Approximating the binomial with the normal
A machine makes $600$ parts and $5\,\%$ come out defective: what is the probability that there are between $25$ and $35$ defective parts? It is a binomial $B(600,\ 0{,}05)$, but calculating that probability by adding up each case by hand with $\binom{600}{k}$ would be madness. When $n$ is large, the binomial can be approximated by a normal distribution, and the problem is solved with the table you already know.
Why approximate a binomial by a normal
As $n$ grows, the histogram of a binomial $B(n,p)$ looks more and more like a bell curve: the bars become narrow and their outline traces the same symmetric shape as the normal.
A basketball free-throw shooter has a success probability of $p = 0,8$. If they take $n = 100$ shots, we want to know if we can approximate the binomial variable $X$…
A basketball free-throw shooter has a success probability of $p = 0,8$. If they take $n = 100$ shots, we want to know if we can approximate the binomial variable $X$ (number of successful shots) using a normal distribution.
- a) Check whether conditions $n \cdot p \ge 5$ and $n \cdot q \ge 5$ are satisfied.
- b) If so, state what the parameters $\mu$ and $\sigma$ of the new normal distribution $X'$ would be.
Conditions and parameters of the approximation
It cannot always be done: if $p$ is very close to $0$ or $1$ and $n$ is not very large, the histogram is very skewed and the bell curve does not fit well.
A fair coin ($p = 0,5$) is tossed $400$ times. Calculate the probability of getting fewer than $190$ heads using the normal approximation to the binomial distribution…
A fair coin ($p = 0,5$) is tossed $400$ times. Calculate the probability of getting fewer than $190$ heads using the normal approximation to the binomial distribution (without applying the continuity correction yet).
Step 1: Calculate the mean $\mu$ and standard deviation $\sigma$.
Step 2: Standardise the value $X = 190$ and look it up in the table.
Yates's continuity correction
The binomial only takes integer values ($0,1,2,3,\dots$), but the normal is continuous: for an exact value, $P[X'=65]$ would equal $0$ in the normal (the area of a single line is zero). That is why, before standardising, each integer value $k$ is "widened" into an interval $[k-0{,}5,\ k+0{,}5]$.
On a streaming platform, $15\%$ of users cancel their subscription before the end of the trial month. If $400$ new users signed up this month, calculate, using the…
On a streaming platform, $15\%$ of users cancel their subscription before the end of the trial month. If $400$ new users signed up this month, calculate, using the normal approximation with continuity correction, the probability that exactly $65$ users cancel their subscription.
Solving complete problems: "fewer than", "more than", "between"
The complete procedure chains together everything above: check the conditions, find $\mu$ and $\sigma$, apply the continuity correction according to what is asked, standardise and read the table.
A study on recycling habits states that $35\%$ of households in a city correctly separate plastic. If a random sample of $120$ households is selected: a) Justify why…
A study on recycling habits states that $35\%$ of households in a city correctly separate plastic. If a random sample of $120$ households is selected:
- a) Justify why this experiment can be approximated by a normal distribution.
- b) Calculate the probability that, in that sample, between $40$ and $50$ households (both inclusive) separate plastic.
A study on reading habits in teenagers reveals that 70% of 15- to 16-year-olds prefer digital format over paper. If we select a random sample of 400 teenagers: a)…
A study on reading habits in teenagers reveals that 70% of 15- to 16-year-olds prefer digital format over paper. If we select a random sample of 400 teenagers:
- a) Justify whether it is possible to approximate the distribution of the variable "number of teenagers who prefer digital format" using a normal distribution.
- b) Using the above approximation and applying the continuity correction, calculate the probability that between 270 and 290 teenagers (inclusive) prefer digital format.
5% of the parts manufactured by a high-precision machine turn out to be defective. If they are packed in boxes of 200 units, calculate the probability that in a box: a)…
5% of the parts manufactured by a high-precision machine turn out to be defective. If they are packed in boxes of 200 units, calculate the probability that in a box:
- a) There are exactly 10 defective parts.
- b) There are more than 15 defective parts.
- c) The number of defective parts is less than or equal to 5.
Discuss whether the results would be significantly different if the continuity correction were not applied.
More demanding problems: extremes and percentiles
Some problems combine the approximation with ideas from the previous section on the normal: the empirical rule $\mu\pm2\sigma$, or a percentile (finding the value that leaves a given probability on one side).
In a multiple-choice exam with 100 questions, each with 4 answer options (only one correct), a student answers all the questions at random. Let $X$ be the number of…
In a multiple-choice exam with 100 questions, each with 4 answer options (only one correct), a student answers all the questions at random. Let $X$ be the number of questions they get right. You can approximate the distribution of $X$ by a normal distribution with the same mean and the same standard deviation.
- a) What is the probability that they get more than 30 questions right?
- b) If at least 40 correct answers are needed to pass the exam, what is the probability that the student passes by pure chance?
- c) Calculate the interval $(np - 2\sigma,\ np + 2\sigma)$, where $n$ is the number of questions, $p$ is the probability of getting a question right at random and $\sigma=\sqrt{np(1-p)}$ is the standard deviation of $X$. Interpret its meaning in this context.
A gym chain knows that 20% of its registered members never visit the facilities. If a specific gym has 600 registered members: a) Calculate the probability that the…
A gym chain knows that 20% of its registered members never visit the facilities. If a specific gym has 600 registered members:
- a) Calculate the probability that the number of members who never visit is between 110 and 135 inclusive.
- b) If the gym decides to launch a motivational campaign for these members if the number of "inactive" members exceeds 140, what is the probability that they will have to launch this campaign?
- c) What is the maximum number of inactive members expected with a probability of 90%?
Digital lab: exact calculation versus approximation
In Python 3, using only the standard library, compare the exact probability of between 40 and 50 successes in 120 independent trials with probability 0,35 against its normal approximation. Predict the order of magnitude before running it:
from math import comb, sqrt
from statistics import NormalDist
n, p = 120, 0.35
masa = [comb(n,k)*p**k*(1-p)**(n-k) for k in range(n+1)]
modelo = NormalDist(n*p, sqrt(n*p*(1-p)))
exacta = sum(masa[40:51])
aprox = modelo.cdf(50.5)-modelo.cdf(39.5)
print(exacta, aprox, abs(exacta-aprox))
print('Control:', sum(masa))
print('Puntual en 42:', masa[42])
print('Acumulada hasta 42:', sum(masa[:43]))
print('Percentil normal 90:', modelo.inv_cdf(0.90))
Compare the approximation with the table in the previous example: the small difference is because the program does not round z before calculating. The sum of all the masses should come out close to 1. The percentile of the normal does not have to be an integer: to find the binomial percentile, look for the smallest integer k whose cumulative probability is at least 0,90.
Record the exact value, the approximation and the absolute error; change p to 0,02 and compare again. Explain why a large n on its own is not enough. To estimate an unknown probability from frequencies, repeat an experiment or simulation with 100 and 1 000 trials: more data usually stabilises the estimate, but does not guarantee getting closer at every step.
Back to the beginning: For the machine with $B(600,\ 0{,}05)$, we have $\mu=30$ and $\sigma=\sqrt{28{,}5}\approx5{,}34$; applying Yates, $P[25\leq X\leq35]\to P[24{,}5\leq X'\leq35{,}5]=P[-1{,}03\leq Z\leq1{,}03]=2\cdot0{,}8485-1=0{,}6970$ (that is, $69{,}7\,\%$).