Answer: 0.00494.
Step-by-step explanation:
Given : The proportion of the company's orders come from first-time customers. [tex]p=0.45[/tex]
Sample size : n= 166
Now , the probability that the sample proportion is between 0.24 and 0.35 would be :-
[tex]P(0.24<p<0.35)=P\left({\dfrac{0.24-0.45}{\sqrt{\dfrac{0.45(1-0.45)}{166}}}<\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}<\dfrac{0.35-0.45}{\sqrt{\dfrac{0.45(1-0.45)}{166}}}}\right)\\\ \ \left[\because z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}\right][/tex]
[tex]=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[/tex]
Hence, the probability that the sample proportion is between 0.24 and 0.35 is 0.00494.
