Question

The president of Doerman Distributors, Inc., believes that 28% of the firm's orders come from first-time customers. A simple random sample of 100 orders will be used to estimate the proportion of first-time customers. Use z-table.

What is the probability that the sample proportion will be between 0.16 and 0.40 (to 4 decimals)?

What is the probability that the sample proportion will be between 0.21 and 0.35 (to 4 decimals)?

Answer 1

1)

for normal distribution z score =(p̂-p)/σp
here population proportion=     p=	0.280
sample size       =n=	100
std error of proportion=σp=√(p*(1-p)/n)=	0.0449

probability =

P(0.16<X<0.4)

=

P(-2.67<Z<2.67)=

0.9962-0.0038=

0.9924

2)

probability =

P(0.21<X<0.35)

=

P(-1.56<Z<1.56)=

0.9406-0.0594=

0.8812