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Algebra 2 · Mathematics
Probability · The normal or Gaussian distribution

Continuous Probability Distributions and the Normal Curve

Before you start, make sure you know: Concepts (Math 7)
Before you start, make sure you know: Intervals, half-lines and neighborhoods (Algebra 1)

You plug in your phone and it says "about 87 minutes to a full charge." But exactly how long will it take: 87 minutes, 87.2, 87.2314...? Time can be any number in an interval, so we can't list the possible values and hand out probabilities one by one. Here's the question for this page: what is the probability that it takes exactly 87 minutes? The answer will surprise you, and it leads to the most important curve in statistics.

Continuous Random Variables and Density Curves

Until now, our random variables took separate values (0, 1, 2, ... goals). A continuous random variable can take any value in an interval.

Separate bars for a discrete variable next to a smooth curve with no gaps for a continuous variable. discrete: you count goals scored in a game continuous: you measure time to charge a phone

Between 1 goal and 2 goals there is nothing. Between 87 and 88 minutes there are infinitely many possible times.

Density curves

With infinitely many values we can't give each one its own probability. Instead we draw a curve, the probability density function $f(x)$, and probability becomes area.

A density curve with the area between a and b shaded, representing the probability that X is between a and b. a b P(a ≤ X ≤ b) = shaded area total area = 1

The simplest density is a rectangle: uniform density, where every value in an interval is equally likely.

A rectangle of height 0.2 over the interval from 0 to 5 with the part between 1 and 3.5 shaded. f(x) = 0.2 0 1 3.5 5 P(1 ≤ X ≤ 3.5) = 2.5 × 0.2 = 0.5

One exact value has probability zero

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It is known that the time it takes for a smartphone model to charge its battery from 0% to 100% follows a continuous probability distribution whose probability density…

It is known that the time it takes for a smartphone model to charge its battery from 0% to 100% follows a continuous probability distribution whose probability density function is constant in the interval $[80, 120]$ minutes, and is $0$ at any other point.

  1. a) Determine the value of the probability density function in this interval for it to be valid.
  2. b) Calculate the probability that the phone takes less than 90 minutes to charge.
  3. c) Represent the probability density function graphically.
80 90 100 120 Time (min) 0.01 0.02 0.025 f(x) 25% 75% f(x) = 0.025 P(X < 90)
Solve

Density Shapes: Triangles and Pieces

Many real situations need a density that is not a rectangle. A waiting time may be most likely in the middle, giving a triangle. When $f$ is defined piece by piece, you find the area with geometry: rectangles, triangles and trapezoids.

A triangular density from 0 to 6 with peak height one third at 3 and the part between 1 and 3 shaded as a trapezoid. 0 1 3 6 peak height 1/3 P(1 ≤ X ≤ 3) = area of the shaded trapezoid
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The time it takes a group of young people to completely disconnect after turning off their cell phones before sleeping is being studied. The probability density function…

The time it takes a group of young people to completely disconnect after turning off their cell phones before sleeping is being studied. The probability density function of this continuous random variable, $x$, expressed in minutes, is given by the following graph:

0 15 30 k x (min) f(x)

Calculate, justifying your answer:

  1. a) The value of the maximum ordinate $k$ for it to be a valid probability density function.
  2. b) The probability that a young person takes between 15 and 20 minutes to disconnect.
  3. c) What is the probability that they take exactly 22 minutes?
Solve
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Consider the piecewise continuous probability function of a random variable $X$, which represents the time (in hours) that a group of teenagers spends studying math over…

Consider the piecewise continuous probability function of a random variable $X$, which represents the time (in hours) that a group of teenagers spends studying math over a weekend:

$f(x) = \begin{cases} kx & \text{if } 0 \le x < 2 \\ k(4 - x) & \text{if } 2 \le x \le 4 \\ 0 & \text{otherwise} \end{cases}$

  1. a) Determine logically the value of the constant $k$ so that $f(x)$ is a probability density function.
  2. b) Calculate the probability that a student chosen at random studies between 1.5 and 3 hours.
  3. c) What is the probability that they study exactly 2 hours? Justify your answer based on the nature of the variable.
0 1 1.5 2 3 4 2k x [h] f(x) f(x) = kx f(x) = k(4-x) P(1.5 ≤ X ≤ 3)
Solve

The Normal Distribution

The most important density in statistics is the normal distribution, whose graph is the bell curve. Heights, test scores, and the weights of products from the same machine tend to cluster near the middle and thin out symmetrically on both sides.

Properties of every normal curve:

  • It is bell-shaped and symmetric about $x = \mu$.

  • Its highest point is at $x = \mu$, so the mean, median and mode are equal.

  • The total area under the curve is $1$, and the curve never touches the $x$-axis.

  • A bigger $\sigma$ gives a wider, flatter bell. A smaller $\sigma$ gives a narrower, taller bell. The mean only slides the curve left or right.

Try it. Slide the mean and the standard deviation and watch what changes.

Here is a very useful pattern, called the 68–95–99.7 rule: in any normal distribution, about 68% of the values are within $1\sigma$ of the mean, about 95% within $2\sigma$, and about 99.7% within $3\sigma$.

