The number of parking tickets issued in a certain city on any given weekday has a Poisson distribution with parameter $\lambda = 50$. What is the approximate probability that

a. Between 35 and 70 tickets are given out on a particular day? (Hint: When $\lambda$ is large, a Poisson rv has approximately a normal distribution.)

b. The total number of tickets given out during a 5-day week is between 225 and 275?

#### Solution

Let $X$ denote the number of parking tickets issued in a certain city on any given weekday. Given that $\lambda = 50$. $X$ follows Poisson distribution, i.e., $X\sim P(50)$.

Since $\lambda= 50$ is large enough, we use normal approximation to Poisson distribution. That is $\frac{X-\lambda}{\sqrt{\lambda}}\to N(0,1)$ for large $\lambda$.

(a) The approximate probability that between 35 and 70 tickets are given out on a particular day is

`$$ \begin{aligned} P(35< X< 70)&= P\big(\frac{35-50}{\sqrt{50}}<\frac{X-\lambda}{\sqrt{\lambda}}<\frac{70-50}{\sqrt{50}}\big)\\ &= P(-2.1213< Z<2.8284)\\ & = P(Z < 2.8284) - P(Z < -2.1213)\\ &= 0.9977-0.0169\\ &= 0.9808 \end{aligned} $$`

(b) During a 5-day week the total number of tickets given out follows Poisson distribution with paramater $5*\lambda = 250$.

The approximate probability that total number of tickets given out during a 5-day week is between 225 and 275 tickets is

`$$ \begin{aligned} P(225< X < 275)&= P\big(\frac{225-250}{\sqrt{250}}<\frac{X-\lambda}{\sqrt{\lambda}}<\frac{275-250}{\sqrt{250}}\big)\\ &= P(-1.5811< Z< 1.5811)\\ & = P(Z < 1.5811) - P(Z < -1.5811)\\ &= 0.9431-0.0569\\ &= 0.8862 \end{aligned} $$`

#### Further Reading

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