Answer:
a) 22.663%
b) 44%
c) 38.3%
Explanation:
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 41 months and a standard deviation of 4 months. On the basis of this information, determine the percentage of scanners that can be expected to fail in the following time periods:
We solve the above question using z score formula
z = (x-μ)/σ, where
x is the raw score
μ is the population mean = 41 months
σ is the population standard deviation = 4 months
a. Before 38 months of service
Before in z score score means less than 38 months
Hence,
z = 38 - 41/4
z = -0.75
Probability value from Z-Table:
P(x<38) = 0.22663
Converting to percentage = 0.22663 × 100
= 22.663%
b. Between 40 and 45 months of service
For x = 40 months
z = 40 - 41/4
z = -0.2
Probability value from Z-Table:
P(x = 40) = 0.40129
For x = 45
z = 45 - 41/4
z = 1
Probability value from Z-Table:
P(x = 45) = 0.84134
Between 40 and 45 months of service
= 0.84134 - 0.40129
= 0.44005
Converting to Percentage
= 0.44005 × 100
= 44.005%
= 44%
c. Within ± 2 months of the mean life
+ 2 months = 41 months + 2 months
= 43 months
- 2 months = 41 months - 2 months
= 39 months
For x = 43
z = 43 - 41 /4
z = 0.5
P-value from Z-Table:
P(x = 43) = 0.69146
For x = 39
z = 39 - 41/4
z = -2/4
z = -0.5
Probability value from Z-Table:
P(x = 39) = 0.30854
Within ± 2 months of the mean life
= 0.69146 - 0.30854
= 0.38292
= 38.3%
