Problem 4S-16 An office manager has received a report from a consultant that inc

Question

Problem 4S-16

An office manager has received a report from a consultant that includes a section on equipment replacement. The report indicates that scanners have a service life that is normally distributed with a mean of 50 months and a standard deviation of 5 months. On the basis of this information, determine the proportion of scanners that can be expected to fail in the following time periods:

Use Table A and Table B.

a. Before 43 months of service. (Round z values to 2 decimal places, probability calculations to 4 decimal places and your final answer to 4 decimal places.)

    
b. Between 49 and 53 months of service. (Round z values to 2 decimal places, probability calculations to 4 decimal places and your final answer to 4 decimal places.)

    
c. Within ± 2 months of the mean life. (Round z values to 2 decimal places, probability calculations to 4 decimal places and your final answer to 4 decimal places.)

Explanation / Answer

Solution:
Given in the question
Mean = 50
SD = 5
Solution(1)
P(Xbar<43)
Z = (43-50)/5 = -7/5 = -1.4
From Z table we found that there is 0.0808 or 8.08% probability of less than 43 months of service
Solution(2)
P(49<Xbar<53) = P(Xbar<53)-P(Xbar<49)
Z = (49-50)/5 = -1/5 or -0.2
Z = (53-50)/5 = 3/5 = 0.6
From z table we found
P(49<Xbar<53) = 0.7257 - 0.4207 = 0.305
so there is 30.5% probability between 49 and 53 months of service.
Solution(3)
P(48<Xbar<52) = P(Xbar<52) - P(Xbar<48)
Z = (52-50)/5 = 2/5 or 0.4
Z = (48-50)/5 = -/5 or -0.4
From Z table we found
P(48<Xbar<52) = 0.6554 -0.3446 = 0.3108 or 31.08%
So there is 31.08% probability within +/- 2 months of the mean life

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