Answer to Question #272812 in Statistics and Probability for nick

Question

Draw 25 samples from a distribution of IQ scores that are normally distributed with a mean of 100 and standard deviation of 15. What is the probability that the mean of 25 randomly drawn IQ scores will exceed 103 points?

Answer 1

$z=\frac{\overline{x}-\mu}{\sigma/\sqrt n}=\frac{103-100}{15/\sqrt{25}}=1$

$P(\overline{x}>103)=P(z>1)=1-P(z<1)=1-0.8413=0.1587$