# Normal Approximation Calculator

Created by Tibor Pál, PhD candidate
Reviewed by Rijk de Wet
Last updated: Feb 02, 2023

The normal approximation calculator (more precisely, normal approximation to binomial distribution calculator) helps you to perform normal approximation for a binomial distribution.

## What is normal approximation to binomial distribution?

The normal approximation of binomial distribution is a process where we apply the normal distribution curve to estimate the shape of the binomial distribution. The fundamental basis of the normal approximation method is that the distribution of the outcome of many experiments is at least approximately normally distributed. If you are not familiar with that typically bell-shaped curve, check our normal distribution calculator.

## How to calculate normal approximation — normal approximation calculator with steps

You need to set the following variables to run the normal approximation to binomial calculator. Before using the tool, however, you may want to refresh your knowledge of the concept of probability with our probability calculator.

#### 1. Problem setup

First, tell the normal approximation calculator about the probabilistic problem.

• Number of occurrences or trials ($N$);

• Probability of success ($p$) or the probability of failure ($q = 1-p$);

• Number of successes ($n$); and

• Select the probability you would like to approximate at the event restatement.

• $P(x = n)$ — the probability for an exact discrete value of $n$;
• $P(x > n)$ — the probability for events corresponding to a value greater than $n$;
• $P(x ≤ n)$ — the probability for events occurring at most $n$;
• $P(x < n)$ — the probability for events corresponding to a value lesser than $n$; or
• $P(x ≥ n)$ — the probability for events occurring at least $n$ times.

#### 2. Results

After specifying the problem, you can immediately read both the final and partial results.

• The mean ($μ$);
• The variance($σ^2$);
• The standard deviation ($σ$);
• The problem statement;
• The continuity correction;
• The Z-score;
• The Z-value; and
• The approximated probability.

## How do I calculate normal approximation to binomial distribution?

If you want to compute the normal approximation to binomial distribution by hand, follow the below steps.

1. Find the sample size (the number of occurrences or trials, $N$) and the probabilities $p$ and $q$ — which can be the probability of success ($p$) and probability of failure ($q = 1 - p$), for example.

2. Check if you can apply the normal approximation to the binomial. If $N \times p$ and $N \times q$ are both larger than $5$, then you can use the approximation without worry.

3. Find the mean ($μ$) by multiplying $n$ with $p$, i.e. $μ = N \times p$.

4. Compute the variance ($σ^2$) by multiplying $N$, $p$ and $q$, as $σ^2 = N \times p \times q$.

5. Determine the standard deviation ($\text{SD}$ or $σ$) by taking the square root of the variance: $\sqrt{N \times p \times q}$.

6. State the problem (the number of successes, $n$) using the continuity correction factor according to the below table. (You can learn about this concept more if you open our continuity correction calculator). This is necessary because the normal distribution is a continuous probability distribution, and the binomial distribution is a discrete probability distribution.

Problem statement

Continuity correction

x = n

n-0.5 < x < n+0.5

x ≤ n

x < n + 0.5

x < n

x < n − 0.5

x ≥ n

x > n − 0.5

x > n

x > n + 0.5

1. Find the Z-score with $(x - μ) / σ$.
2. Check the Z-value in the Z-table.
3. Determine the probability associated with the Z-values according to the below table, which will be the normal approximation of binomial distribution.

Problem statement

Probability

x = n

difference of Z-values for n+0.5 and n-0.5

x ≤ n

Z-value

x < n

Z-value

x ≥ n

1 − Z-value

x > n

1 − Z-value

## Normal approximation to the binomial — Example #1

Assume you have a fair coin and want to know the probability that you would get 40 heads after tossing the coin 100 times.

1. Gather information from the above problem.

• $N = 100$ (number of occurrences or trials);
• $n = 40$ (number of successes); and
• $p = 0.5$ (probability of success on a given trial).
2. Verify that the sample size is large enough to use the normal approximation.

• $N \times p = 50 ≥ 5$
• $N \times (1-p) = 50 ≥ 5$

We're good!

3. State the problem using the continuity correction factor.

• $x ≤ 40 \therefore P(x < 40.5)$
4. Find the mean ($μ$) and standard deviation ($σ$) of the binomial distribution.

