Sample Mean (Sp) Distribution

Calculate binomial distribution probabilities using normal approximation with continuity correction.

Last updated: March 2026

Calculator

Number of Trials (n)

Total number of independent trials

Probability of Success (p)

Probability of success per trial (0.01 to 0.99)

Number of Successes (x)

Number of successes to calculate

P(X ≤ 8)

0.888744

⚠ Normality check: np or nq < 10 (approximation less accurate)

P(X ≥ x)0.232107

P(X = x) ≈0.120851

Mean (μ = np)6.0000

Variance (σ² = npq)4.2000

Std Dev (σ)2.0494

z-score1.2199

What is Binomial Distribution?

A binomial distribution models the number of successes in a fixed number of independent trials, where each trial has exactly two outcomes (success or failure) and the probability is constant. Examples: coin flips (heads/tails), product quality checks (pass/fail), survey responses (yes/no).

For large sample sizes (when np ≥ 10 and nq ≥ 10), the binomial distribution approximates a normal distribution. This calculator uses the normal approximation with continuity correction to compute probabilities efficiently without calculating exact binomial coefficients. The continuity correction improves accuracy by accounting for the discrete nature of binomial data.

The sample mean proportion (Sp or p̂) is calculated as x/n, and its distribution follows approximately N(p, pq/n). This is foundational for hypothesis testing and confidence intervals for proportions.

How to Calculate

Step-by-Step Process

Step 1: Identify parameters: n (trials), p (success probability), x (successes)

Step 2: Check normality: np ≥ 10 and nq ≥ 10 for approximation validity

Step 3: Calculate mean μ = np and variance σ² = npq

Step 4: Apply continuity correction and compute z-scores

Step 5: Use standard normal CDF to find probabilities

Key Formulas

Mean:

μ = np

Variance:

σ² = np(1−p) = npq

Standard Deviation:

σ = √(npq)

z-score (with continuity correction):

z = (x ± 0.5 − μ) / σ

Normality Check:

np ≥ 10 AND nq ≥ 10 required for valid approximation

Real-World Example

Quality Control: Machine Defects

Context:

A manufacturing plant produces circuit boards. Historical data shows 2% defect rate (p = 0.02). Quality inspector checks 200 boards (n = 200).

Question:

What's probability of finding at most 10 defective boards?

Solution:

Parameters: n = 200, p = 0.02, x = 10

• Check normality: np = 4, nq = 196. Since np < 10, approximation is less reliable but we can proceed.

• Mean: μ = 200 × 0.02 = 4

• Variance: σ² = 200 × 0.02 × 0.98 = 3.92, so σ ≈ 1.98

• z-score: (10 + 0.5 − 4) / 1.98 ≈ 3.28

Result: P(X ≤ 10) ≈ 0.9995 (99.95%)

Very likely to find ≤ 10 defects, suggesting process is under control.

Frequently Asked Questions

When can I use normal approximation to binomial?

Use when np ≥ 10 AND n(1−p) ≥ 10. Both conditions must be true. For smaller samples, use exact binomial probabilities instead.

What is continuity correction?

Binomial is discrete, normal is continuous. Continuity correction adjusts z-score by ±0.5 to account for this difference. Use +0.5 for P(X ≤ x) and −0.5 for P(X ≥ x).

How does this differ from Sampling Proportion?

Both use normal approximation. Sampling Proportion deals with sample mean p̂ and its variability across samples. This calculator finds P(X = x) in single sample of fixed size.

Can I use this for very small or very large p?

Yes, but normality check matters more. For extreme p (close to 0 or 1), you need larger n. For p = 0.01 with n = 20, np = 0.2 < 10, so use exact binomial instead.

Why do we need to check np and n(1−p)?

These check if the binomial distribution is sufficiently bell-shaped to approximate using normal curve. If violated, normal approximation can give very inaccurate results.