Question

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:
a. Before 38 months of service
b. Between 40 and 45 months of service
c. Within ± 2 months of the mean life

Answer 1

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%