Question

A normal population has a mean of 0.1 and standard deviation 2.1. Find the
probability that the mean of a sample of size 900 will be negative

Answer 1

Given

$$\mu=0.1, \sigma=2.1, n=900$$

The standard normal variate is

$$z=(xmean - \mu)/(\sigma/sqrt(n))$$ $$z=(xmean - 0.1)/(2.1/sqrt(900))=(xmean-0.1)/0.07$$ $$xmean=0.1+0.07z$$

The required probability, that the sample mean is negative is given by

$$P(xmean<0)=P(0.1+0.07z<0)=P(z<-1/0.7)=P(z<-1.42857)=0.076564$$