Question

Given a population mean weight for baggage's of 62 kg and a standard deviation of 8 kg. a sample of 50 baggage's is taken. what is the probability that the sample mean is at most 61 kg?

Answer 1

$P(X\le61) = P(Z \le \frac {61 - \mu_X} {\sigma_X}) = P(Z \le \frac {61 - 62} {8/ \sqrt50}) = P(Z\le -0.8839) = 0.1884$