Respuesta :
Answer:
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:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
Problems of normally distributed distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
Approximately 0.0746 of the tax returns filed are fraudulent or will contain errors.
This means that [tex]p = 0.0746[/tex]
Random sample of 318 independent returns
This means that [tex]n = 318[/tex]
Mean and standard deviation:
[tex]\mu = E(X) = np = 318*0.0746 = 23.7228[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{318*0.0746*0.9254} = 4.6854[/tex]
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 [tex]P(X \geq 23 - 0.5) = P(X \geq 22.5)[/tex], which is 1 subtracted by the p-value of Z when X = 22.5. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{22.5 - 23.7228}{4.6854}[/tex]
[tex]Z = -0.26[/tex]
[tex]Z = -0.26[/tex] 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
