Normal approximation to Poisson distribution Examples

In this tutorial we will discuss some numerical examples on Poisson distribution where normal approximation is applicable. For large value of the $\lambda$ (mean of Poisson variate), the Poisson distribution can be well approximated by a normal distribution with the same mean and variance.

Normal approximation to Poisson distribution Examples

Let $X$ be a Poisson distributed random variable with mean $\lambda$.

The mean of Poisson random variable $X$ is $\mu=E(X) = \lambda$ and variance of $X$ is $\sigma^2=V(X)=\lambda$.

The general rule of thumb to use normal approximation to Poisson distribution is that $\lambda$ is sufficiently large (i.e., $\lambda \geq 5$).

For sufficiently large $\lambda$, $X\sim N(\mu, \sigma^2)$. That is $Z=\frac{X-\mu}{\sigma}=\frac{X-\lambda}{\sqrt{\lambda}} \sim N(0,1)$.

Formula for continuity corrections

Poisson distribution is a discrete distribution, whereas normal distribution is a continuous distribution. When we are using the normal approximation to Poisson distribution we need to make correction while calculating various probabilities.

• $P(X=A)=P(A-0.5 < X < A+0.5)$
• $P(X < A)=P(X < A-0.5)$
• $P(X\leq A)=P(X < A+0.5)$
• $P(A< X\leq B)=P(A-0.5 < X < B+0.5)$
• $P(A\leq X< B)=P(A-0.5 < X < B-0.5)$
• $P(A\leq X\leq B)=P(A-0.5 < X < B+0.5)$

Normal approximation to Poisson Distribution Calculator

How to calculate probabilities of Poisson distribution approximated by Normal distribution?

Step 1 - Enter the Poisson Parameter $\lambda$

Step 2 - Select appropriate probability event

Step 3 - Enter the values of $A$ or $B$ or Both

Step 4 - Click on "Calculate" button to get normal approximation to Poisson probabilities

Step 5 - Gives output for mean of the distribution

Step 6 - Gives the output for variance of the distribution

Step 7 - Calculate the required probability

Normal approximation to Poisson distribution Example 1

The mean number of kidney transplants performed per day in the United States in a recent year was about 45. Find the probability that on a given day,

a. exactly 50 kidney transplants will be performed,
b. at least 65 kidney transplants will be performed, and
c. no more than 40 kidney transplants will be performed.

Solution

Let $X$ denote the number of kidney transplants per day. The mean number of kidney transplants performed per day in the United States in a recent year was about 45. $\lambda = 45$. $X$ follows Poisson distribution, i.e., $X\sim P(45)$.

Since $\lambda= 45$ is large enough, we use normal approximation to Poisson distribution. That is $Z=\dfrac{X-\lambda}{\sqrt{\lambda}}\to N(0,1)$ for large $\lambda$. (We use continuity correction)

a. The probability that on a given day, exactly 50 kidney transplants will be performed is

$$ \begin{aligned} P(X=50) &= P(49.5< X < 50.5)\\ & \quad\quad (\text{Using continuity correction})\\ &= P\bigg(\frac{49.5-45}{\sqrt{45}} < \frac{X-\lambda}{\sqrt{\lambda}} < \frac{50.5-45}{\sqrt{45}}\bigg)\\ &= P(0.67 < Z < 0.82)\\ & = P(Z < 0.82) - P(Z < 0.67)\\ &= 0.7939-0.7486\\ & \quad\quad (\text{Using normal table})\\ &= 0.0453 \end{aligned} $$

b. The probability that on a given day, at least 65 kidney transplants will be performed is

$$ \begin{aligned} P(X\geq 65) &= 1-P(X\leq 64)\\ &= 1-P(X < 64.5)\\ & \quad\quad (\text{Using continuity correction})\\ &= 1-P\bigg(\frac{X-\lambda}{\sqrt{\lambda}} < \frac{64.5-45}{\sqrt{45}}\bigg)\\ &= 1-P(Z < 3.06)\\ &= 1-0.9989\\ & \quad\quad (\text{Using normal table})\\ &= 0.0011 \end{aligned} $$

c. The probability that on a given day, no more than 40 kidney transplants will be performed is

$$ \begin{aligned} P(X < 40) &= P(X < 39.5)\\ & \quad\quad (\text{Using continuity correction})\\ &= P\bigg(\frac{X-\lambda}{\sqrt{\lambda}} < \frac{39.5-45}{\sqrt{45}}\bigg)\\ &= P(Z < -0.82)\\ & = P(Z < -0.82) \\ &= 0.2061\\ & \quad\quad (\text{Using normal table}) \end{aligned} $$

Normal approximation to Poisson distribution Example 2

A radioactive element disintegrates such that it follows a Poisson distribution. If the mean number of particles ($\alpha$) emitted is recorded in a 1 second interval as 69, evaluate the probability of:

a. Less than 60 particles are emitted in 1 second.
b. Between 65 and 75 particles inclusive are emitted in 1 second.

Solution

Let $X$ denote the number of particles emitted in a 1 second interval. The mean number of $\alpha$-particles emitted per second $69$. Thus $\lambda = 69$ and given that the random variable $X$ follows Poisson distribution, i.e., $X\sim P(69)$.

