Exercises: Normal approximation to the binomial distribution

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.

1. a) Check whether conditions $n \cdot p \ge 5$ and $n \cdot q \ge 5$ are satisfied.
2. b) If so, state what the parameters $\mu$ and $\sigma$ of the new normal distribution $X'$ would be.

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.

For each of the three situations presented below, analyze the data and choose, in a reasoned manner, the most appropriate calculation method between the two options: the exact binomial formula $P(X = k) = \binom{n}{k}p^k(1-p)^{n-k}$ or the normal approximation with Yates's continuity correction. If in any case you consider that applying the normal distribution is not acceptable, justify it using the technical criteria covered in the unit.

1. Situation A: In a small Correos post office with 15 employees, the probability that a worker is absent on medical leave on a specific day is $p = 0{,}04$. We want to calculate the probability that at most 1 employee is absent on that day.

2. Situation B: In a DGT sobriety checkpoint on a highway, $5\,\%$ of drivers test positive. 60 vehicles are stopped and we want to calculate the probability that 5 or more drivers test positive.

3. Situation C: A CIS survey indicates that $40\,\%$ of Spanish households have at least one pet. A sample of 200 households is taken and we want to calculate the probability that between 75 and 90 households (both inclusive) have a pet.

In each situation, evaluate $n\cdot p$ and $n\cdot q$ and argue why it is or is not advisable to use the normal approximation instead of the exact binomial calculation.

At a vehicle inspection station (ITV) in Andalusia, it is known that $30\,\%$ of the passenger cars inspected are hybrid or electric. During one morning, $n = 100$ vehicles arrive. The goal is to determine the probability that more than 35 are hybrid or electric.

Hugo, an 11th-grade student, writes the following solution on his exam:

"I identify a binomial $X \equiv B(100;\ 0{,}30)$. I calculate its mean $\mu = 100 \cdot 0{,}30 = 30$ and its standard deviation $\sigma = \sqrt{100 \cdot 0{,}30 \cdot 0{,}70} = \sqrt{21} \approx 4{,}58$.

Since they ask for more than 35, I standardize directly:

$$P(X > 35) = P\left(Z > \frac{35 - 30}{4{,}58}\right) = P(Z > 1{,}09) = 1 - \Phi(1{,}09) = 1 - 0{,}8621 = 0{,}1379$$"

1. a) Hugo made two mistakes: a prior methodological one (not checking whether the approximation is valid) and a serious conceptual one when transferring a strict inequality from a discrete variable to a continuous normal variable. Explain in detail why each integer value $k$ of the binomial must be represented as the interval $[k - 0{,}5,\ k + 0{,}5]$ and what correction he should have applied when asking for "more than 35".
2. b) Correct Hugo's exercise: check the conditions $n\cdot p \geq 5$ and $n\cdot q \geq 5$, set up the standardization with the appropriate Yates' continuity correction, and calculate the correct probability.

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:

1. a) Justify why this experiment can be approximated by a normal distribution.
2. b) Calculate the probability that, in that sample, between $40$ and $50$ households (both inclusive) separate plastic.

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.

A survey on sports habits has been carried out, and it is known that $30\%$ of Spanish teenagers play some federated team sport. If we select a random sample of $200$ teenagers:

1. a) Justify whether it is possible to approximate the distribution of the variable "number of teenagers who play a federated team sport" by a normal distribution. If so, state the parameters of the approximating normal distribution.

1. b) Using that approximation (with Yates' continuity correction), calculate the probability that more than $70$ teenagers in the sample play a federated team sport.

According to the Survey on Equipment and Use of Information Technologies by the INE, $40\,\%$ of workers in a certain professional sector in Spain habitually telework. To analyze the demand for coworking spaces, a random sample of 150 professionals from this sector is selected. Let $X$ be the number of teleworking professionals.

1. a) Model the variable $X$ using its binomial distribution and calculate its mean $\mu$ and its standard deviation $\sigma$. Without looking at any normal distribution table or standardizing yet, use the property that the central interval $(\mu - 2\sigma,\ \mu + 2\sigma)$ contains approximately $95\,\%$ of the probability to estimate between which two numbers of teleworkers the results will be concentrated with that level of confidence.
2. b) Based on your previous estimate, what qualitative or approximate value would you expect to obtain for $P(48 \leq X \leq 72)$? Justify your prediction before performing any numerical calculations.
3. c) Formally verify whether the conditions for approximating by the normal distribution are met, apply Yates's continuity correction to the interval $[48,\ 72]$, and calculate the exact probability value by standardizing with the $N(0,1)$ table. Compare the result with your initial estimate and assess whether it was accurate.

An audit of the Cercanías Renfe service in an urban area determines that the probability of a train arriving at its destination on time is $85\,\%$. Over the course of a working day, 400 trains run independently.

