Normal Approximation to Binomial Calculator with Examples

Normal Approximation to Binomial Calculator with examples

Let $X$ be a Binomial random variable with number of trials $n$ and probability of success $p$.

The mean of $X$ is $\mu=E(X) = np$ and variance of $X$ is $\sigma^2=V(X)=np(1-p)$.

The general rule of thumb to use normal approximation to binomial distribution is that the sample size $n$ is sufficiently large if $np \geq 5$ and $n(1-p)\geq 5$.

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

Continuity Correction for normal approximation

Binomial distribution is a discrete distribution, whereas normal distribution is a continuous distribution. When we are using the normal approximation to Binomial 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 Binomial Calculator

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

Step 1 - Enter the number of trials $n$

Step 2 - Enter the probability of success $p$

Step 3 - Select appropriate probability event

Step 4 - Enter the values of $A$ or $B$ or Both

Step 5 - Click on "Calculate" button to get normal approximation to Binomial probabilities

Step 6 - Gives output for mean of the distribution

Step 7 - Gives the output for variance of the distribution

Step 8 - Calculate the required probability

Normal Approximation to Binomial Example 1

In a large population 40% of the people travel by train. If a random sample of size $n=20$ is selected, then find the approximate probability that

a. exactly 5 persons travel by train,
b. at least 10 persons travel by train,
c. between 5 and 10 (inclusive) persons travel by train.

Solution

Let $X$ denote the number of persons travelling by train out of $20$ selected persons and let $p$ be the probability that a person travel by train.

Given that $n =20$ and $p=0.4$. Thus $X\sim B(20, 0.4)$.

Here $n*p = 20\times 0.4 = 8 > 5$ and $n*(1-p) = 20\times (1-0.4) = 12>5$, we use Normal approximation to Binomial distribution.

Mean of $X$ is

$$ \begin{aligned} \mu&= n*p \\ &= 20 \times 0.4 \\ &= 8. \end{aligned} $$

and standard deviation of $X$ is

$$ \begin{aligned} \sigma &= \sqrt{n*p*(1-p)} \\ &= \sqrt{20 \times 0.4 \times (1- 0.4)}\\ &=2.1909. \end{aligned} $$

a. Using the continuity correction, $P(X=5)$ can be written as $P(5-0.5 < X < 5+0.5)=P(4.5 < X < 5.5)$.

The $Z$-scores that corresponds to $4.5$ and $5.5$ are respectively

$$ \begin{aligned} z_1&=\frac{4.5-\mu}{\sigma}\\ &=\frac{4.5-8}{2.1909}\\ &\approx-1.6 \end{aligned} $$
and

$$ \begin{aligned} z_2&=\frac{5.5-\mu}{\sigma}\\ &=\frac{5.5-8}{2.1909}\\ &\approx-1.14 \end{aligned} $$

Thus the probability that exactly $5$ persons travel by train is

$$ \begin{aligned} P(X= 5) & = P(4.5 < X < 5.5)\\ &=P(z_1 < Z < z_2)\\ &=P(-1.6 < Z < -1.14)\\ &=P(Z < -1.14)-P(Z < -1.6)\\ & = 0.1271-0.0548\\ & \qquad (\text{from normal table})\\ & = 0.0723 \end{aligned} $$

b. Using the continuity correction, the probability that at least $10$ persons travel by train i.e., $P(X\geq 10)$ can be written as $P(X\geq10)=P(X\geq 10-0.5)=P(X\geq9.5)$.

The $Z$-score that corresponds to $9.5$ is

$$ \begin{aligned} z&=\frac{9.5-\mu}{\sigma}\\ &=\frac{9.5-8}{2.1909}\\ &\approx0.68 \end{aligned} $$

Thus, the probability that at least 10 persons travel by train is

$$ \begin{aligned} P(X\geq 10) &= P(X\geq9.5)\\ &= 1-P(X < 9.5)\\ &= 1-P(Z < 0.68)\\ & = 1-0.7517\\ & \qquad (\text{from normal table})\\ & = 0.2483 \end{aligned} $$

c. Using the continuity correction, the probability that between $5$ and $10$ (inclusive) persons travel by train i.e., $P(5\leq X\leq 10)$ can be written as $P(5-0.5 < X < 10+0.5)=P(4.5 < X < 10.5)$.

