Answer to Question #148241 in Statistics and Probability for manisha maraseni

Question

Based upon past experience, 2% of the telephone bills mailed to suburban households are incorrect. If a sample
of 20 bills is selected, find the probability that at least one bill will be incorrect. Do this problem using two
probability distributions (the binomial and Poisson) and briefly compare and explain your results.

Answer 1

n = 20

p = 2% = 0.02

We need to find $P(X≥1)$ , where X is the number of incorrect telephone bills mailed to households.

i) using binomial distribution

$$P(X=x)=C(n,x)\cdot p^x\cdot (1-p)^{n-x}$$

$P(X\ge1)=1-P(X=0)$

$P(X\ge1)=1-\frac{20!}{0!20!}\cdot0.02^0\cdot0.98^{20}=1-0.6676=0.3324$

ii) using Poisson distribution

The average rate of incorrect bill is $λ = np = 20*0.02 = 0.4$

$$P(X=x)=\frac{e^{-\lambda}\lambda^x}{x!}$$

$P(X\ge1)=1-P(X=0)$

$P(X\ge1)=1-\frac{e^{-0.4}0.4^0}{0!}=1-0.6703=0.3297$

The probability that at least one bill will be incorrect in 20 bills is 0.3324 (by using binomial distribution) and 0.3297 (by using Poisson distribution). The binomial distribution and Poisson distribution give approximately the same result. Since one is a limiting case of the other there is not that much variation is observed in finding the probability of at least one bill will be incorrect.