probability that exactly 20 cruise ships will have leftover rooms using the binomial method?
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
The speed limit in the state of Ohio is 55 miles per hour. On a certain stretch of interstate highway the mean speed of traffic (as checked by radar) is 58 miles per hour with a standard deviation of 4 miles per hour. The speeds of vehicles on this section of highway are normally distributed.
a) What percentage of vehicles are exceeding the speed limit?
z(55) = (55-58)/4 = -3/4
P(x > 55 mph) = P(z> -3/4) = -.7734
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b) Above what speed are the top 5% of the speeders travelling?
invNorm(0.95) = 1.645
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x = zs+u = 1.645*4+58 = 64.5 mph
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A cruise ship has 74% chance of having leftover rooms the week before the cruise. There are 25 cruise ships setting sail this month.
a) Is it possible to approximate this binomial distribution using a normal approximation?
Check the critera in your text.
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b) What is the mean and the standard deviation for this approximation?
mean = np = 25*0.74 = 18.5
std = sqrt(npq) = sqrt[18.5*0.26] = 2.183
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c) What is the probability that exactly 20 cruise ships will have leftover rooms using the normal approximation?
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z(19.5) = (19.5-18.5)/2.183 = 0.4581
z(20.5) = (20.5-18.5)/2.183 = 0.9162
P(x = 20) = P(0.4581< z< 0.9162) = 0.1437
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d) How different is the normal approximation from the binomial distribution value? Calculate the probability that exactly 20 cruise ships will have leftover rooms using the binomial method?
Ans: 25C20(0.74)^20(0.26)^5 = 0.1531
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Cheers,
Stan H.