Answer: b.) 0.9706
Step-by-step explanation:
Given : Proportion of the 5,000 payroll vouchers issued since January 1, 2000, have irregularities : p=0.10
Now , using standard normal z-value table , we have
The probability that Pinky's random sample of 200 vouchers will have a sample proportion greater than .06 will be :-
[tex]P(z>\dfrac{0.6-0.10}{\sqrt{\dfrac{0.10(1-0.10)}{200}}})\ \ [\because z=\dfrac{\hat{p}-p}{\sqrt\dfrac{p(1-p)}{n}}}]\\\\=P(z> -1.89)\\\\=P(z<1.89)\ \ [\because P(Z>-z)=P(Z<z)]\\\\=0.9706[/tex] [By using z-value table]
Hence, the probability that Pinky's random sample of 200 vouchers will have a sample proportion greater than .06 is b.) 0.9706.