Answer to Question #149403 in Statistics and Probability for jee

Question

A factory’s worker productivity is normally distributed.
One worker produces an average of 75 units
per day with a standard deviation of 20. Another
worker produces at an average rate of 65 per day
with a standard deviation of 21. What is the probability
that in 1 week (5 working days), worker 1
will outproduce worker 2?

Answer 1

We have that

$\mu_1=75, \ \sigma_1=20$

$\mu_2=65, \ \sigma_2=21$

$n = 5$

Let's consider that $X_1$ and $X_2$ represent the production by worker 1 and worker 2 respectively. We need to find the probability that in 5 working days worker 1 will outproduce worker 2.

That is $P(\bar X_1>\bar X_2)$

$\bar X_1-\bar X_2$ follows normal distribution with the mean $\mu=\mu_1-\mu_2=75-65=10$

and the standard deviation $\sigma=\sqrt{\frac{\sigma_1^2}{n}+\frac{\sigma_2^2}{n}}=\sqrt{\frac{20^2}{5}+\frac{21^2}{5}}=12.97$

$P(\bar X_1>\bar X_2)=P(\bar X_1-\bar X_2>0)=P(Z>\frac{0-\mu}{\sigma})=P(Z>\frac{-10}{12.97})=$

$=P(Z>-0.77)=1-P(Z<-0.77)=1-0.2206=0.7794$

Answer: the probability that in 5 working days worker1 will outproduce worker2 is 0.7794.