Mathematics, 02.07.2021 21:30 djdjd11

According to estimates by the office of the Treasury Inspector General of IRS, approximately 0.0746 of the tax returns filed are fraudulent or will contain errors that are purposely made to cheat the IRS. In a random sample of 318 independent returns from this year, what is the probability that at least 23 will be fraudulent or will contain errors that are purposely made to cheat the IRS

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0.6026 = 60.26% probability that at least 23 will be fraudulent or will contain errors that are purposely made to cheat the IRS

Step-by-step explanation:

We use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x successes 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 distributions 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 $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

Approximately 0.0746 of the tax returns filed are fraudulent or will contain errors.

This means that p = 0.0746

Random sample of 318 independent returns

This means that n = 318

Mean and standard deviation:

$\mu = E(X) = np = 318*0.0746 = 23.7228$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{318*0.0746*0.9254} = 4.6854$

What is the probability that at least 23 will be fraudulent or will contain errors that are purposely made to cheat the IRS?

Using continuity correction, this is $P(X \geq 23 - 0.5) = P(X \geq 22.5)$ , which is 1 subtracted by the p-value of Z when X = 22.5. So

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

$Z = \frac{22.5 - 23.7228}{4.6854}$

Z = -0.26

Z = -0.26 has a p-value of 0.3974.

1 - 0.3974 = 0.6026

0.6026 = 60.26% probability that at least 23 will be fraudulent or will contain errors that are purposely made to cheat the IRS

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According to estimates by the office of the Treasury Inspector General of IRS, approximately 0.0746...