Answer to Question #202280 in Statistics and Probability for Julian Chris

Question

The weights of 1000 children, in average is 61kg with a standard deviation of 18kg. Suppose the weights ate normally distributed, how many children weigh less than 35kg?

Answer 1

Solution:

Given, $\mu=61,\sigma=18$

$X\sim N(\mu,\sigma) \\P(X<35)=P(z<\dfrac{35-61}{18})=P(z<-1.44)=0.07493$

Number of required children$=1000\times0.07493=74.93\approx75$