Since $\lambda= 69$ is large enough, we use normal approximation to Poisson distribution. That is $Z=\dfrac{X-\lambda}{\sqrt{\lambda}}\to N(0,1)$ for large $\lambda$. (We use continuity correction)

a. The probability that less than 60 particles are emitted in 1 second is

$$ \begin{aligned} P(X < 60) &= P(X < 59.5)\\ & \quad\quad (\text{Using continuity correction})\\ &= P\bigg(\frac{X-\lambda}{\sqrt{\lambda}} < \frac{59.5-69}{\sqrt{69}}\bigg)\\ &= P(Z < -1.14)\\ & = P(Z < -1.14) \\ &= 0.1271\\ & \quad\quad (\text{Using normal table}) \end{aligned} $$

b. The probability that between $65$ and $75$ particles (inclusive) are emitted in 1 second is

$$ \begin{aligned} P(65\leq X\leq 75) &= P(64.5 < X < 75.5)\\ & \quad\quad (\text{Using continuity correction})\\ &= P\bigg(\frac{64.5-69}{\sqrt{69}} < \frac{X-\lambda}{\sqrt{\lambda}} < \frac{75.5-69}{\sqrt{69}}\bigg)\\ &= P(-0.54 < Z < 0.78)\\ &= P(Z < 0.78)- P(Z < -0.54) \\ &= 0.7823-0.2946\\ & \quad\quad (\text{Using normal table})\\ &= 0.4877 \end{aligned} $$

Normal approximation to Poisson distribution Example 3

The number of a certain species of a bacterium in a polluted stream is assumed to follow a Poisson distribution with a mean of 200 cells per ml. if a one ml sample is randomly taken, then what is the probability that this sample contains 225 or more of this bacterium?

Solution

Let $X$ denote the number of a certain species of a bacterium in a polluted stream. The mean number of certain species of a bacterium in a polluted stream per ml is $200$. Thus $\lambda = 200$ and given that the random variable $X$ follows Poisson distribution, i.e., $X\sim P(200)$.

Since $\lambda= 200$ is large enough, we use normal approximation to Poisson distribution. That is $Z=\dfrac{X-\lambda}{\sqrt{\lambda}}\to N(0,1)$ for large $\lambda$. (We use continuity correction)

The probability that one ml sample contains 225 or more of this bacterium is

$$ \begin{aligned} P(X\geq 225) &= 1-P(X\leq 224)\\ &= 1-P(X < 224.5)\\ & \quad\quad (\text{Using continuity correction})\\ &= 1-P\bigg(\frac{X-\lambda}{\sqrt{\lambda}} < \frac{224.5-200}{\sqrt{200}}\bigg)\\ &= 1-P(Z < 1.8)\\ &= 1-0.9641\\ & \quad\quad (\text{Using normal table})\\ &= 0.0359 \end{aligned} $$

Normal approximation to Poisson distribution Example 4

The vehicles enter to the entrance at an expressway follow a Poisson distribution with mean vehicles per hour of 25. Find the probability that in 1 hour the vehicles are between 23 and 27 inclusive, using Normal approximation to Poisson distribution.

Solution

Let $X$ denote the number of vehicles enter to the expressway per hour. The mean number of vehicles enter to the expressway per hour is $25$. Thus $\lambda = 25$ and given that the random variable $X$ follows Poisson distribution, i.e., $X\sim P(25)$.

Since $\lambda= 25$ is large enough, we use normal approximation to Poisson distribution. That is $Z=\dfrac{X-\lambda}{\sqrt{\lambda}}\to N(0,1)$ for large $\lambda$. (We use continuity correction)

The probability that in 1 hour the vehicles are between $23$ and $27$ (inclusive) is

$$ \begin{aligned} P(23\leq X\leq 27) &= P(22.5 < X < 27.5)\\ & \quad\quad (\text{Using continuity correction})\\ &= P\bigg(\frac{22.5-25}{\sqrt{25}} < \frac{X-\lambda}{\sqrt{\lambda}} < \frac{27.5-25}{\sqrt{25}}\bigg)\\ &= P(-0.5 < Z < 0.5)\\ &= P(Z < 0.5)- P(Z < -0.5) \\ &= 0.6915-0.3085\\ & \quad\quad (\text{Using normal table})\\ &= 0.383 \end{aligned} $$

Normal approximation to Poisson distribution Example 5

Assuming that the number of white blood cells per unit of volume of diluted blood counted under a microscope follows a Poisson distribution with $\lambda=150$, what is the probability, using a normal approximation, that a count of 140 or less will be observed?

Solution

Let $X$ denote the number of white blood cells per unit of volume of diluted blood counted under a microscope. Given that the random variable $X$ follows Poisson distribution, i.e., $X\sim P(150)$.

Since the parameter of Poisson distribution is large enough, we use normal approximation to Poisson distribution. That is $Z=\dfrac{X-\lambda}{\sqrt{\lambda}}\to N(0,1)$ for large $\lambda$. (We use continuity correction)

The probability that a count of 140 or less will be observed is

$$ \begin{aligned} P(X \leq 140) &= P(X < 140.5)\\ & \quad\quad (\text{Using continuity correction})\\ &= P\bigg(\frac{X-\lambda}{\sqrt{\lambda}} < \frac{140.5-150}{\sqrt{150}}\bigg)\\ &= P(Z < -0.78)\\ &= 0.2177\\ & \quad\quad (\text{Using normal table}) \end{aligned} $$

Conclusion

In this tutorial, you learned about how to calculate probabilities of Poisson distribution approximated by normal distribution using continuity correction. You also learned about how to solve numerical problems on normal approximation to Poisson distribution.

To read more about the step by step tutorial about the theory of Poisson Distribution and examples of Poisson Distribution Calculator with Examples. This tutorial will help you to understand Poisson distribution and its properties like mean, variance, moment generating function.

To learn more about other probability distributions, please refer to the following tutorial:

Probability distributions

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