You may need to use the appropriate appendix table or technology to answer this question.
The president of Howahkan Distributors, Inc., believes that 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 . Describe the sampling distribution of for this study by specifying
its distribution form and parameters? (Round your answer for to four decimal places.)

(b) What is the probability that the sample proportion will be between 0.20 and 0.40 ? (Round your answer to four
decimal places.)
distribution – Select— ) appropriate in this case.
© What is the probability that the sample proportion will be between 0.25 and 0.35 ? (Round your answer to four
decimal places.)

Ask by Beck Bird. in the United States

Mar 16,2025

Question

You may need to use the appropriate appendix table or technology to answer this question.
The president of Howahkan Distributors, Inc., believes that 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 . Describe the sampling distribution of for this study by specifying
its distribution form and parameters? (Round your answer for to four decimal places.)

(b) What is the probability that the sample proportion will be between 0.20 and 0.40 ? (Round your answer to four
decimal places.)
distribution – Select— ) appropriate in this case.
© What is the probability that the sample proportion will be between 0.25 and 0.35 ? (Round your answer to four
decimal places.)

Ask by Beck Bird. in the United States

Mar 16,2025

UpStudy AI · Accepted Answer

(a) Since the sample size is $$ n = 100 $$ and the true proportion is $$ p = 0.30 $$, the sampling distribution for the sample proportion $$ \bar{p} $$ is approximately normal with

$$
E(\bar{p}) = p = 0.30
$$

and standard error

$$
\sigma_{\bar{p}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.30 \times 0.70}{100}} = \sqrt{0.0021} \approx 0.0458.
$$

Thus, the distribution is

$$
\bar{p} \sim N(0.30, \, (0.0458)^2).
$$

(b) To find the probability that the sample proportion lies between 0.20 and 0.40, we standardize these values.

For $$ \bar{p} = 0.20 $$:

$$
z = \frac{0.20 - 0.30}{0.0458} \approx \frac{-0.10}{0.0458} \approx -2.183.
$$

For $$ \bar{p} = 0.40 $$:

$$
z = \frac{0.40 - 0.30}{0.0458} \approx \frac{0.10}{0.0458} \approx 2.183.
$$

Thus, the probability is

$$
P(0.20 \leq \bar{p} \leq 0.40) = P(-2.183 \leq Z \leq 2.183).
$$

Using the standard normal distribution table or appropriate technology, we find that

$$
P(Z \leq 2.183) \approx 0.9854 \quad \text{and} \quad P(Z \leq -2.183) \approx 0.0146.
$$

Therefore,

$$
P(0.20 \leq \bar{p} \leq 0.40) \approx 0.9854 - 0.0146 = 0.9708.
$$

(c) Similarly, to find the probability that the sample proportion is between 0.25 and 0.35, we standardize these values.

For $$ \bar{p} = 0.25 $$:

$$
z = \frac{0.25 - 0.30}{0.0458} \approx \frac{-0.05}{0.0458} \approx -1.091.
$$

For $$ \bar{p} = 0.35 $$:

$$
z = \frac{0.35 - 0.30}{0.0458} \approx \frac{0.05}{0.0458} \approx 1.091.
$$

Thus, the probability is

$$
P(0.25 \leq \bar{p} \leq 0.35) = P(-1.091 \leq Z \leq 1.091).
$$

From standard normal tables or using technology,

$$
P(Z \leq 1.091) \approx 0.8621 \quad \text{and} \quad P(Z \leq -1.091) \approx 0.1379.
$$

Therefore,

$$
P(0.25 \leq \bar{p} \leq 0.35) \approx 0.8621 - 0.1379 = 0.7242.
$$
(a) The sampling distribution of $$ \bar{p} $$ is approximately normal with $$ E(\bar{p}) = 0.30 $$ and $$ \sigma_{\bar{p}} \approx 0.0458 $$. (b) The probability that $$ \bar{p} $$ is between 0.20 and 0.40 is approximately 0.9708. (c) The probability that $$ \bar{p} $$ is between 0.25 and 0.35 is approximately 0.7242.

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