At a collection agency, 80% of the accounts are eventually paid in full. Over the next month, the collection agency gets 120 new accounts. What is the probability that at least 75% of them are eventually paid in full?

#### Solution

Let X be the number of accounts paid in full in the sample of 120 and $p$ be the probability that the account is paid in full.

Given that $n =120$ and $p=0.8$. Thus $X\sim B(120, 0.8)$.

The distribution of X is approximately Normal with mean = $\mu= n*p = 120 \times 0.8 = 96$ and standard deviation

` $$ \begin{aligned} \sigma&= \sqrt{n*p*(1-p)}\\ &= \sqrt{120 \times 0.8 \times (1- 0.8)}\\ &=4.38 \end{aligned} $$ `

We need to find the probability that 75% of 120 (i.e., 90) accounts are eventually paid in full.

Thus the probability that at least 75% of them are eventually paid in full is $P(X\geq 90)$.

` $$ \begin{aligned} P(X\geq 90) & = P(X > 90.5)\\ &\quad \quad (\text{ using continuity correction})\\ & = 1- P(X < 90.5)\\ &= 1-P\bigg(\frac{X-\mu}{\sigma} < \frac{90.5- 96}{4.38}\bigg)\\ & =1-P(Z < -1.26)\\ & = 1-0.1046\\ & = 0.8954 \end{aligned} $$ `

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators