Mean and Standard Deviation of a Binomial Distribution
Emma makes 80% of her free throws. At practice she shoots 15 of them. How many should she expect to make? And how far from that number would a normal day stray before you'd say "something's off"? You can answer both without listing a single probability.
The formulas: μ = np and σ = √(np(1−p))
You already know how to recognize a binomial setting: each trial has two outcomes (success or failure), the trials are independent, the number of trials $n$ is fixed, and the probability of success $p$ is the same on every trial. If $X$ counts the successes, we say $X$ is binomial with parameters $n$ and $p$, written $X \sim B(n, p)$.
Computing every $P(X = k)$ and adding up $x \cdot P(x)$ would take forever. For a binomial there is a shortcut.
A recent study on digital habits indicates that $85\%$ of American teenagers between 15 and 16 years old use two-factor authentication on their video game accounts. If…
A recent study on digital habits indicates that $85\%$ of American teenagers between 15 and 16 years old use two-factor authentication on their video game accounts. If we select a random sample of $20$ teenagers:
- a) Identify the parameters $n$ and $p$ of the binomial distribution that models this situation.
- b) Calculate the expected number (mean) of teenagers who use this security system in the sample.
In a vocational training center, $40\%$ of the students use public transportation to get to class. A random sample of $50$ students from the center is taken. Calculate…
In a vocational training center, $40\%$ of the students use public transportation to get to class. A random sample of $50$ students from the center is taken. Calculate the mean and standard deviation of the number of students who use public transportation in this sample.
In a smartphone components factory, it is known that $2\%$ of the touch screens are produced with a minor cosmetic defect. The screens are sent to stores in boxes of…
In a smartphone components factory, it is known that $2\%$ of the touch screens are produced with a minor cosmetic defect. The screens are sent to stores in boxes of $250$ units.
- a) Calculate the mean number of defective screens we will find in a box.
- b) Calculate the standard deviation of the variable "number of defective screens per box".
Interpreting μ and σ
The mean tells you what to expect on average. The standard deviation tells you how far a typical result lands from that average. The values within one standard deviation of the mean, from $\mu - \sigma$ to $\mu + \sigma$, are the typical outcomes. In a roughly symmetric distribution, about two-thirds of results fall there.
Here is Emma's practice session: $n = 15$, $p = 0.8$, so $\mu = 15(0.8) = 12$ and $\sigma = \sqrt{15(0.8)(0.2)} = \sqrt{2.4} \approx 1.55$.
The shaded bars (11, 12, and 13 baskets) fall inside $(\mu - \sigma,\ \mu + \sigma) = (10.45,\ 13.55)$. Together they have probability about $0.67$.
To interpret $\sigma$ in context, use this sentence: "The number of ___ typically varies from the mean of ___ by about ___."
The most likely value sits near the mean
Look again at Emma's graph: the tallest bar is at 12, right at the mean. That is no accident. The most probable value of a binomial (its mode) is the whole number $k$ that satisfies
$$(n+1)p - 1 \le k \le (n+1)p$$This always puts $k$ within 1 of $np$. So the "most typical" values, the ones in $(\mu - \sigma,\ \mu + \sigma)$, are the whole numbers clustered around the mean. If $(n+1)p$ is itself a whole number, two neighboring values tie for the highest bar.
A music streaming platform has proven that the probability of a user "skipping" a song recommended by the algorithm before it finishes is $0,3$. If one afternoon you…
A music streaming platform has proven that the probability of a user "skipping" a song recommended by the algorithm before it finishes is $0,3$. If one afternoon you listen to a playlist of $12$ recommended songs:
- a) Determine the mean and variance of the number of songs you are expected to skip.
- b) Calculate the standard deviation and interpret what information it provides about the dispersion of the data in this context.
Two basketball players, Ashley and Brandon, practice free throws. Ashley has a probability of scoring of $0,8$ and takes $50$ shots. Brandon has a probability of $0,6$…
Two basketball players, Ashley and Brandon, practice free throws. Ashley has a probability of scoring of $0,8$ and takes $50$ shots. Brandon has a probability of $0,6$ but takes $80$ shots.
- a) Which of the two players has a higher expected number of baskets?
