Question

The average amount of water a person in America drinks on a daily basis is 55 ounces with a standard deviation of 18 ounces. Assume this data is normally distributed.
(a) Find the probability that a person in America drinks less than 40 ounces of water.

(b) Find the probability that a person in America drinks more than 64 ounces of water.

(c) In a sample of 36 people in America, what is the probability that this sample will have drank a mean within ±8 ounces of the population mean?

Answer 1

for normal distribution z score =(X-μ)/σx
here mean=       μ=	55
std deviation   =σ=	18.000

a)

probability =P(X<40)=(Z<(40-55)/18)=P(Z<-0.83)=0.2033

b)

probability =P(X>64)=P(Z>(64-55)/18)=P(Z>0.5)=1-P(Z<0.5)=1-0.6915=0.3085

c)

sample size       =n=	36
std error=σx̅=σ/√n=	3.0000

probability =P(47<X<63)=P((47-55)/3)<Z<(63-55)/3)=P(-2.67<Z<2.67)=0.9962-0.0038=0.9924

answered by: anonymous