Normal Approximation to Binomial Distribution Calculator with Examples

Normal Approximation to Binomial Distribution Calculator

Normal Approximation Calculator is used to calculate the probabilities of events from discrete distributions, such as the Binomial and Poisson distributions, using the normal distribution. The normal distribution is a continuous probability distribution that is used to model data that is approximately normally distributed.

How to Use Normal Approximation for Binomial Distribution Calculator?

Enter the Number of Trails (n)
Enter the Probability of Success (p)
Enter the Mean value
Enter the Standard Deviation
Select the Probability
Click on “Calculate” button to use Normal Approximation Calculator
Calculate Required approximate Probability

Normal Approximation Calculator

Use this normal approximation calculator to binomial distribution to calculate approximate probability using number of trials, probability of success,mean,standard deviation.

Normal Approximation to Binomial Distribution Calculator
Number of Trials ($n$)
Probability of Success ($p$)
Mean ($\mu=np$)
Standard deviation ($\sqrt{np(1-p)}$)
P(X = A)
P(X < A)
P(X ≤ A)
P(A< X ≤ B)	and
P(A ≤ X < B)	and

Normal Approximate Calculator Results
Required Probability :

Here’s the brief explanation about each fields in normal approximation calculator.

NUmber of Trials ($n$) : This field represents the total number of independent trials or experiments. In the context of binomal distribution, it’s the number of times an event with two possible outcomes (success or failure) will be repeated. For example, if you are flipping coin 20 times to see how many times it will show heads up, the total number of trials would be 20.
Probability of Success ($p$) : This field represents the probability of a single trial resulting in a “Success”. For example, If you expects heads 50% of the time when you are flip a coin, the probability of success (heads) would be 0.5.
Mean ($\mu=np$) : This field represents the the mean or average value of the binomial distribution.
Standard Deviation ($\sqrt{np(1-p)}$) : This field represents the variability of the binomial distribution.
Select the Probability : This fields allows you to choose the specific probability you want to calculate using the normal approximation. You can select on of the following options:
1. P(X = A): Probability that a specific number of successes (A) occur.
2. P(X < A): Probability that the number of successes is less than A.
3. P(X ≤ A): Probability that the number of successes is less than or equal to A.
4. P(A < X ≤ B): Probability that the number of successes falls within a range defined by A and B.
5. P(A ≤ X < B): Probability that the number of successes falls within a range defined by A and B.
Required Probability: This field displays the calculated probability based on the selected probability scenario (e.g., P(X = A), P(X < A), P(X ≤ A), P(A < X ≤ B), P(A ≤ X < B)). This probability represents the likelihood of the specified number of successes (A) falling within the chosen range.

Normal Approximation to Binomial Distribution Formula

Let $X$ be a binomially distributed 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 to binomial distribution are as follows:

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

Example 1 - Normal Approximation to Binomial Distribution

In a certain Binomial distribution with probability of success $p=0.20$ and number of trials $n = 30$.

Calculate approximate probability that

a. the probability of getting 5 successes,

b. the probability of getting at least 5 successes,

c. the probability of getting between 5 and 10 (inclusive) successes.

Solution

Let $X$ denote the number of successes in 30 trials and let $p$ be the probability of success.

Given that $n =30$ and $p=0.2$. Thus $X\sim B(30, 0.2)$.

Here $n*p = 30\times 0.2 = 6>5$ and $n*(1-p) = 30\times (1-0.2) = 24>5$ , we use Normal approximation to Binomial distribution.