- b) Which of the two players shows a greater dispersion (standard deviation) in their results? Interpret what this means in terms of consistency in the game.
It has been proven that $40\%$ of the students at a language center take the official B2 level exam. If there are $100$ students enrolled this year: a) Determine the…
It has been proven that $40\%$ of the students at a language center take the official B2 level exam. If there are $100$ students enrolled this year:
- a) Determine the parameters of the variable $X$ that defines the number of students who take the exam.
- b) In binomial distributions, it is considered that the most "typical" or probable values are found in the interval $(\mu - \sigma, \mu + \sigma)$. Calculate this interval for this specific case.
How far is too far? Unusual results
A result is unusual when it is far from the mean, compared with the usual spread. A common rule of thumb is to call a value unusual when it is more than $2\sigma$ from $\mu$. Companies use the same idea as an alarm threshold.
A logistics company uses drones to deliver parcels in urban centers. The probability of a drone suffering a delay due to weather conditions is $5\%$. If $200$…
A logistics company uses drones to deliver parcels in urban centers. The probability of a drone suffering a delay due to weather conditions is $5\%$. If $200$ independent flights are made in a day:
- a) Calculate the value of the mean and the variance of the number of flights that arrive late.
- b) The maintenance protocol is activated if the number of delays in a day exceeds the value of $\mu + 2\sigma$. Calculate this threshold value.
Expected value in context
Because $\mu$ is the expected number of successes, you can use it to predict totals such as costs, scores, or counts. It also lets you compare a goal with what you'd expect. If a goal sits far above $\mu$ (relative to $\sigma$), reaching it is unlikely. If it sits far below, it is likely. A mean below a pass mark does not make passing impossible; it just makes it less likely.
A fair six-sided die is rolled a total of $30$ times and the number of times the number "6" appears is recorded. a) Define the random variable and its distribution. b)…
A fair six-sided die is rolled a total of $30$ times and the number of times the number "6" appears is recorded.
- a) Define the random variable and its distribution.
- b) Calculate the expected value of the variable and its standard deviation. Round the results to two decimal places.
A multiple-choice exam consists of $10$ questions with three possible answers each, of which only one is correct. If a student answers all the questions at random: a)…
A multiple-choice exam consists of $10$ questions with three possible answers each, of which only one is correct. If a student answers all the questions at random:
- a) What is the mean number of correct answers?
- b) If the student needs at least $5$ correct answers to pass, is the mean value above or below the pass mark? Justify your answer.
On a well-known social network, it has been estimated that the probability of a user interacting with a cultural content post is $p = 0,18$. If an "influencer" launches…
On a well-known social network, it has been estimated that the probability of a user interacting with a cultural content post is $p = 0,18$. If an "influencer" launches a campaign that reaches a group of $n = 150$ followers:
- a) Calculate the expected number of followers who will interact with the post and its standard deviation.
- b) If the cost of the campaign is a fixed $\$50$ plus $\$2$ for each expected interaction (based on the mean), what is the expected budget for this advertising action?
Working backward from the mean
Sometimes you are told $\mu$ and have to rebuild a missing parameter. Since $\mu = np$, you can solve for either one:
$$p = \frac{\mu}{n} \qquad\qquad n = \frac{\mu}{p}$$Consider a discrete random variable $X$ that follows a binomial distribution $B(n, p)$. If we know that the number of trials is $n = 100$ and that the mean of the…
Consider a discrete random variable $X$ that follows a binomial distribution $B(n, p)$. If we know that the number of trials is $n = 100$ and that the mean of the distribution is $\mu = 60$:
- a) Find the value of the probability of success $p$.
- b) Calculate the variance of this distribution.
The quality control department of an LED bulb factory knows that the probability of a unit being defective is $p = 0,02$. The bulbs are packed in boxes of $n$ units. a)…
The quality control department of an LED bulb factory knows that the probability of a unit being defective is $p = 0,02$. The bulbs are packed in boxes of $n$ units.
- a) If it is desired that the mean number of defective bulbs per box is exactly $1,6$, how many bulbs must each box contain?
- b) With that value of $n$, calculate the probability that there are no defective bulbs in a box.
