Answer: 0.7030
Step-by-step explanation:
Given : The population proportion for company's orders come from first-time customers : p=0.43
Sample size : n= 98
The test statistic for population proportion:-
[tex]z=\dfrac{\hat{p}-p}{\sqrt{\dfrac{p(1-p)}{n}}}[/tex]
For , [tex]\hat{p}=0.33[/tex]
[tex]z=\dfrac{0.33-0.43}{\sqrt{\dfrac{0.43(1-0.43)}{98}}}\approx-2.00[/tex]
For , [tex]\hat{p}=0.46[/tex]
[tex]z=\dfrac{0.46-0.43}{\sqrt{\dfrac{0.43(1-0.43)}{98}}}\approx0.60[/tex]
[tex]\text{The p-value =}P(-2.00<z<0.60)=P(z<0.60)-P(z<-2.00)\\\\=0.7257469-0.0227501=0.7029968\approx0.7030[/tex]
Hence, the probability that the sample proportion is between 0.33 and 0.46 is 0.7030.