Binomial Distribution vs Normal Distribution

A binomial distribution counts successes across a fixed number of trials that are binary, independent, and identical in success probability. Write it $X \sim B(n, p)$: $n$ trials, success probability $p$ on each one, and $X$ the number of successes, which can be any whole number from 0 to $n$. The probability of exactly $k$ successes is $P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}$, where $\binom{n}{k}$ (n choose k) counts the orders in which those $k$ successes could fall. The mean is $\mu_X = np$ and the standard deviation is $\sigma_X = \sqrt{np(1-p)}$.

Flip a fair coin 10 times, so $n = 10$ and $p = 0.5$. Then $P(X = 6) = \binom{10}{6}(0.5)^6(0.5)^4 = \frac{210}{1024} = 0.2051$. The mean is $\mu_X = 10(0.5) = 5$ heads and the standard deviation is $\sigma_X = \sqrt{10(0.5)(0.5)} = \sqrt{2.5} \approx 1.581$ heads.

Now the error that costs the most marks: "so the probability of 6 or more heads is 0.2051." It is not. That number is the probability of exactly 6. At least 6 means 6, 7, 8, 9 or 10, and adding those five probabilities gives 0.3770, nearly double. Circle the words at least, at most, more than and fewer than before any arithmetic starts, because the formula answers only the exactly question and every other question is assembled out of it.

Independence is the condition that breaks most often in practice. Deal 10 cards from a deck without replacement and the chance of a heart shifts with every card removed, so the count of hearts is not binomial. It is close enough to treat as binomial when the sample is a small fraction of the population, which is what the 10 percent condition checks.

The binomial distribution is topic 2.10 of Unit 2, Probability, Random Variables, and Probability Distributions.

A normal distribution is a continuous model written $N(\mu, \sigma)$: a density curve, symmetric about its mean $\mu$ (mu), single-peaked, and spread out by its standard deviation $\sigma$ (sigma). Two numbers fix the whole curve. Probability is area underneath it, the total area is exactly 1, and half of that area sits on each side of $\mu$. Because the area over a single point is zero, every normal question is really a question about an interval.

Suppose adult male heights are approximately $N(69, 2.8)$ in inches. To find the share above 72 inches, standardize: $z = \frac{72 - 69}{2.8} \approx 1.07$. The area to the left of $z = 1.07$ is 0.8577, so about 0.142, roughly 14 percent, are taller than 72 inches. The same curve puts about 71.6 percent of men between 66 and 72 inches.

The claim that ruins the most work is "the sample is large, so the data are normal." Sample size does not change the shape of the variable being measured. A right-skewed variable such as household income stays right-skewed however many households you collect. What a large sample buys is that the sampling distribution of $\bar{x}$ (x-bar, the sample mean) is close to normal, and that is a statement about the average of a sample, not about the individual values inside it.

A normal curve also runs on forever in both directions, so a normal model always assigns some probability to values the real variable cannot reach, negative heights included. For heights that leftover is far too small to matter. Where the model genuinely fails is shape: strongly skewed, hard-bounded, or clearly bimodal data should not be pushed through a normal calculation, and no amount of extra data repairs that.

The normal distribution is topic 2.11 in Unit 2, Probability, Random Variables, and Probability Distributions. The empirical rule gives the quick version of its areas and the z-table gives the exact ones.

There is a calculus reason the areas have to come from a table or a calculator rather than from an antiderivative. The bell-shaped function $e^{-x^{2}}$ has no elementary antiderivative at all, so no amount of algebra produces a formula for the area under it between two bounds. CalcLearn works through that function's behaviour in the limit of e to the minus x squared.