Question

Based on historical data, your manager believes that 45% of the company's orders come from first-time customers. A random sample of 166 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is between 0.24 and 0.35?

Answer 1

Answer: 0.00494.

Explanation:

Given : The proportion of the company's orders come from first-time customers.
`p=0.45`

Sample size : n= 166

Now , the probability that the sample proportion is between 0.24 and 0.35 would be :-

`P(0.24<p<0.35)=P\left({\frac{0.24-0.45}{\sqrt{(0.45(1-0.45))/(166)}}<\frac{\hat{p}-p}{\sqrt{(p(1-p))/(n)}}<\frac{0.35-0.45}{\sqrt{(0.45(1-0.45))/(166)}}}\right)\\\ \ \left[\because z=\frac{\hat{p}-p}{\sqrt{(p(1-p))/(n)}}\right]`

`=P(-5.44<z<-2.58)=P(z<-2.58)-P(z<-5.44)\\\\=1-P(z<2.58)-(1-P(z<5.44))\\\\=P(z<5.44)-P(z<2.58)\\\\=1-0.99506\ [\text{By z-value table}]\\\\=0.00494`

Hence, the probability that the sample proportion is between 0.24 and 0.35 is 0.00494.