## Normal Approximation to Binomial Distribution Calculator

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 $Z=\frac{X-\mu}{\sigma}=\frac{X-np}{\sqrt{np(1-p)}} \sim N(0,1)$.

## Normal Approximation Calculator

Normal Approximation to Binomial Distribution Calculator | |
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No. 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 |

Results | |

Required Probability : |

### How to use Normal Approximation for Binomial Distribution Calculator?

Step 1 - Enter the Number of Trails (n)

Step 2 - Enter the Probability of Success (p)

Step 3 - Enter the Mean value

Step 4 - Enter the Standard Deviation

Step 5 - Select the Probability

Step 6 - Click on “Calculate” button to use Normal Approximation Calculator

Step 7 - Calculate Required Probability

## Normal Approximation to Binomial Distribution Formula

Continuity correction for normal approximation to binomial distribution

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

## Normal Approximation to Binomial Distribution Example 1

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

Compute

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, $P(X=5)$ 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
`$$ \begin{aligned} z_2=\frac{10.5-\mu}{\sigma}=\frac{10.5-6}{2.1909}\approx2.05 \end{aligned} $$`

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

## Normal Approximation to Binomial Distribution Example 2

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

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

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

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

## Normal Approximation Calculator Example 3

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

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

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