Answer:
![\hat p \sim N( p, \sqrt{\frac{p (1-p)}{n}})](https://tex.z-dn.net/?f=%20%5Chat%20p%20%5Csim%20N%28%20p%2C%20%5Csqrt%7B%5Cfrac%7Bp%20%281-p%29%7D%7Bn%7D%7D%29)
And we can use the z score formula given by:
![z = \frac{\hat p -\mu_p}{\sigma_p}](https://tex.z-dn.net/?f=%20z%20%3D%20%5Cfrac%7B%5Chat%20p%20-%5Cmu_p%7D%7B%5Csigma_p%7D)
And if we find the parameters we got:
![\mu_p = 0.26](https://tex.z-dn.net/?f=%20%5Cmu_p%20%3D%200.26)
![\sigma_p = \sqrt{\frac{0.26(1-0.26)}{158}} = 0.0349](https://tex.z-dn.net/?f=%20%5Csigma_p%20%3D%20%5Csqrt%7B%5Cfrac%7B0.26%281-0.26%29%7D%7B158%7D%7D%20%3D%200.0349)
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](https://tex.z-dn.net/?f=%20z%20%3D%20%5Cfrac%7B0.4-0.26%7D%7B0.0349%7D%3D%204.0119)
And we can find this probability:
![P(z>4.0119) = 1-P(z](https://tex.z-dn.net/?f=%20P%28z%3E4.0119%29%20%3D%201-P%28z%3C4.0119%29)
And if we use the normal standard table or excel we got:
![P(z>4.0119) = 1-P(z](https://tex.z-dn.net/?f=%20P%28z%3E4.0119%29%20%3D%201-P%28z%3C4.0119%29%3D1-0.99997%20%3D%200.00003)
Step-by-step explanation:
For this case we have the following info given:
represent the proportion of the company's orders come from first-time customers
represent the sample size
And we want to find the following probability:
![p(\hat p >0.4)](https://tex.z-dn.net/?f=%20p%28%5Chat%20p%20%3E0.4%29)
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}})](https://tex.z-dn.net/?f=%20%5Chat%20p%20%5Csim%20N%28%20p%2C%20%5Csqrt%7B%5Cfrac%7Bp%20%281-p%29%7D%7Bn%7D%7D%29)
And we can use the z score formula given by:
![z = \frac{\hat p -\mu_p}{\sigma_p}](https://tex.z-dn.net/?f=%20z%20%3D%20%5Cfrac%7B%5Chat%20p%20-%5Cmu_p%7D%7B%5Csigma_p%7D)
And if we find the parameters we got:
![\mu_p = 0.26](https://tex.z-dn.net/?f=%20%5Cmu_p%20%3D%200.26)
![\sigma_p = \sqrt{\frac{0.26(1-0.26)}{158}} = 0.0349](https://tex.z-dn.net/?f=%20%5Csigma_p%20%3D%20%5Csqrt%7B%5Cfrac%7B0.26%281-0.26%29%7D%7B158%7D%7D%20%3D%200.0349)
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](https://tex.z-dn.net/?f=%20z%20%3D%20%5Cfrac%7B0.4-0.26%7D%7B0.0349%7D%3D%204.0119)
And we can find this probability:
![P(z>4.0119) = 1-P(z](https://tex.z-dn.net/?f=%20P%28z%3E4.0119%29%20%3D%201-P%28z%3C4.0119%29)
And if we use the normal standard table or excel we got:
![P(z>4.0119) = 1-P(z](https://tex.z-dn.net/?f=%20P%28z%3E4.0119%29%20%3D%201-P%28z%3C4.0119%29%3D1-0.99997%20%3D%200.00003)