# Solved:Dottie’s Tax Service specializes in federal tax returns for professional clients, such as physicians

#### ByDr. Raju Chaudhari

Sep 20, 2020

Dottie's Tax Service specializes in federal tax returns for professional clients, such as physicians, dentists, accountants, and lawyers. A recent audit by the IRS of the returns she prepared indicated that an error was made on 11% of the returns she prepared last year. Assuming this rate continues into this year and she prepares 53 returns, what is the probability that she makes errors on:

a) More than 6 returns? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)
b) At least 6returns? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)
c) Exactly 6 returns? (Round z-score computation to 2 decimal places and your final answer to 4 decimal places.)

### Solution

Here $X$ denote the number of returns with errors out of 53 returns.

Let $p$ be the probability of making error in returns.

Given that $p=0.11$ and $n =53$. Thus $X\sim B(53, 0.11)$.

We use Normal approximation to Binomial distribution.

Mean is

 \begin{aligned} \mu&= n*p \\ &= 53 \times 0.11 \\ &= 5.83. \end{aligned}

and standard deviation is
 \begin{aligned} \sigma &= \sqrt{n*p*(1-p)} \\ &= \sqrt{53 \times 0.11 \times (1- 0.11)}\\ &=2.2779. \end{aligned}

a. The probability that she makes errors on more than 6 returns is

 \begin{aligned} P(X > 6) &= P(X > 6.5)\\ &=P\bigg(\frac{X-\mu}{\sigma} > \frac{6.5- 5.83}{2.2779}\bigg)\\ &= P(Z >0.29)\\ & =1-P(Z < 0.29)\\ & = 0.3859 \end{aligned}

b. The probability that she makes errors on at least 6 returns is

 \begin{aligned} P(X \geq 6) &= P(X \geq 5.5)\\ &=P\bigg(\frac{X-\mu}{\sigma} \geq \frac{5.5- 5.83}{2.2779}\bigg)\\ &= P(Z \geq-0.14)\\ & =1-P(Z \leq -0.14)\\ & = 0.5557 \end{aligned}

c. The probability that she makes errors on exactly 6 returns is

 \begin{aligned} P(X=6) &= P(X\geq 6) - P(X>6)\\ & = 0.5557-0.3859\\ & = 0.1698 \end{aligned}