In a survey on reading habits carried out on $60$ teenagers, it is known that the expected number of regular readers of digital press is $21$. a) Find the probability…
In a survey on reading habits carried out on $60$ teenagers, it is known that the expected number of regular readers of digital press is $21$.
- a) Find the probability $p$ that a randomly chosen teenager is a regular reader of digital press.
- b) Calculate the standard deviation of this distribution.
- c) If we select another sample where the number of teenagers is doubled ($n=120$), how does the standard deviation vary with respect to the original sample? Justify your answer with the calculation.
Working backward from μ and σ
When both the mean and the variance are known, divide one by the other:
$$\frac{\sigma^2}{\mu} = \frac{np(1-p)}{np} = 1 - p$$When $n$ and $p$ are known, you can find an exact probability with the binomial formula:
$$P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}$$On a TI-84, this is binompdf(n, p, k).
A discrete random variable follows a binomial distribution $X \sim B(n, p)$. It is known that its mean is $\mu = 12$ and its standard deviation is $\sigma = 3$. a)…
A discrete random variable follows a binomial distribution $X \sim B(n, p)$. It is known that its mean is $\mu = 12$ and its standard deviation is $\sigma = 3$.
- a) Determine the values of the parameters $n$ (number of trials) and $p$ (probability of success).
- b) Justify whether this distribution could be approximated by a normal distribution, based on the values obtained.
A logistics company uses drones to deliver parcels in smart cities. It is known that the variable $X$, which defines the number of failed deliveries in a batch of $n$…
A logistics company uses drones to deliver parcels in smart cities. It is known that the variable $X$, which defines the number of failed deliveries in a batch of $n$ shipments, follows a binomial distribution. Following an internal study, it has been proven that the mean number of failures is $4$ and the variance is $3,2$.
- a) Calculate the parameters $n$ and $p$ of the distribution.
- b) If a single shipment is made, what is the probability that the delivery is successful?
- c) What is the probability that in a batch of $n$ shipments none fail?
A biased coin is flipped $n$ times. It is known that the probability of getting heads is $p$. If the mean number of heads obtained is $15$ and the standard deviation is…
A biased coin is flipped $n$ times. It is known that the probability of getting heads is $p$. If the mean number of heads obtained is $15$ and the standard deviation is $\sqrt{10}$, calculate:
- a) The probability $p$ of getting tails.
- b) The number of flips $n$ made.
- c) The probability of getting exactly $2$ heads in only $5$ flips.
When only σ is known: two possible values of p
If you know $n$ and $\sigma$ but not $\mu$, the equation $np(1-p) = \sigma^2$ has the unknown $p$ twice, so it becomes a quadratic equation:
$$p^2 - p + \frac{\sigma^2}{n} = 0 \quad\Longrightarrow\quad p = \frac{1 \pm \sqrt{1 - 4\sigma^2/n}}{2}$$The two roots always add to 1: they are $p$ and $1 - p$. Success and failure play symmetric roles in $\sigma^2$, so both give the same $\sigma$ (but different means). Use information from the problem to choose.
The shape also depends on $p$. If $p < 0.5$, the distribution is skewed right (positively skewed), because there is a long tail of larger counts. If $p > 0.5$, it is skewed left, as in Emma's graph. If $p = 0.5$, it is symmetric.
Let $X$ be a random variable that follows a binomial distribution. If the number of trials is $n = 25$ and the variance is $Var(X) = 4$: a) Calculate the two possible…
Let $X$ be a random variable that follows a binomial distribution. If the number of trials is $n = 25$ and the variance is $Var(X) = 4$:
- a) Calculate the two possible values of the probability of success $p$.
- b) For each value of $p$ found, calculate its corresponding mean $\mu$.
- c) If we know that the distribution is positively skewed (i.e., success is less probable than failure), which value of $p$ would you choose? Justify your answer.
In a study on the use of sustainable transport, it has been observed that the standard deviation of the number of students who go to school by electric scooter in groups…
In a study on the use of sustainable transport, it has been observed that the standard deviation of the number of students who go to school by electric scooter in groups of $50$ people is $\sigma = 3$.
- a) Find the two possible values of the probability $p$ that a student uses an electric scooter.
- b) If it is known that the scooter is a minority mode of transportation ($p < 0,5$), calculate the mean number of users in a sample of $100$ students.
