Question

1. Consider the following sample spot speed data. The speeds are in miles per hour Speed Grou Lower Limit Upper Limit Mid-point Frequenc 27.5 32.5 37.5 42.5 47.5 52.5 57.5 62.5 67.5 72.5 77.5 82.5 87.5 92.5 25 30 35 40 45 50 22.6 27.6 32.6 37.6 42.6 47.6 52.6 57.6 62.6 67.6 72.6 77.6 82.6 87.6 2 14 25 60 65 70 75 80 85 90 4 (a) Calculate the sample mean speed (b) Calculate the sample median speed (c) Calculate the sample modal speed (d) Calculate the sample variance (e) If the speed limit on this highway was 70 MPH, what percentage of drivers was exceeding this speed limit? (f) Determine the 50th percentile speed Assume that the population standard deviation of speeds on this highway is known to be 13 mph, conduct the following analysis: (g) Find the 95 percent two-sided confidence interval on the mean speed (h) Find the 90 percent lower-sided confidence interval on the mean speed (i) Find the 90 percent upper-sided confidence interval on the mean speed (j) IfPCX 57 mph)-?. What is ??

Answer 1

a) Sample mean =

x = midpoint

f= frequency

b) $\frac{N}{2}=\frac{142}{2}=121$

2060 3 3 4 2 2 2 4 1 uu 3 4 ue 3 | | | 261122|1|1 4851-84-8642 5-0-5-0-5-0-5 2-3-3-4-4-5-5 6-7-7-8-8-9 し55555555555555 r772 | 7 | 2 | 7 | 2 | 7 | 2 | 7 | 2 | 7-2-7-2 e23344556677889 SL 66666666666666 r2 7|2|7|2|7|2|7|2|7-2-7-2-7 e|2|2|3|3 4 | 4 | 5 | 5 | 6 | 6-7-7-8-8

is the median group

(n/2) -B X w Estimated Median = L+ where: .L is the lower class boundary of the group containing the median . n is the total number of values . B is the cumulative frequency of the groups before the median group G is the frequency of the median group .w is the group width
$median= 67.6+\frac{121-108}{14}(4.9)=72.15$

c)

fm-m-1 Estimated Mode = L+ m-1 where: the modal group (the group with the highest frequency) .L is the lower dass boundary of the modal group m · fm-1 is the frequency of the group before the modal group fm is the frequency of the modal group fm+1 is the frequency of the group after the modal group .w is the group width

is the modal group

$mode=52.6+\frac{25-18}{(25-18)+(25-21)}(4.9)=52.6+3.12=55.72$

d) $S^2=\frac{1}{n-1}[\sum fX^2-\frac{(\sum fX)^2}{n}]=\frac{1}{142-1}[512550-\frac{(8267)^2}{142}]=221.70$