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3 years ago

What is the percent of 0.25% out of 50

Mathematics

1 answer:

kicyunya [14]3 years ago

4 0

It would be 0.125 because by converting percent to a decimal you must first multiply 0.0025 x 50.

Hope that helps

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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

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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

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