A normal approximation calculator is used to estimate probabilities for binomial distributions when certain conditions are met. It’s particularly useful when dealing with large sample sizes or when exact binomial probabilities are difficult to compute.
This calculator applies the central limit theorem, which states that for sufficiently large sample sizes, the distribution of a sum of independent random variables approaches a normal distribution, regardless of the underlying distribution of the variables.
Normal Approximation Example Chart
| n | p | np | nq | √(npq) |
|---|---|---|---|---|
| 1000 | 0.33 | 330 | 670 | 4.582 |
| 2000 | 0.48 | 960 | 1040 | 6.935 |
| 5000 | 0.21 | 1050 | 3950 | 8.940 |
| 1500 | 0.40 | 600 | 900 | 12.247 |
| 2500 | 0.60 | 1500 | 1000 | 12.247 |
| 3000 | 0.25 | 750 | 2250 | 20.412 |
| 4000 | 0.10 | 400 | 3600 | 60.000 |
| 6000 | 0.75 | 4500 | 1500 | 30.000 |
- np (Expected number of successes):
- Formula:
np = n * p- nq (Expected number of failures):
- Formula:
nq = n * (1 - p)- Note:
q = 1 - p- √(npq) (Standard deviation of the approximated normal distribution):
- Formula:
√(npq) = √(n * p * (1 - p))- This can also be expressed as:
√(npq) = √(np * nq)
Normal Approximation Formula
The normal approximation to a binomial distribution uses the following formulas:
μ = np
σ = √(npq)Where:
- μ is the mean of the normal distribution
- σ is the standard deviation
- n is the number of trials
- p is the probability of success
- q is (1 – p)
If you have 100 coin flips (n = 100) with a fair coin (p = 0.5):
μ = 100 * 0.5 = 50
σ = √(100 0.5 0.5) ≈ 5This tells us that the number of heads in 100 flips can be approximated by a normal distribution with a mean of 50 and a standard deviation of 5.
How to Find Normal Approximation?
To find the normal approximation:
- Check if np ≥ 5 and nq ≥ 5 to ensure the approximation is valid.
- Calculate μ and σ using the formulas above.
- Use the standard normal distribution (z-score) to find probabilities.
In a factory, 20% of products are defective. What’s the probability of finding 15 or fewer defective products in a batch of 100?
Check validity: np = 100 0.2 = 20 and nq = 100 0.8 = 80 (both ≥ 5)Calculate: μ = 100 0.2 = 20, σ = √(100 0.2 * 0.8) = 4
Find z-score: z = (15.5 – 20) / 4 = -1.125
Use a standard normal table or calculator to find the probability: approximately 0.1303 or 13.03%