Question

The Center for Medicare and Medical Services reported that there were 295,000 appeals for hospitalization and other Part A Medicare service. For this group, 40% of first-round appeals were successful (The Wall Street Journal, October 22, 2012). Suppose ten first-round appeals have just been received by a Medicare appeals office. Round your answers to four decimal places.
(a) Compute the probability that none of the appeals will be successful.
(b) Compute the probability that exactly one of the appeals will be successful.
(c) What is the probability that at least two of the appeals will be successful?
(d) What is the probability that more than half of the appeals will be successful?

Answer 1

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;

$P(X=r)= \binom{n}{r}p^{r}(1-p)^{n-r} ; x = 0,1,2,3,....$

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 ~ $Binom(n=10,p=0.40)$

(a) Probability that none of the appeals will be successful = P(X = 0)

     P(X = 0) = $\binom{10}{0}0.40^{0}(1-0.40)^{10-0}$

                   = $1*0.6^{10}$ = 0.00605

(b) Probability that exactly one of the appeals will be successful = P(X = 1)

     P(X = 1) = $\binom{10}{1}0.40^{1}(1-0.40)^{10-1}$

                  = $10*0.4^{1} *0.6^{10-1}$ = 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 - $\binom{10}{0}0.40^{0}(1-0.40)^{10-0} - \binom{10}{1}0.40^{1}(1-0.40)^{10-1}$

                     = 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($\mu = n*p=4,\sigma^{2}= n*p*q = 2.4$)

  and standard normal z has distribution as;

      Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

  P(X > 0.5) = P( $\frac{X-\mu}{\sigma}$ > $\frac{0.5-4}{\sqrt{2.4} }$ ) = P(Z > -2.26) = P(Z < 2.26) = 0.98809