Answer: 0.1289
Step-by-step explanation:
Given : The proportion of all students at a large university are absent on Mondays. : [tex]p=0.15[/tex]
Sample size : [tex]n=12[/tex]
Mean : [tex]\mu=np=12\times0.15=1.8[/tex]
Standard deviation = [tex]\sigma=\sqrt{np(1-p)}[/tex]
[tex]\Rightarrow\ \sigma=\sqrt{12(0.15)(1-0.15)}=1.23693168769\approx1.2369[/tex]
Let x be a binomial variable.
Using the standard normal distribution table ,
[tex]P(x=4)=P(x\leq4)-P(x\leq3)[/tex] (1)
Z score fro normal distribution:-
[tex]z=\dfrac{x-\mu}{\sigma}[/tex]
For x=4
[tex]z=\dfrac{4-1.8}{1.2369}\approx1.78[/tex]
For x=3
[tex]z=\dfrac{3-1.8}{1.2369}\approx0.97[/tex]
Then , from (1)
[tex]P(x=4)=P(z\leq1.78)-P(z\leq0.97)\\\\=0.962462-0.8339768\approx0.1289[/tex]
Hence, the probability that four students are absent = [tex]0.1289[/tex]
