Answer: Our required probability is 0.99338.
Explanation:
Since we have given that
n = 100
p = 0.200
So,
![np=100* 0.2=20>5](https://img.qammunity.org/2020/formulas/mathematics/college/ezj65lkiumvx385jt0qjf6y2q1o75s0hiz.png)
So, we can apply normal approximation.
Mean = 20
Standard deviation =
![√(npq)=√(100* 0.2* 0.8)=√(16)=4=\sigma](https://img.qammunity.org/2020/formulas/mathematics/college/19336z0ydoowt8ozd1xry7uqj03wypooci.png)
Since
![z=\frac{\bar{X}-\mu}{\sigma}](https://img.qammunity.org/2020/formulas/mathematics/college/uhiqy2sm6cuovotblzl74m3j5o1ymk08r9.png)
So, Probability that fewer than 30 but a packet after testing a free sample is given by
![P(X<30)\\\\=P(z<(30-20)/(4))\\\\=P(z<(10)/(4))\\\\=P(z<2.5)\\\\=0.9938](https://img.qammunity.org/2020/formulas/mathematics/college/2jr283furg8i00q8do09dov4tit1t7lb1h.png)
Hence, our required probability is 0.99338.