Respuesta :
Answer:
(a) 0.00605
(b) 0.0403
(c) 0.9536
(d) 0.98809
Step-by-step explanation:
We are given that 40% of first-round appeals were successful (The Wall Street Journal, October 22, 2012) and suppose ten first-round appeals have just been received by a Medicare appeals office.
This situation can be represented through Binomial distribution as;
[tex]P(X=r)= \binom{n}{r}p^{r}(1-p)^{n-r} ; x = 0,1,2,3,....[/tex]
where, n = number of trials (samples) taken = 10
r = number of success
p = probability of success which in our question is % of first-round
appeals that were successful, i.e.; 40%
So, here X ~ [tex]Binom(n=10,p=0.40)[/tex]
(a) Probability that none of the appeals will be successful = P(X = 0)
P(X = 0) = [tex]\binom{10}{0}0.40^{0}(1-0.40)^{10-0}[/tex]
= [tex]1*0.6^{10}[/tex] = 0.00605
(b) Probability that exactly one of the appeals will be successful = P(X = 1)
P(X = 1) = [tex]\binom{10}{1}0.40^{1}(1-0.40)^{10-1}[/tex]
= [tex]10*0.4^{1} *0.6^{10-1}[/tex] = 0.0403
(c) Probability that at least two of the appeals will be successful = P(X>=2)
P(X >= 2) = 1 - P(X = 0) - P(X = 1)
= 1 - [tex]\binom{10}{0}0.40^{0}(1-0.40)^{10-0} - \binom{10}{1}0.40^{1}(1-0.40)^{10-1}[/tex]
= 1 - 0.00605 - 0.0403 = 0.9536
(d) Probability that more than half of the appeals will be successful = P(X > 0.5)
For this probability we will convert our distribution into normal such that;
X ~ N([tex]\mu = n*p=4,\sigma^{2}= n*p*q = 2.4[/tex])
and standard normal z has distribution as;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)
P(X > 0.5) = P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{0.5-4}{\sqrt{2.4} }[/tex] ) = P(Z > -2.26) = P(Z < 2.26) = 0.98809
