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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 σp to four decimal places.)σp =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 proportionp 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 :

Answer:

a)0.04582576

b)0.9707

c)0.7243

Step-by-step explanation:

a)the sampling distribution of  for this study

Normal distribution with p=0.3 and standard deviation=sqrt(p*(1-p)/n)

=sqrt(0.3*0.7/100)

=0.04582576

b)A normal distribution because np and n(1-p) are both greater than 5

So the probability that the sample proportion will be between .20 and .40 is

P(0.2<phat<0.4)

= P((0.2-0.3)/sqrt(0.3*0.7/100) <(phat-p)/sqrt(p*(1-p)/n) <(0.4-0.3)/sqrt(0.3*0.7/100))

=P(-2.18<Z<2.18)

=0.9707 (from standard normal table)

c)So the probability that the sample proportion will be between .25 and .35 is

P(0.25<phat<0.35)

=P((0.25-0.3)/sqrt(0.3*0.7/100) <Z<(0.35-0.3)/sqrt(0.3*0.7/100))

=P(-1.09<Z<1.09)

=0.7243 (from standard normal table)

MrRoyal

The probability that the sample proportion will be between .25 and .35 is 0.724

(a) The sampling distribution

The given parameters are:

n = 100 and p = 0.3

So, the sampling distribution of the study is:

[tex]\sigma =\sqrt{\frac{p*(1-p)}n}[/tex]

This gives

[tex]\sigma =\sqrt{\frac{0.3*(1-0.3)}{100}}[/tex]

[tex]\sigma =0.046[/tex]

The probability that the sample proportion will be between .20 and .40

To do this, we make use of:

[tex]p = P(0.2 < \bar p < 0.4)[/tex]

The z values at p = 0.2 and p = 0.4 for the distribution are -2.18 and 2.18.

So, we have:

[tex]p = P(-2.18 < z < 2.18)\\[/tex]

Using the standard normal table, the result is:

[tex]p =0.971[/tex]

Hence, the probability that the sample proportion will be between .20 and .40 is 0.971

The probability that the sample proportion will be between .20 and .40

To do this, we make use of:

[tex]p = P(0.25 < \bar p < 0.35)[/tex]

The z values at p = 0.25 and p = 0.35 for the distribution are -1.09 and 1.09.

So, we have:

[tex]p = P(-1.09 < z < 1.09)[/tex]

Using the standard normal table, the result is:

[tex]p =0.724[/tex]

Hence, the probability that the sample proportion will be between .25 and .35 is 0.724

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