Answer: Our required probability is 0.99338.
Step-by-step explanation:
Since we have given that
n = 100
p = 0.200
So, [tex]np=100\times 0.2=20>5[/tex]
So, we can apply normal approximation.
Mean = 20
Standard deviation = [tex]\sqrt{npq}=\sqrt{100\times 0.2\times 0.8}=\sqrt{16}=4=\sigma[/tex]
Since [tex]z=\dfrac{\bar{X}-\mu}{\sigma}[/tex]
So, Probability that fewer than 30 but a packet after testing a free sample is given by
[tex]P(X<30)\\\\=P(z<\dfrac{30-20}{4})\\\\=P(z<\dfrac{10}{4})\\\\=P(z<2.5)\\\\=0.9938[/tex]
Hence, our required probability is 0.99338.