Question

The president of Doerman Distributors, Inc., believes that 30% of the firm's orders come from first-time customers. A random sample of 100 orders will be used to estimate the proportion of first-time customers.

(a)

Assume that the president is correct and

p = 0.30.

What is the sampling distribution of

p

for n = 100? (Round your answer for

σ_p

to four decimal places.)

σ_p

=

E(p)

=

Since np =  and n(1 − p) =  , approximating the sampling distribution with a normal distribution  ---Select--- is is not appropriate in this case.

(b)

What is the probability that the sample proportion

p

will be between 0.20 and 0.40? (Round your answer to four decimal places.)

(c)

What is the probability that the sample proportion will be between 0.25 and 0.35? (Round your answer to four decimal places.)

Answer 1

Solution:

Given that,

n = 100

= 0.30

1 - = 1 - 0.30 = 0.70

So,

a ) =   = 0.30

Expected value = 0.30

= ( 1 - ) / n

=  0.30 * 0.70 / 100

= 0.0458

= 0.0458

Standard error = 0.0458

b ) p ( 0.20 < < 0.40 )

p ( 0.20 - 0.30 / 0.0458 ) < ( -    / ) < ( 0.30 - 0.30 / 0.0458 )

p ( - 0.10 /0.0458 < z < 0.10 / 0.0458 )

p ( -2.18 < z < 2.18 )

p ( z < 2.18 ) - p ( z < -2.18)

Using z table

= 0.9854 - 0.0146

= 0.9708

Probability = 0.9708

c ) p ( 0.25 < < 0.35 )

p ( 0.25 - 0.30 / 0.0458 ) < ( -    / ) < ( 0.35 - 0.30 / 0.0458 )

p ( - 0.05 /0.0458 < z < 0.05 / 0.0458 )

p ( -1.09 < z < 1.09)

p ( z < 1.09 ) - p ( z < -1.09)

Using z table

= 0.8621 - 0.1379

= 0.7242

Probability = 0.7242