Mean of $X$ is
$$ \begin{aligned} \mu&= n*p \\ &= 30 \times 0.2 \\ &= 6. \end{aligned} $$

and standard deviation of $X$ is $$ \begin{aligned} \sigma &= \sqrt{n*p*(1-p)} \\ &= \sqrt{30 \times 0.2 \times (1- 0.2)}\\ &=2.1909. \end{aligned} $$

a. Using the continuity correction calculator, $P(X=5)$ 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

$$ \begin{aligned} z_1=\frac{4.5-\mu}{\sigma}=\frac{4.5-6}{2.1909}\approx-0.68 \end{aligned} $$ and $$ \begin{aligned} z_2=\frac{5.5-\mu}{\sigma}=\frac{5.5-6}{2.1909}\approx-0.23 \end{aligned} $$

Thus the probability of getting exactly 5 successes is $$ \begin{aligned} P(X= 5) & = P(4.5<X<5.5)\\ &=P(z_1< Z< z_2)\\ &=P(-0.68<Z<-0.23)\\ &=P(Z<-0.23)-P(Z<-0.68)\\ & = 0.409-0.2483\\ & = 0.1607 \end{aligned} $$

b. Using the continuity correction of normal binomial distribution, the probability of getting at least 5 successes i.e., $P(X\geq 5)$ can be written as $P(X\geq5)=P(X\geq 5-0.5)=P(X\geq4.5)$.

The $Z$-score that corresponds to $4.5$ is

$$ \begin{aligned} z=\frac{4.5-\mu}{\sigma}=\frac{4.5-6}{2.1909}\approx-0.68 \end{aligned} $$ Thus, the probability of getting at least 5 successes is

$$ \begin{aligned} P(X\geq 5) &= P(X\geq4.5)\\ &= 1-P(X<4.5)\\ &= 1-P(Z<-0.68)\\ & = 1-0.2483\\ & = 0.7517 \end{aligned} $$

c. Using the continuity correction for normal distribution approximation, the probability of getting between 5 and 10 (inclusive) successes is $P(5\leq X\leq 10)$ 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

$$ \begin{aligned} z_1=\frac{4.5-\mu}{\sigma}=\frac{4.5-6}{2.1909}\approx-0.68 \end{aligned} $$ and $$ \begin{aligned} z_2=\frac{10.5-\mu}{\sigma}=\frac{10.5-6}{2.1909}\approx2.05 \end{aligned} $$

$$ \begin{aligned} P(5\leq X\leq 10) &= P(5-0.5 < X <10+0.5)\\ &= P(4.5 < X <10.5)\\ &=P(-0.68\leq Z\leq 2.05)\\ &=P(Z\leq 2.05)-P(Z\leq -0.68)\\ &=0.9798-0.2483\\ &=0.7315 \end{aligned} $$

Example 2 - Normal Approximation Calculator

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 the Binomial distribution calculation as below:

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 calculation, $P(X=5)$ 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

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

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

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

c. Using the continuity correction normal binomial distribution, 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

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

The probability that between $5$ and $10$ (inclusive) persons travel by train is

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

Example 3 - Normal Approximation Calculator

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 for normal approximation to binomial distribution, $P(X=215)$ 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 $$ \begin{aligned} P(X= 215) & = P(214.5 < X < 215.5)\\ &=P(z_1 < Z < z_2)\\ &=P(1.32 < Z < 1.41)\\ &=P(Z < 1.41)-P(Z < 1.32)\\ & = 0.9207-0.9066\\ & = 0.0141 \end{aligned} $$

b. By continuity correction normal approximation distribution,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

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

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 approximate probability that at most $215$ drivers wear a seat belt is

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

d. Using the continuity correction calculator, 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

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

The approximate probability that between $210$ and $220$ (inclusive) drivers wear seat belt is

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

Example 4 - Normal Approximation to Binomial Distribution

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 approximate 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 calculator, 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 approximate probability that at least 150 persons travel by train is

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

Example 5 - Normal Approximation to Binomial Distribution

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 approximate probability that at least $20$ eagle will survive their first flight is

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

Example 6 - Continuity Correction Examples

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 calculation

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. Normal Approximation 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 calculator.

Using the continuity correction, the approximate 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

Hope you like Normal Approximation to Binomial Distribution Calculator and step by step guide with examples and calculator. Click on Theory button to read more about Normal approximation to bionomial distribution.