The $Z$-scores that corresponds to $4.5$ and $10.5$ are respectively

$$ \begin{aligned} z_1&=\frac{4.5-\mu}{\sigma}\\ &=\frac{4.5-8}{2.1909}\\ &\approx-1.6 \end{aligned} $$
and

$$ \begin{aligned} z_2&=\frac{10.5-\mu}{\sigma}\\ &=\frac{10.5-8}{2.1909}\\ &\approx1.14 \end{aligned} $$

$$ \begin{aligned} P(5\leq X\leq 10) &= P(5-0.5 < X <10+ 0.5)\\ &= P(4.5 < X < 10.5)\\ &=P(-1.6\leq Z\leq 1.14)\\ &=P(Z\leq 1.14)-P(Z\leq -1.6)\\ &=0.8729-0.0548\\ & \qquad (\text{from normal table})\\ &=0.8181 \end{aligned} $$

Normal Approximation to Binomial Example 2

Suppose that only 40% of drivers in a certain state wear a seat belt. A random sample of 500 drivers is selected.
Approximate the probability that

a. exactly 215 drivers wear a seat belt,
b. at least 220 drivers wear a seat belt,
c. at the most 215 drivers wear a seat belt,
d. between 210 and 220 drivers wear a seat belt.

Solution

Let $X$ denote the number of drivers who wear seat beltout of 500 selected drivers and let $p$ be the probability that a driver wear seat belt.

Given that $n =500$ and $p=0.4$. Thus $X\sim B(500, 0.4)$.

As $n*p = 500\times 0.4 = 200 > 5$ and $n*(1-p) = 500\times (1-0.4) = 300 > 5$, we use Normal approximation to Binomial distribution.

Mean of $X$ is

$$ \begin{aligned} \mu&= n*p \\ &= 500 \times 0.4 \\ &= 200. \end{aligned} $$

and standard deviation of $X$ is

$$ \begin{aligned} \sigma &= \sqrt{n*p*(1-p)} \\ &= \sqrt{500 \times 0.4 \times (1- 0.4)}\\ &=10.9545. \end{aligned} $$

a. Using the continuity correction, $P(X=215)$ can be written as $P(215-0.5 < X < 215+0.5)=P(214.5 < X < 215.5)$.

The $Z$-scores that corresponds to $214.5$ and $215.5$ are respectively

$$ \begin{aligned} z_1&=\frac{214.5-\mu}{\sigma}\\ &=\frac{214.5-200}{10.9545}\\ &\approx1.32 \end{aligned} $$
`
and

$$ \begin{aligned} z_2&=\frac{215.5-\mu}{\sigma}\\ &=\frac{215.5-200}{10.9545}\\ &\approx1.41 \end{aligned} $$

Thus the probability that exactly $215$ drivers wear a seat belt is

$$ \begin{aligned} P(X= 215) & = P(214.5 < X < 215.5)\\ &=P(z_1 < Z < z_2)\\ &=P(1.32 < Z < 1.41)\\ &=P(Z < 1.41)-P(Z < 1.32)\\ & = 0.9207-0.9066\\ & \qquad (\text{from normal table})\\ & = 0.0141 \end{aligned} $$

b. By continuity correction the probability that at least 220 drivers wear a seat belt i.e., $P(X\geq 220)$ can be written as $P(X\geq220)=P(X\geq 220-0.5)=P(X\geq219.5)$.

The $Z$-score that corresponds to $219.5$ is

$$ \begin{aligned} z&=\frac{219.5-\mu}{\sigma}\\ &=\frac{219.5-200}{10.9545}\\ &\approx1.78 \end{aligned} $$
Thus, the probability that at least $220$ drivers wear a seat belt is

$$ \begin{aligned} P(X\geq 220) &= P(X\geq219.5)\\ &= 1-P(X < 219.5)\\ &= 1-P(Z < 1.78)\\ & = 1-0.9625\\ & \qquad (\text{from normal table})\\ & = 0.0375 \end{aligned} $$

c. By continuity correction the probability that at most $215$ drivers wear a seat belt i.e., $P(X\leq 215)$ can be written as $P(X\leq215)=P(X\leq 215-0.5)=P(X\leq214.5)$.

The $Z$-score that corresponds to $214.5$ is

$$ \begin{aligned} z&=\frac{214.5-\mu}{\sigma}\\ &=\frac{214.5-200}{10.9545}\\ &\approx1.32 \end{aligned} $$
Thus, the probability that at most $215$ drivers wear a seat belt is

$$ \begin{aligned} P(X\leq 215) &= P(X\leq214.5)\\ &= P(X < 214.5)\\ &= P(Z < 1.32)\\ & \qquad (\text{from normal table})\\ &=0.9066 \end{aligned} $$

d. Using the continuity correction, the probability that between $210$ and $220$ (inclusive) drivers wear seat belt is $P(210\leq X\leq 220)$ can be written as $P(210-0.5 < X < 220+0.5)=P(209.5 < X < 220.5)$.

