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 45 months and a standard deviation of 6 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 36 months of service. (Round z values to 2 decimal places, probability calculations to 4 decimal places and your final answer to 4 decimal places.) Proportion of scanners <36 b. Between 43 and 48 months of service. (Round z values to 2 decimal places, probability calculations to 4 decimal places and your final answer to 4 decimal places.) Proportion of scanners 43<T<48 C. Within ± 3 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.) Proportion of scanners Within t 3 months of the mean life

Answer 1

Solution:- Given that mean = 45, standard deviation = 6

a. P(T < 36) = P((T-μ)/σ < (36-45)/6)
= P(Z < -1.50)
= 0.0668

b. P(43 < T < 48) = P((43-45)/6 < (T-μ)/σ < (48-45)/6)
= P(-0.33 < Z < 0.5)
= 0.3208

c. P(42 < T < 48) = P((42-45)/6 < (T-μ)/σ < (48-45)/6))
= P(-0.5 < Z < 0.5)
= 0.383