Question

A hospital finds that 22% of its accounts are at least 1 month in arrears. A random sample of 425 accounts was taken. What is the probability that fewer than 82 accounts in the sample were at least 1 month in arrears

Answer 1

Answer:

8.85% probability that fewer than 82 accounts in the sample were at least 1 month in arrears

Step-by-step explanation:

For each account, there are only two possible outcomes. Either they are at least 1 month in arrears, or they are not. The probability of an account being at least 1 month in arrears is independent from other accounts. So the binomial probability distribution is used to solve this question.

However, we are working with a large sample. So i am going to aproximate this binomial distribution to the normal.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

In this problem, we have that:

$p = 0.22, n = 425$

So

$\mu = E(X) = np = 425*0.22 = 93.5$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{425*0.22*0.78} = 8.54$

What is the probability that fewer than 82 accounts in the sample were at least 1 month in arrears

This probability is the pvalue of Z when X = 82. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{82 - 93.5}{8.54}$

$Z = -1.35$

$Z = -1.35$ has a pvalue of 0.0885.

8.85% probability that fewer than 82 accounts in the sample were at least 1 month in arrears