Question

1. Over a very long period of time, a sheriff determines the mean vehicle speed on a road is 67 miles per hour, the standard deviation is 3 miles per hour. Treat this as the population, not a sample.
   1. If the speed limit is 60 miles per hour, what percentage of the vehicles exceed the speed limit?
   2. The sheriff cracks down on speeders. His deputies ignore any vehicles going less than 63 mph, and issue warnings to those going between 63 and 65, and issue tickets to those going faster than 65 mph.
      1. What percentage will get warnings?
      2. What percentage will get tickets?

Following the crackdown, the sheriff takes a random sample (n=84) of vehicle speeds on the road way. His sample data: mean is 63 mph, sample SD is 4 mph.

3. Calculate the standard error of the sample mean using the sample standard deviation, and applying the +/- 2 rule of thumb, calculate the 95% confidence interval around the sample mean
4. A county commissioner claims that the crackdown had no effect, and the average speed is still 67 mph. Is this commissioner correct?
   1. How likely is it that the sample of came from a population where the average speed is 67?
5. What if the sheriff’s post-crackdown random sample was not n=84 but in fact n=30? Assume the sample mean and standard deviation were unchanged. Use a t-table to determine the 95% confidence interval around that sample mean.
   1. Would the county commissioner be accurate in this case?

Answer 1

a) Since μ=67 and σ=3 we have:
P ( X>60 )=P ( X−μ>60−67 )=P ((X−μ)/σ>(60−67)/3)
Since Z=(x−μ)/σ and (60−67)/3=−2.33 we have:
P ( X>60 )=P ( Z>−2.33 )
Use the standard normal table to conclude that:
P (Z>−2.33)=0.9901= 99.01%

b i) P ( 63<X<65 )=P ( 63−67< X−μ<65−67 )=P ((63−67)/3<(X−μ)/σ<(65−67)/3)
Since Z=(x−μ)/σ , (63−67)/3=−1.33 and (65−67)/3=−0.67 we have:
P ( 63<X<65 )=P ( −1.33<Z<−0.67 )
Use the standard normal table to conclude that:
P ( −1.33<Z<−0.67 )=0.1596

15.96% will get warnings.

ii) P ( X>65 )=P ( X−μ>65−67 )=P ( (X−μ)/σ>(65−67)/3)
Since Z=(x−μ)/σ and (65−67)/3=−0.67 we have:
P ( X>65 )=P ( Z>−0.67 )
Use the standard normal table to conclude that:
P (Z>−0.67)=0.7486

74.86% will get tickets.

c) Standard error= sample standard deviation/sqrt(n)

= 4/sqrt(84)

=4/9.165

= 0.44

95% confidence interval is

mean +/-2*S.E

63 +/-2*0.44

63 +/-0.88

(62.12,63.88) is 95% confidence interval.

NOTE: AS WE ARE ALLOWED TO SOLVE First four. I have done it. Please repost question d and e. Thank you