Answer to Question #342979 in Statistics and Probability for Carol

Question

QUESTION 24

According to a report from the research for Studying Health System Change, 20% of South Africans

delay or go without medical care because of concerns about cost. Suppose that 8 individuals are

randomly selected.

(a) What is the probability that two individuals will delay or go without medical care? (2)

(b) What is the probability that at most two individuals will delay or go without medical care? (3)

(c) What is the probability that at least seven individuals will delay or go without medical care?

(3)

[8]

Answer 1

Here we have a binomial distribution:

$P(x)=\frac {n!} {(n-x)!x!}p^xq^{n-x}$

where $n=8, p=1/5 \:(20\%),q=1-p=4/5$

So,

(a) $P(2)=\frac {8!} {(8-2)!2!}(\frac 1 5)^2(\frac 4 5 )^{8-2}=1.12*0.26=0.294=29.4\%$

(b) $P(\le2)=P(0)+P(1)+P(2)=$

$=\frac {8!} {(8-0)!0!}(\frac 1 5)^0(\frac 4 5 )^{8-0}+\frac {8!} {(8-1)!1!}(\frac 1 5)^1(\frac 4 5 )^{8-1}+\frac {8!} {(8-2)!2!}(\frac 1 5)^2(\frac 4 5 )^{8-2}=0.168+0.335+0.294=0.7975=79.75\%$

c) $n=20$

$P(\ge7)=P(7)+P(8)=$

$=\frac {8!} {(8-7)!7!}(\frac 1 5)^7(\frac 4 5 )^{8-7}+\frac {8!} {(8-8)!8!}(\frac 1 5)^8(\frac 4 5 )^{8-8}=$

$0.00008192+0.00000256=0.00008448=0.008448\%$