A random sample of 50 employees at a company was obtained. The one-way distance
ID: 3134197 • Letter: A
Question
A random sample of 50 employees at a company was obtained. The one-way distance from home to work was recorded for each employee in the sample. Suppose the mean of the sample was 15.2 miles and the standard deviation was 4.1 miles.
a. Compute the standard score (z - score) for a distance of 27 miles. Round your answer to the nearest hundredth and interpret the meaning of your answer as it pertains to this problem.
b. What is the one-way distance from home to work for an employee in the sample who had a standard score (z - score) of -1.50?
c. At least what percent of the distances in this sample should we expect to find within 2.5 standard deviations of the mean? Give your answer to the nearest percent.
) d. At most what percent of the distances in this sample should we expect to find less than 2.9 miles and more than 27.5 miles? Give your answer to the nearest percent.
e. At least what percent of the distances in this sample should we expect to find between 9.9 and 20.5 miles? Give your answer to the nearest percent.
Explanation / Answer
a) FRom information given, Xbar=15.2, s=4.1, Xi=27. Substitute the values to compute the z score.
z=(Xi-Xbar)/s=(27-15.2)/4.1=2.88
The z score correspond to raw score of 27, as a process of changing value scales.
b) The one way distance corresponding to z score of -1.50 is as follows;
-1.50=(Xi-15.2)/4.1
Xi=-6.15+15.2=9.05 [ans]
c) Xbar+2.5s=15.2+2.5*4.1=25.45 [ans]
d) Find z scores corresponding to Xi=2.9 and 27.5 respectively.
z1=(2.9-15.2)/4.1=-3 and z2=(27.5-15.2)/4.1=3
The areas corresponding to z scores are 0.0014 and 0.4986. Therefore, required percent is (0.0014+0.4986)*100=50%
e) Find z scores corresponding to Xi=9.9 and 20.5.
z1=(9.9-15.2)/4.1=-1.29 and z2=(20.5-15.2)/4.1=1.29
Required percentage is (0.0985+0.4015)*100=50%.
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