I don't know what ten times 10 is?

A center for medical services reported that there were 295,000 appeals for hospitalization and other services. For this group, 4

pshichka [43]

Answer:

a) 0.0025 = 0.25% probability that none of the appeals will be successful.

b) 0.0207 = 2.07% probability that exactly one of the appeals will be successful.

c) 0.9768 = 97.68% probability that at least two of the appeals will be successful

Step-by-step explanation:

For each appeal, there are only two possible outcomes. Either it is succesful, or it is not. The probability of an appeal being succesful is independent of other appeals, so we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

45% of first-round appeals were successful.

This means that $p = 0.45$

Suppose 10 first-round appeals have just been received by a Medicare appeals office.

This means that $n = 10$

(a) Compute the probability that none of the appeals will be successful.

This is P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{10,0}.(0.45)^{0}.(0.55)^{10} = 0.0025$

0.0025 = 0.25% probability that none of the appeals will be successful.

(b) Compute the probability that exactly one of the appeals will be successful.

This is P(X = 1).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 1) = C_{10,1}.(0.45)^{1}.(0.55)^{9} = 0.0207$

0.0207 = 2.07% 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

This is $P(X \geq 2)$

Either less than two appeals are succesful, or at least two are. The sum of the probabilities of these events is decimal 1. So

$P(X < 2) + P(X \geq 2) = 1$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.0025 + 0.0207 = 0.0232$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.0232 = 0.9768$

0.9768 = 97.68% probability that at least two of the appeals will be successful

6 0

3 years ago

I need help on this one plizz

andrew11 [14]

I'm probably wrong but here you go!

4 0

3 years ago

Which of the following is a statistical question?

Kazeer [188]

I think D because you can get a lot of different answers and that's what a statistical question is.

8 0

3 years ago

Read 2 more answers

I think I used everyone can someone explain this to me

nirvana33 [79]

Answer:

4

Step-by-step explanation:

Terms appear as components of a formula or equation.

Ex. 2n²

So the terms in that equation are:

2n

5

3p

4n

Hope this helps

7 0

3 years ago

Find the intergration as 0 approaches 3

Nuetrik [128]

You have to integrate two equations, one for the case where the value in the absolute signs is negative and one for the case where the value is positive...

So you will integrate one on the interval 0 to 1 and the other on the interval from 1 to 3...

⌠-x+1 dx, x=[0,1] + ⌠x-1 dx, x=[1,3]

(-x^2/2+x)[0,1] + (x^2/2-x)[1,3]

(-0.5+1)+(4.5-3-0.5+1)

(0.5)+(2)

2.5 u^2

6 0

3 years ago