Vehicle speeds at a certain highway location are normally distributed with a mea
ID: 3125010 • Letter: V
Question
Vehicle speeds at a certain highway location are normally distributed with a mean of 70 mph and standard deviation of 10 mph. For each of the following questions, fill in the blank with the appropriate speeds. You may apply the Empirical Rule where appropriate. Hint: Pay careful attention to whether you are dealing with a distribution of individual observations or a distribution of sample means. The standard deviation that you use will differ for each (i.e., standard deviation versus standard error of the mean).
a. One vehicle is randomly selected; there is about a 68% chance that the vehicle’s speed will be between ___ and ___.
b. One vehicle is randomly selected; there is about a 99.7% chance that the vehicle’s speed will be between ___ and ___.
c. The speeds of randomly selected samples of 25 vehicles will be recorded. For samples of n=25 vehicles, there is about a 68% chance that a sample’s mean speed will be between ___ and ___.
d. The speeds of randomly selected samples of 25 vehicles will be recorded. For samples of n=25 vehicles, there is about a 99.7% chance that a sample’s mean speed will be between ___ and ___.
Explanation / Answer
(a) About 68% of the area under the normal curve is within one standard deviation of the mean. i.e.
(u ± 1s.d)
So to the given normal distribution about 68% of the observations lie in between
= (70 ± 10)
= [70-10, 70+10]
= [70-10, 70+10]
= [ 60,80]
(b)
Practically 99.7% of the area under the normal curve is within three standard deviations of the mean. i.e.
(u ± 3*s.d)
So to the given normal distribution about 99.7% of the observations lie in between
=[70 ± 2 * 10]
= [ 50, 90]
(c) For sample size n=25, mean = 70, s.d = s.d/Sqrt(n) = 10/Sqrt(25) = 10/5 = 2
About 68% of the area under the normal curve is within one standard deviation of the mean. i.e.
(u ± 1s.d)
= [70± 2]
= [70-2, 70+2]
= [ 68, 82]
d)
(c) For sample size n=25, mean = 70, s.d = s.d/Sqrt(n) = 10/Sqrt(25) = 10/5 = 2
About 99.7% of the area under the normal curve is within one standard deviation of the mean. i.e.
(u ± 3s.d)
= [70± 3*2]
= [70-6, 70+6]
= [ 64, 76]
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