Question

A mall supervisor has found that mall sellers, on the average, complete a certain task in 10 minutes. If the times required to complete the task are approximately normally distributed with a standard deviation of 3 minutes, find: (3 marks each)
a. The proportion of mall sellers completing the task in less than 4 minutes
b. The proportion of mall sellers requiring more than 5 minutes to complete the task
c. The probability that a mall seller who has just been assigned the task will complete it within 3 minutes

Answer 1

Given that the tasks are approximately normally distributed,

Population mean $\mu$ = 10

Standard deviation $\sigma$ = 3

x = time taken by the mall supervisors to complete a task

The formula for standard normal variable Z is given by,

Z = $\frac {x-\mu}{\sigma}$

Answer a)

To find the proportion of mall sellers completing the task in less than 4 minutes is given by,

p(x<4) = $p(\frac {x-\mu}{\sigma}<\frac {4-\mu}{\sigma})=p(Z<\frac {4-10}{3})=p(Z<-2)$

From the standard normal distribution table we get the probability for p(x<4) = 0.0228

Answer b)

To find the proportion of mall sellers requiring more than 5 minutes to complete the task

p(x>5) = $p(\frac {x-\mu}{\sigma}>\frac {5-\mu}{\sigma})=p(Z>\frac {5-10}{3})=p(Z>-1.6666)$

From the standard normal distribution table we get the probability for p(x>5) = 1 - 0.0485 = 0.9522

Answer b)

To find the probability that a mall seller who has just been assigned the task will complete it within 3 minutes

p(x<3) = $p(\frac {x-\mu}{\sigma}<\frac {3-\mu}{\sigma})=p(Z<\frac {3-10}{3})=p(Z<-2.333)$

From the standard normal distribution table we get the probability for p(x<3) = 0.0099