1. a) Precisely define the random variable $X$ (number of on-time trains) and its binomial model. Check the conditions that allow approximating $X$ with a normal variable $X' \equiv N(\mu,\sigma)$, and state the values of its mean and standard deviation.
2. b) The operator's quality protocol requires activating a user compensation plan if the number of delayed trains during the day is greater than 70. Translate this condition into an equivalent event concerning the number of trains that were indeed on time, apply the corresponding Yates's continuity correction, and calculate the probability that the company must activate the compensations on that day.
3. c) Suppose that at the end of a particularly difficult day, only 320 on-time trains are recorded. Using the statistical normality interval $(\mu - 2\sigma,\ \mu + 2\sigma)$, explain whether this result can be attributed to the usual random variability of the service or if it constitutes a significant statistical anomaly.

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:

1. a) Justify whether it is possible to approximate the distribution of the variable "number of teenagers who prefer digital format" using a normal distribution.
2. 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:

1. a) Exactly 200 heads.
2. b) More than 215 heads.
3. c) Fewer than 185 heads.

Note: Use the normal approximation with Yates' continuity correction.

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:

1. a) There are exactly 10 defective parts.
2. b) There are more than 15 defective parts.
3. 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 questions they get right. You can approximate the distribution of $X$ by a normal distribution with the same mean and the same standard deviation.

1. a) What is the probability that they get more than 30 questions right?
2. b) If at least 40 correct answers are needed to pass the exam, what is the probability that the student passes by pure chance?
3. 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:

1. a) Calculate the probability that the number of members who never visit is between 110 and 135 inclusive.
2. 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?
3. 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 $2.500$ 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:

1. a) There are exactly $12$ viral posts.
2. b) There are between $8$ and $15$ viral posts (both inclusive).

A sociological research center wants to estimate the level of support for a new energy transition law in Spain. Preliminary surveys show that $20\,\%$ of citizens are fully in favor of the measure ($p = 0{,}20$). The center's management requires designing a sample size $n$ of people sufficient to guarantee, with a probability of at least $95\,\%$ ($0{,}95$), that at least 50 favorable responses will be recorded.

1. a) Express the mean $\mu$ and the standard deviation $\sigma$ as functions of the unknown $n$. Apply Yates's continuity correction to transform the discrete event $P(X \geq 50)$ into a condition on the approximate continuous normal variable $X' \equiv N(\mu,\sigma)$.
2. b) In a standard normal distribution $N(0,1)$, the critical value that leaves an area of $0{,}95$ to its right is $z = -1{,}645$. Set up the equation associated with the probability condition that relates $n$, $49{,}5$, and the critical value $-1{,}645$.
3. c) Apply the change of variable $u = \sqrt{n}$ to convert the obtained expression into a quadratic equation in $u$. Solve it, determine the minimum integer sample size $n$ that satisfies the study's requirement, and check whether this $n$ verifies a posteriori the empirical criteria $n\cdot p \geq 5$ and $n\cdot q \geq 5$.

In the fraud prevention department of a financial institution in Spain, operations carried out with bank cards are analyzed. It is estimated that only $0{,}5\,\%$ of daily transactions ($p = 0{,}005$) turn out to be fraudulent. During a special audit, a sample of $n = 1000$ independent operations is taken.

Two analysts discuss the appropriateness of using the normal approximation to calculate the probability of finding no fraudulent operations, $P(X = 0)$:

• Carmen states: "Since $n\cdot p = 1000 \cdot 0{,}005 = 5$ and $n\cdot q = 1000 \cdot 0{,}995 = 995 \geq 5$, the minimum requirements are formally met. Therefore, approximating with the normal distribution and Yates's correction will provide a reliable value."

• David disagrees: "The criterion $np \geq 5$ is merely a guideline. Since $p$ is so close to 0, the binomial distribution is highly skewed, and calculating the probability at the extreme $k = 0$ using the normal distribution will accumulate an unacceptable relative error compared to the exact binomial formula."

1. a) Calculate the exact value of the probability of recording no fraud, $P(X = 0)$, using the formula for the binomial distribution $B(1000;\ 0{,}005)$.
2. b) Model the normal approximation $X' \equiv N(\mu,\sigma)$ with Yates's correction to calculate $P(X = 0) = P(-0{,}5 \leq X' \leq 0{,}5)$ by standardizing in $N(0,1)$. Then, calculate the percentage relative error committed by the normal approximation with respect to the exact binomial value: $E_r = \dfrac{|P_{\text{normal}} - P_{\text{binomial}}|}{P_{\text{binomial}}} \cdot 100\,\%$.
3. c) In view of the obtained error, determine in a reasoned manner which of the two analysts is right in this specific case and explain why, in highly skewed distributions, it is recommended to require a stricter threshold ($n\cdot p \geq 10$) or avoid the normal approximation in the tails.