Before you start, make sure you know: Mean and variance (High School Year 1 (semesters 1 and 2))
A factory in Querétaro produces 600 parts a day 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 adding up every case by hand with $\binom{600}{k}$ would be crazy. When $n$ is large, the binomial can be approximated by a normal, and the problem is solved with the table you already know.
Why approximate a binomial with 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.
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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 $n \times p \ge 5$ and $n \times q \ge 5$ are satisfied.
b) If so, state what the parameters $\mu$ and $\sigma$ of the new normal distribution $X'$ would be.
It is not always possible to approximate: 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.
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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.
The binomial takes only integer values ($0, 1, 2, \dots$), but the normal is continuous: in the normal, $P[X' = 65]$ would be $0$ (the area of a single line is zero). That is why, before standardizing, each integer $k$ is "widened" to the interval $[k - 0.5,\ k + 0.5]$. This is the continuity correction (Yates's correction).
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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.
The complete procedure chains together everything above.
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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 survey on sports habits has been conducted, showing that $30\%$ of Mexican teenagers play a federated team sport. If we select a random sample of $200$ teenagers: a)…
A survey on sports habits has been conducted, showing that $30\%$ of Mexican teenagers play a federated team sport. If we select a random sample of $200$ teenagers:
a) Justify whether it is possible to approximate the distribution of the variable "number of teenagers who play a federated team sport" using a normal distribution.
b) Using this approximation (with Yates' continuity correction), calculate the probability that more than $70$ teenagers in the sample play a federated team sport.
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.
A biased coin, with a probability of landing on heads of 0.4, is tossed 500 times. Calculate the probability of getting: a) Exactly 200 heads. b) More than 215 heads. c)…
A biased coin, with a probability of landing on heads of 0.4, is tossed 500 times. Calculate the probability of getting:
a) Exactly 200 heads.
b) More than 215 heads.
c) Fewer than 185 heads.
Note: Use the normal approximation with Yates' continuity correction.
More challenging problems: typical intervals and percentiles
Some problems combine the approximation with the $\mu \pm 2\sigma$ rule or with a percentile (the value that leaves a given probability on one side).
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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.
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%?
On a microblogging social network, the probability that a published post goes "viral" (more than $10,000$ shares) is $0.004$. A data analyst studies a random sample of…
On a microblogging social network, the probability that a published post goes "viral" (more than $10,000$ shares) is $0.004$. A data analyst studies a random sample of $2,500$ posts published in one day.
Using the normal approximation to the binomial distribution (with the corresponding continuity correction), calculate the probability that in that sample:
a) There are exactly $12$ viral posts.
b) There are between $8$ and $15$ viral posts (both inclusive).
Back to the start: for the factory, $B(600, 0.05)$ gives $\mu = 30$ and $\sigma = \sqrt{28.5} \approx 5.34$. With the correction, $P[25 \leq X \leq 35] \to P[24.5 \leq X' \leq 35.5] = P[-1.03 \leq Z \leq 1.03] = 2(0.8485) - 1 = 0.6970$, that is, about $69.7\,\%$.
Let's practice! 3 medium exercises in this section