• $μ = N \times p = 100 \times 0.5 = 50$

• $σ² = N \times p \times (1-p) = 100 \times 0.5 \times (1-0.5) = 25$

• $σ = \sqrt{25} = 5$

5. Find the Z-score using the mean and standard deviation.

• $z = (x - μ) / σ = (40.5 - 50) / 5 = -9.5 / 5 = -1.9$
6. Find the Z-value and determine the probability.

• $z = 0.4713$

Thus, the probability that a coin lands on heads less than or equal to 40 times during 100 flips is $0.0287$ or $2.8717\\%$.

## Normal approximation to the binomial — Example #2

Assume you have reliable data stating that 60% of the working people commute by public transport to work in a given city. If a random sample of size 30 is selected (all working persons), what is the probability that precisely 10 persons will travel from those by public transport?

1. Gather information from the above statement.

• $N = 30$ (number of occurrences or trials);
• $n = 10$ (number of successes); and
• $p = 0.6$ (probability of success on a given trial).
2. Verify that the sample size is large enough to use the normal approximation.

• $N \times p = 18 ≥ 5$
• $N \times (1-p) = 12 ≥ 5$

As these numbers are nice and large, we're good to go!

3. State the problem using the continuity correction factor.

• $x = 10 \therefore P(9.5 < x < 10.5)$
4. Find the mean ($\mu$) and standard deviation ($\sigma$) of the binomial distribution.

• $μ = N \times p = 30 \times 0.6 = 18$

• $σ^2 = N \times p \times (1-p) = 30 \times 0.6 \times (1-0.6) = 7.2$

• $σ = \sqrt{7.2} = 2.6833$

5. Find the two z-scores using the mean and standard deviation.

• $z_1 = (x - μ) / σ = (9.5 - 18) / 2.68 = −8.5 / 2.68 = -3.168$

• $z_2 = (x - μ) / σ = (10.5 - 18) / 2.68 = −7.5 / 2.68 = -2.795$

6. Find the Z-value and determine the probability.

• $P = z_2 - z_1 = 0.0026 - 0.0008 = 0.0018$

Thus, the probability that precisely 10 people travel by public transport out of the 30 randomly chosen people is $0.0018$ or $0.18\\%$.

## FAQ

### Can I use normal approximation if the product of the trials and the probability of the event is less than five?

No. The number of trials (or occurrences, N) relative to its probabilities (p and 1−p) must be sufficiently large (N×p ≥ 5 and N×(1−p) ≥ 5) to apply the normal distribution in order to approximate the probabilities related to the binomial distribution.

### What is normal approximation to binomial distribution?

The normal approximation to the binomial distribution is a process by which we approximate the probabilities related to the binomial distribution.

### What is the z-value of 60.5 occurrences when the mean is 50 and standard deviation is 5?

The z-value is 2.3 for the event of 60.5 (x = 60.5) occurrences with the mean of 50 (μ = 50) and standard deviation of 5 (σ = 5). The computation takes the form of (x – μ) / σ = (60.5 – 50) / 5 = 11.5 / 5 = 2.3).

### What are the main steps for the normal approximation to binomial distribution?

You should take the following steps to proceed with the normal approximation to binomial distribution.

1. Find the number of occurrences or trials (N) with its probabilities (p).
2. Check if the number of trials is sufficiently high (N×p ≥ 5 and N×(1-p) ≥ 5).
3. Apply a continuity correction by adding or subtracting 0.5 from the discrete x-value.
4. Find the mean (μ) and standard deviation (σ).
5. Find the z-score (z = (x – μ) / σ).
6. Find the probability associated with the z-score.
Tibor Pál, PhD candidate
Problem setup
Probability of success (0<p<1)
Probability of failure (0<q<1)
Number of occurrences (N)
Number of successes (n)
Event restatement
P(x ≥ n)
Results
Mean (μ)
Variance (σ²)
25
Standard deviation (σ)
5
 Problem statement P(x ≥ 40) Continuity correction P(x > 39.5) Z-score P(z ≥ -2.1) Z-value 0.0179 Approximated probability 1 - 0.0179 = 0.9821
The approximated probability for the event to happen is 0.9821 or 98.2136%.
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