The $Z$-scores that corresponds to $209.5$ and $220.5$ are respectively

$$ \begin{aligned} z_1&=\frac{209.5-\mu}{\sigma}\\ &=\frac{209.5-200}{10.9545}\\ &\approx0.87 \end{aligned} $$
and

$$ \begin{aligned} z_2&=\frac{220.5-\mu}{\sigma}\\ &=\frac{220.5-200}{10.9545}\\ &\approx1.87 \end{aligned} $$

$$ \begin{aligned} P(210\leq X\leq 220) &= P(210-0.5 < X < 220+0.5)\\ &= P(209.5 < X < 220.5)\\ &=P(0.87\leq Z\leq 1.87)\\ &=P(Z\leq 1.87)-P(Z\leq 0.87)\\ &=0.9693-0.8078\\ & \qquad (\text{from normal table})\\ &=0.1615 \end{aligned} $$

Normal Approximation to Binomial Example 3

When telephone subscribers call from the National Magazine Subscription Company, 18% of the people who answer stay on the line for more than one minute. If 800 people are called in a day, find the probability that

a. at least 150 stay on the line for more than one minute. (Use normal approximation to binomial).
b. more than 200 stay on the line. (Use normal approximation to Binomial).

Solution

Let $X$ denote the number of people who answer stay online for more than one minute out of 800 people called in a day and let $p$ be the probability people who answer stay online for more than one minute.

Given that $n =800$ and $p=0.18$. Thus $X\sim B(800, 0.18)$.

As $n*p = 800\times 0.18 = 144 > 5$ and $n*(1-p) = 800\times (1-0.18) = 656>5$, we use Normal approximation to Binomial distribution.

Mean of $X$ is

$$ \begin{aligned} \mu&= n*p \\ &= 800 \times 0.18 \\ &= 144. \end{aligned} $$

and standard deviation of $X$ is

$$ \begin{aligned} \sigma &= \sqrt{n*p*(1-p)} \\ &= \sqrt{800 \times 0.18 \times (1- 0.18)}\\ &=10.8665. \end{aligned} $$

a. By continuity correction the probability that at least 150 people stay online for more than one minute i.e., $P(X\geq 150)$ can be written as $P(X\geq150)=P(X\geq 150-0.5)=P(X \geq 149.5)$.

The $Z$-score that corresponds to $149.5$ is

$$ \begin{aligned} z&=\frac{149.5-\mu}{\sigma}\\ &=\frac{149.5-144}{10.8665}\\ &\approx0.51 \end{aligned} $$

Thus, the probability that at least $150$ people stay online for more than one minute is

$$ \begin{aligned} P(X\geq 150) &= P(X\geq149.5)\\ &= 1-P(X < 149.5)\\ &= 1-P(Z < 0.51)\\ & = 1-0.695\\ & \qquad (\text{from normal table})\\ & = 0.305 \end{aligned} $$

b. Using the continuity correction, the probability that more than $150$ people stay online for more than one minute i.e., $P(X > 150)$ can be written as $P(X\geq150)=P(X\geq 150-0.5)=P(X\geq149.5)$.

The $Z$-score that corresponds to $149.5$ is

$$ \begin{aligned} z&=\frac{149.5-\mu}{\sigma}\\ &=\frac{149.5-144}{10.8665}\\ &\approx0.51 \end{aligned} $$

Thus, the probability that at least 150 persons travel by train is

$$ \begin{aligned} P(X\geq 150) &= P(X\geq149.5)\\ &= 1-P(X < 149.5)\\ &= 1-P(Z < 0.51)\\ & = 1-0.695\\ & \qquad (\text{from normal table})\\ & = 0.305 \end{aligned} $$

Normal Approximation to Binomial Example 4

60% of all young bald eagles will survive their first flight. If 30 randomly selected young bald eagles are observed, what is the probability that at least 20 of them will survive their first flight?

Solution

Let $X$ denote the number of bald eagles who survive their first flight out of 30 observed bald eagles and let $p$ be the probability that young bald eagle will survive their first flight.

Given that $n =30$ and $p=0.6$. Thus $X\sim B(30, 0.6)$.

As $n*p = 30\times 0.6 = 18 > 5$ and $n*(1-p) = 30\times (1-0.6) = 12 > 5$, we use Normal approximation to Binomial distribution.