A strategy video game has a probability $p$ of obtaining a "legendary item" when opening a chest. A player opens $200$ chests and observes that the standard deviation of…
A strategy video game has a probability $p$ of obtaining a "legendary item" when opening a chest. A player opens $200$ chests and observes that the standard deviation of their reward is twice what it would be if the probability of success were $0,01$.
- a) Set up an equation to find the value of $p$.
- b) Determine the expected mean number of legendary items in the $200$ chests with that $p$.
What makes σ big or small?
For a fixed $n$, the spread depends on $p(1-p)$. That product is largest when success and failure are equally likely.
A professional basketball player has a free throw success rate of $80\%$. If she takes a series of $15$ shots in a training session: a) What is the mean number of…
A professional basketball player has a free throw success rate of $80\%$. If she takes a series of $15$ shots in a training session:
- a) What is the mean number of baskets she is expected to score?
- b) Calculate the standard deviation. What would happen to the standard deviation if her probability of success were $50\%$? Show this by calculation.
In a quality control check at an LED bulb factory, the number of defective units in boxes of a certain size is analyzed. The variance of defective bulbs per box is…
In a quality control check at an LED bulb factory, the number of defective units in boxes of a certain size is analyzed. The variance of defective bulbs per box is $3,75$ and it is known that the probability of a bulb being defective is $5\%$.
- a) How many bulbs does each box contain?
- b) If the manufacturer wants to halve the standard deviation, how should the number of bulbs per box change while maintaining the same manufacturing quality ($p$)?
Let $X$ be a random variable that follows a binomial distribution $B(n, p)$. It is known that its mean is $\mu = 30$. a) What is the minimum value its standard deviation…
Let $X$ be a random variable that follows a binomial distribution $B(n, p)$. It is known that its mean is $\mu = 30$.
- a) What is the minimum value its standard deviation can take? Justify your answer by reasoning about the possible values of $p$.
- b) If the standard deviation is $\sigma = 5$, calculate the probability of obtaining any success in a sample of only $3$ individuals from that same population.
Putting it together: exact probabilities and "at least one"
Once you know $n$ and $p$, you can combine the shortcuts with exact binomial probabilities. Here is a quick check that the formulas agree with the story: if $X \sim B(20,\ 0.3)$, then $\mu = 6$, and the probability of getting exactly the mean is $P(X = 6) = \binom{20}{6}(0.3)^6(0.7)^{14} \approx 0.192$ (on a calculator, binompdf(20, 0.3, 6)). Even the single most likely value happens less than 20% of the time.
Questions with "at least one" are easiest through the complement. The opposite of "at least one success" is "no successes":
$$P(X \ge 1) = 1 - P(X = 0) = 1 - (1-p)^n$$A content creator on a social network knows that the probability of a new user subscribing to their channel after watching a "short" is $p = 0,12$. The algorithm informs…
A content creator on a social network knows that the probability of a new user subscribing to their channel after watching a "short" is $p = 0,12$. The algorithm informs them that, in their last campaign, the mean number of new subscribers was $18$ people.
- a) Determine the sample size $n$ (number of views) that the algorithm has analyzed.
- b) Calculate the standard deviation of this distribution.
- c) What is the probability that the number of subscribers is exactly equal to the mean?
A cybersecurity analyst is studying phishing* attacks. The probability of an employee clicking on a malicious link is $0,005$. The analyst wants to send a test campaign…
A cybersecurity analyst is studying phishing attacks. The probability of an employee clicking on a malicious link is $0,005$. The analyst wants to send a test campaign to a group of employees of size $n$.
Calculate the minimum number of employees the email must be sent to so that the probability of at least one of them clicking on the link is greater than $90\%$.
Back to the start: Emma's 15 free throws at 80% give $\mu = 12$ made baskets and $\sigma \approx 1.55$. A typical day is 11 to 13 baskets, and 12 is the single most likely result. The unusual-result lines are $\mu \pm 2\sigma$, about $8.9$ and $15.1$. Making 8 or fewer would be worth noticing, since $8 < 8.9$. Making 9 is a below-average day but still not unusual. Making all 15 is possible but happens only about 3.5% of the time.