Question

Use the normal approximation to the binomial distribution to answer this question. Fifteen percent of all students at a large university are absent on Mondays. If a random sample of 12 names is called on a Monday, what is the probability that four students are absent?

Answer 1

Answer: 0.1289

Step-by-step explanation:

Given : The proportion of all students at a large university are absent on Mondays. : $p=0.15$

Sample size : $n=12$

Mean : $\mu=np=12\times0.15=1.8$

Standard deviation = $\sigma=\sqrt{np(1-p)}$

$\Rightarrow\ \sigma=\sqrt{12(0.15)(1-0.15)}=1.23693168769\approx1.2369$

Let x be a binomial variable.

Using the standard normal distribution table ,

$P(x=4)=P(x\leq4)-P(x\leq3)$              (1)

Z score fro normal distribution:-

$z=\dfrac{x-\mu}{\sigma}$

For x=4

$z=\dfrac{4-1.8}{1.2369}\approx1.78$

For x=3

$z=\dfrac{3-1.8}{1.2369}\approx0.97$

Then , from (1)

$P(x=4)=P(z\leq1.78)-P(z\leq0.97)\\\\=0.962462-0.8339768\approx0.1289$    

Hence, the probability that four students are absent = $0.1289$