Question

Given a population mean weight for baggage of 62 kg and a standard deviation of 8 kg. A sample of 50 baggage is taken. What is the probability that the sample mean differs from the population mean by at most 1kg?

Answer 1

$P(61<\bar X<63)=P(\frac{61-62}{\frac{8}{\sqrt{50}}}<Z<\frac{63-62}{\frac{8}{\sqrt{50}}})=P(-0.88<Z<0.88)=$

$=P(Z<0.88)-P(Z<-0.88)=0.6211.$