Question

a retail manager is analyzing the percentage of customers who make a purchase when they visit a store. based on historical data, the manager believes that a customer makes a purchase 35% of the time. to assess this, 115 customer visits were randomly selected. 1. assuming that the managers estimate is correct and that the probability of a customer making a purchase is p = 0.35, compute the corresponding parameters of the sampling - distribution and evaluate whether it is appropriate to approximate the sampling distribution with a normal distribution. (round to four decimal places where applicable) the parameters of the sampling distribution are e(\\(\hat{p}\\)) = and \\(\sigma_{\hat{p}}\\)= and n(1 - p)=. if appropriate to approximate the sampling distribution with a normal distribution because np = b. calculate the probability that the sample proportion \\(\hat{p}\\) will be between 0.25 and 0.45. (round your answer to four decimal places.) c. calculate the probability that the sample proportion \\(\hat{p}\\) will be between 0.3 and 0.4. (round your answer to four decimal places.) click here to expand the distribution tables.

Explanation:

Step1: Calculate mean and standard deviation of sampling - distribution

The mean of the sampling - distribution of the sample proportion $E(\hat{p})=p = 0.35$. The standard deviation $\sigma_{\hat{p}}=\sqrt{\frac{p(1 - p)}{n}}$, where $n = 115$ and $p=0.35$. So, $\sigma_{\hat{p}}=\sqrt{\frac{0.35\times(1 - 0.35)}{115}}=\sqrt{\frac{0.35\times0.65}{115}}=\sqrt{\frac{0.2275}{115}}\approx\sqrt{0.00197826}\approx0.0445$. Also, $np=115\times0.35 = 40.25$ and $n(1 - p)=115\times0.65 = 74.75$. Since $np\geq5$ and $n(1 - p)\geq5$, it is appropriate to approximate the sampling - distribution with a normal distribution.

Step2: Standardize the bounds for part b

For $\hat{p}_1 = 0.25$, the z - score is $z_1=\frac{\hat{p}_1 - E(\hat{p})}{\sigma_{\hat{p}}}=\frac{0.25 - 0.35}{0.0445}=\frac{- 0.1}{0.0445}\approx - 2.25$. For $\hat{p}_2 = 0.45$, the z - score is $z_2=\frac{\hat{p}_2 - E(\hat{p})}{\sigma_{\hat{p}}}=\frac{0.45 - 0.35}{0.0445}=\frac{0.1}{0.0445}\approx2.25$. Then $P(0.25<\hat{p}<0.45)=P(-2.25

Step3: Standardize the bounds for part c

For $\hat{p}_1 = 0.3$, the z - score is $z_1=\frac{\hat{p}_1 - E(\hat{p})}{\sigma_{\hat{p}}}=\frac{0.3 - 0.35}{0.0445}=\frac{-0.05}{0.0445}\approx - 1.12$. For $\hat{p}_2 = 0.4$, the z - score is $z_2=\frac{\hat{p}_2 - E(\hat{p})}{\sigma_{\hat{p}}}=\frac{0.4 - 0.35}{0.0445}=\frac{0.05}{0.0445}\approx1.12$. Then $P(0.3<\hat{p}<0.4)=P(-1.12

Answer:

$E(\hat{p}) = 0.35$, $\sigma_{\hat{p}}\approx0.0445$, $np = 40.25$, $n(1 - p)=74.75$
b. $0.9756$
c. $0.7372$