Question

Suppose X is the time it takes a randomly chosen clerical worker in an office to type and send a standard letter of recommendation. Suppose X has a normal distribution, and assume the mean is 10.5 minutes and the standard deviation 3 minutes. You take a random sample of 50 clerical workers and measure their times. What is the chance that their average time is less than 9.5 minutes?

Answer 1

Answer:

The chance that their average time is less than 9.5 minutes is 0.0094

Step-by-step explanation:

X is the time it takes a randomly chosen clerical worker in an office to type and send a standard letter of recommendation

We are given that X has a normal distribution

Mean $\mu$ = 10.5 minutes

Standard deviation $\sigma$= 3 minutes

n = 50

We are supposed to find the chance that their average time is less than 9.5 minutes i.e. P(X<9.5 minutes)

Formula : $Z = \frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}$

$Z = \frac{9.5-10.5}{\frac{3}{\sqrt{50}}}$

Z = −2.35

Refer the z table for p value

p value=0.0094

So, P(X<9.5)=P(Z<−2.35)=0.0094

Hence  the chance that their average time is less than 9.5 minutes is 0.0094.