Answer:
0.3907
Step-by-step explanation:
We are given that 36% of adults questioned reported that their health was excellent.
Probability of good health = 0.36
Among 11 adults randomly selected from this area, only 3 reported that their health was excellent.
Now we are supposed to find the probability that when 11 adults are randomly selected, 3 or fewer are in excellent health.
i.e. [tex]P(x\leq 3)=P(x=1)+{P(x=2)+P(x=3)[/tex]
Formula :[tex]P(x=r)=^nC_r p^r q ^ {n-r}[/tex]
p is the probability of success i.e. p = 0.36
q = probability of failure = 1- 0.36 = 0.64
n = 11
So, [tex]P(x\leq 3)=P(x=1)+{P(x=2)+P(x=3)[/tex]
[tex]P(x\leq 3)=^{11}C_1 (0.36)^1 (0.64)^{11-1}+^{11}C_2 (0.36)^2 (0.64)^{11-2}+^{11}C_3 (0.36)^3 (0.64)^{11-3}[/tex]
[tex]P(x\leq 3)=\frac{11!}{1!(11-1)!} (0.36)^1 (0.64)^{11-1}+\frac{11!}{2!(11-2)!} (0.36)^2 (0.64)^{11-2}+\frac{11!}{3!(11-3)!} (0.36)^3 (0.64)^{11-3}[/tex]
[tex]P(x\leq 3)=0.390748[/tex]
Hence the probability that when 11 adults are randomly selected, 3 or fewer are in excellent health is 0.3907
