Question

Q2 if Z is a standard normal varianle find the probability

P(-0.73 < Z < 2.27)

a. -0.2211

b. 0.4884

c. 0.7557

d. 0.2211

e. 1.54

Q3 Assume that women ties are normally distributed with a mean of 63.6 inches in a standard deviation of 2.5 inches if 60 women are randomly selected find the probability that they have a mean height between 62.9 inches and 64.0 inches

-assume that women's heights are normally distributed with a mean of 63.6 inches and a standard deviation of 2.5 inches if 60 women are randomly selected find the probability that they have a mean height between 62.9 inches and 64.0 inches

a. 0.9426

b. 0.1738

c. 0.8773

d. 0.0574

e. 0.1228

Answer 1

2)

P(-0.73<Z<2.27)=0.9884-0.2327=0.7557

3)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	63.6
std deviation   =σ=	2.500
sample size       =n=	60
std error=σx̅=σ/√n=	0.32275

probability =P(62.9<X<64)=P((62.9-63.6)/0.323)<Z<(64-63.6)/0.323)=P(-2.17<Z<1.24)=0.8924-0.015=0.8773