Question

Q1. The company management of We Design Cars are concerned that employees at one of its design plants in Texas tend to show up late for work. The company management wants to send a strong message to the employees by reducing their bonus pay if they do not show up for work on time. Suppose management representatives of We Design Cars are coming to examine attendance records at the plant. A random sample of 36 employee attendance records will be drawn and if 85% or more employee show on-time arrival at the plant, the employees will continue to receive regular bonus pay without any penalty. What is the probability that the employees at the plant will have their bonus pay maintained without any penalty even if the overall actual on time arrival rate at the plant is only 75%? Show your detailed calculations along with proper labels here.

Answer 1

Given, p= 0.75, n= 36

So from the Central limit theorem,

p~Normal(0.75, √0.75*0.25/36) ~ N(0.75, 0.0722) distribution

Now, we want, P(p>=0.85)

Using standard normal approximation,

P(z>= 0.85-0.75/0.0722) = P(z>=1.385) = 1- P(z<=1.385)

= 1-0.91695= 0.08305

So the required probability is 0.08305