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Solved: Using the Central Limit Theorem. In Exercises use
Chapter 6, Problem 7BSC(choose chapter or problem)
Problem 7BSC
Using the Central Limit Theorem. In Exercises use this information about the overhead reach distances of adult females:μ = 205.5 cm,σ = 8.6 cm, and overhead reach distances are normally distributed (based on data from the Federal Aviation Administration). The overhead reach distances are used in planning assembly work stations.
a. If 1 adult female is randomly selected, find the probability that her overhead reach is greater than 218.4 cm.
b. If 9 adult females are randomly selected, find the probability that they have a mean overhead reach greater than 204.0 cm.
c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?
Questions & Answers
QUESTION:
Problem 7BSC
Using the Central Limit Theorem. In Exercises use this information about the overhead reach distances of adult females:μ = 205.5 cm,σ = 8.6 cm, and overhead reach distances are normally distributed (based on data from the Federal Aviation Administration). The overhead reach distances are used in planning assembly work stations.
a. If 1 adult female is randomly selected, find the probability that her overhead reach is greater than 218.4 cm.
b. If 9 adult females are randomly selected, find the probability that they have a mean overhead reach greater than 204.0 cm.
c. Why can the normal distribution be used in part (b), even though the sample size does not exceed 30?
ANSWER:Answer:
Step 1 of 2
The overhead reach distances of adult females = 205.5 cm and
= 8.6 cm and overhead reach distances are normally distributed. The overhead reach distances are used in planning assembly work stations.
a. If 1 adult female is randomly selected, the probability that her overhead reach is greater than 218.4 cm.
We are dealing with individual value from a normally distributed population so mean is 205.5 cm and standard deviation is 8.6 cm.
The Test Statistic can be expressed as
Z = 1.5
To find the associated probability, we have
P()
P(Z > 1.5) = 1 - P(Z 1.5)
= 1 - 0.9332 (From Area Under Normal Curve table)
= 0.0668