Mean of $X$ is

$$ \begin{aligned} \mu&= n*p \\ &= 30 \times 0.6 \\ &= 18. \end{aligned} $$

and standard deviation of $X$ is

$$ \begin{aligned} \sigma &= \sqrt{n*p*(1-p)} \\ &= \sqrt{30 \times 0.6 \times (1- 0.6)}\\ &=2.6833. \end{aligned} $$

a. By continuity correction the probability that at least 20 eagle will survive their first flight, i.e., $P(X\geq 20)$ can be written as $P(X\geq20)=P(X\geq 20-0.5)=P(X \geq 19.5)$.

The $Z$-score that corresponds to $19.5$ is

$$ \begin{aligned} z&=\frac{19.5-\mu}{\sigma}\\ &=\frac{19.5-18}{2.6833}\\ &\approx0.56 \end{aligned} $$

Thus, the probability that at least $20$ eagle will survive their first flight is

$$ \begin{aligned} P(X\geq 20) &= P(X\geq19.5)\\ &= 1-P(X < 19.5)\\ &= 1-P(Z < 0.56)\\ & = 1-0.7123\\ & \qquad (\text{from normal table})\\ & = 0.2877 \end{aligned} $$

Normal Approximation to Binomial Example 5

Use normal approximation to estimate the probability of getting 90 to 105 sixes (inclusive of both 90 and 105) when a die is rolled 600 times.

a. Without continuity correction
b. With continuity correction

Solution

Let $X$ denote the number of sixes when a die is rolled 600 times and let $p$ be the probability of getting six.

Given that $n =600$ and $p=0.1667$. Thus $X\sim B(600, 0.1667)$.

Mean of $X$ is

$$ \begin{aligned} \mu&= n*p \\ &= 600 \times 0.1667 \\ &= 100.02. \end{aligned} $$

and standard deviation of $X$ is

$$ \begin{aligned} \sigma &= \sqrt{n*p*(1-p)} \\ &= \sqrt{600 \times 0.1667 \times (1- 0.1667)}\\ &=9.1294. \end{aligned} $$

a. Without continuity correction

The $Z$-scores that corresponds to $90$ and $105$ are respectively

$$ \begin{aligned} z_1&=\frac{90-\mu}{\sigma}\\ &=\frac{90-100.02}{9.1294}\\ &\approx-1.1 \end{aligned} $$

and

$$ \begin{aligned} z_2&=\frac{105-\mu}{\sigma}\\ &=\frac{105-100.02}{9.1294}\\ &\approx0.55 \end{aligned} $$

$$ \begin{aligned} P(90\leq X\leq 105) &=P(-1.1\leq Z\leq 0.55)\\ &=P(Z\leq 0.55)-P(Z\leq -1.1)\\ &=0.7088-0.1357\\ & \qquad (\text{from normal table})\\ &=0.5731 \end{aligned} $$

b. With continuity correction

As $n*p = 600\times 0.1667 = 100.02 > 5$ and $n*(1-p) = 600\times (1-0.1667) = 499.98 > 5$, we use Normal approximation to Binomial distribution.

Using the continuity correction, the probability of getting between $90$ and $105$ (inclusive) sixes i.e., $P(90\leq X\leq 105)$ can be written as $P(90-0.5 < X < 105+0.5)=P(89.5 < X < 105.5)$.

The $Z$-scores that corresponds to $89.5$ and $105.5$ are respectively

$$ \begin{aligned} z_1&=\frac{89.5-\mu}{\sigma}\\ &=\frac{89.5-100.02}{9.1294}\\ &\approx-1.15 \end{aligned} $$

and

$$ \begin{aligned} z_2&=\frac{105.5-\mu}{\sigma}\\ &=\frac{105.5-100.02}{9.1294}\\ &\approx0.6 \end{aligned} $$

$$ \begin{aligned} P(90\leq X\leq 105) &= P(90-0.5 < X < 105+0.5)\\ &= P(89.5 < X < 105.5)\\ &=P(-1.15\leq Z\leq 0.6)\\ &=P(Z\leq 0.6)-P(Z\leq -1.15)\\ &=0.7257-0.1251\\ & \qquad (\text{from normal table})\\ &=0.6006 \end{aligned} $$

Conclusion

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

To read more about the step by step tutorial about the theory of Binomial Distribution and examples of Binomial Distribution Calculator with Examples. This tutorial will help you to understand binomial 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

Let me know in the comments if you have any questions on Normal Approximation to Binomial Distribution and your on thought of this article.