Question

How to solve the following with a TI-84: suppose babies born in a large hospital have a mean weight of 3758 grams, and a variance of 204,304. If 89 babies are sampled at random from the hospital, what is the probability that the mean weight of the sample babies would differ from the population mean by more than 48 grams ?

Answer 1

for normal distribution z score =(X-μ)/σx
here mean=       μ=	3758
std deviation   =σ=	452.000
sample size       =n=	89
std error=σx̅=σ/√n=	47.91190

probability =P(3710<X<3806)=P((3710-3758)/47.912)<Z<(3806-3758)/47.912)=P(-1<Z<1)=0.8418-0.1582=0.6836

(from ti-84: press 1 - 2nd : vars :2:normalcdf( lower: 3710, upper : 3806 , mu =3758 , sigma =47.9119 , paste)