Question

Based on historical data, your manager believes that 26% of the company's orders come from first-time customers. A random sample of 158 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is greater than than 0.4?
Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations.



Answer =


(Enter your answer as a number accurate to 4 decimal places.)

Answer 1

Answer:

$\hat p \sim N( p, \sqrt{\frac{p (1-p)}{n}})$

And we can use the z score formula given by:

$z = \frac{\hat p -\mu_p}{\sigma_p}$

And if we find the parameters we got:

$\mu_p = 0.26$

$\sigma_p = \sqrt{\frac{0.26(1-0.26)}{158}} = 0.0349$

And we can find the z score for the value of 0.4 and we got:

$z = \frac{0.4-0.26}{0.0349}= 4.0119$

And we can find this probability:

$P(z>4.0119) = 1-P(z$

And if we use the normal standard table or excel we got:

$P(z>4.0119) = 1-P(z$

Step-by-step explanation:

For this case we have the following info given:

$p = 0.26$ represent the proportion of the company's orders come from first-time customers

$n=158$ represent the sample size

And we want to find the following probability:

$p(\hat p >0.4)$

And we can use the normal approximation since we have the following two conditions:

1) np = 158*0.26 = 41.08>10

2) n(1-p) = 158*(1-0.26) = 116.92>10

And for this case the distribution for the sample proportion is given by:

$\hat p \sim N( p, \sqrt{\frac{p (1-p)}{n}})$

And we can use the z score formula given by:

$z = \frac{\hat p -\mu_p}{\sigma_p}$

And if we find the parameters we got:

$\mu_p = 0.26$

$\sigma_p = \sqrt{\frac{0.26(1-0.26)}{158}} = 0.0349$

And we can find the z score for the value of 0.4 and we got:

$z = \frac{0.4-0.26}{0.0349}= 4.0119$

And we can find this probability:

$P(z>4.0119) = 1-P(z$

And if we use the normal standard table or excel we got:

$P(z>4.0119) = 1-P(z$