A normal curve with the area within one standard deviation of the mean shaded, labeled with the percent of the area in each band. −3σ −2σ −σ μ σ 2σ 3σ 2.35% 13.5% 34% 34% 13.5% 2.35%
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Observe the following two curves corresponding to normal distributions. Both represent the weight of two different breeds of domestic cats. a) Do both distributions have…

Observe the following two curves corresponding to normal distributions. Both represent the weight of two different breeds of domestic cats.

  1. a) Do both distributions have the same mean $\mu$? Justify your answer by observing the position of the maximum.
  2. b) Which of the two distributions has a larger standard deviation $\sigma$? Explain how this parameter influences the shape of the "bell".
Weight μ σ σ Breed A Breed B
Solve

The Standard Normal Table

There are infinitely many normal curves, one for each pair $(\mu, \sigma)$. Statisticians tabulate just one: the standard normal distribution $Z \sim N(0, 1)$, with mean $0$ and standard deviation $1$. In the next sections you will turn any normal distribution into this one.

The standard normal curve with the area to the left of z shaded, representing P(Z less than z). 0 z P(Z < z) = shaded area to the left

Standard normal table: $P(Z < z)$

$z$ 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359
0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753
0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141
0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517
0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879
0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224
0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549
0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.7852
0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133
0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.8389
1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621
1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.8830
1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015
1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177
1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319
1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441
1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545
1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633
1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706
1.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.9767
2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817
2.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.9857
2.2 0.9861 0.9864 0.9868 0.9871 0.9875 0.9878 0.9881 0.9884 0.9887 0.9890
2.3 0.9893 0.9896 0.9898 0.9901 0.9904 0.9906 0.9909 0.9911 0.9913 0.9916
2.4 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929 0.9931 0.9932 0.9934 0.9936
2.5 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946 0.9948 0.9949 0.9951 0.9952
2.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960 0.9961 0.9962 0.9963 0.9964
2.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970 0.9971 0.9972 0.9973 0.9974
2.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 0.9981
2.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.9986
3.0 0.9987 0.9987 0.9987 0.9988 0.9988 0.9989 0.9989 0.9989 0.9990 0.9990

This table lists positive $z$-values only. For $z \ge 3.5$ the area to the left is practically $1$, and for $z \le -3.5$ it is practically $0$.

Practice★☆☆Not started
Using the standard normal distribution table $Z \sim N(0, 1)$, obtain the following direct probabilities: a) $P(Z < 1,25)$ b) $P(Z \leq 0,47)$ c) $P(Z < 2,03)$ Briefly…

Using the standard normal distribution table $Z \sim N(0, 1)$, obtain the following direct probabilities:

  1. a) $P(Z < 1,25)$
  2. b) $P(Z \leq 0,47)$
  3. c) $P(Z < 2,03)$

Briefly explain what the value obtained in part (a) represents with respect to the total area under the Gaussian curve.

Solve

Symmetry, Complements, and Areas Between Two Values

The table only gives areas to the left of a positive $z$. Symmetry and the fact that the total area is $1$ give you everything else.

A standard normal curve with the left tail below minus z and the right tail above z both shaded, showing they have equal area. −z 0 z same area same area P(Z < −z) = P(Z > z)
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In a standard normal distribution $Z \sim N(0, 1)$, calculate the following probabilities using the properties of symmetry and the complementary event: a) $P(Z > 1,12)$…

In a standard normal distribution $Z \sim N(0, 1)$, calculate the following probabilities using the properties of symmetry and the complementary event:

  1. a) $P(Z > 1,12)$
  2. b) $P(Z < -0,54)$
  3. c) $P(Z \geq -1,96)$

Draw a Gaussian bell curve for part (b), shading the area you are calculating.

0 −0,54 P(Z < −0.54)
Solve
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Calculate the area enclosed in the following intervals for a standard normal distribution $Z \sim N(0, 1)$: a) $P(0,5 < Z < 1,5)$ b) $P(-1,2 < Z < 1,2)$ Represent the…

Calculate the area enclosed in the following intervals for a standard normal distribution $Z \sim N(0, 1)$:

  1. a) $P(0,5 < Z < 1,5)$
  2. b) $P(-1,2 < Z < 1,2)$

Represent the interval from part (b) graphically, indicating why the area is symmetrical with respect to the vertical axis.

Z −1,2 0 1,2 A₁ A₂ 1,2 1,2 A₁ = A₂
Solve

Standardizing: z-Scores

How do we use the single standard table for $N(100, 15)$ or $N(3.5, 1.2)$? We ask how many standard deviations a value is from its mean.

Why does it work? Subtracting $\mu$ moves the center of the bell to $0$. Dividing by $\sigma$ rescales the spread to $1$. The shape does not change, so areas are preserved: $P(X < x) = P(Z < z)$.

A normal curve with mean 100 and standard deviation 15 showing the x scale from 70 to 130 above the matching z scale from minus 2 to 2. 70 85 100 115 130 −2 −1 0 1 2 x: z:
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The score in a popular mental agility test for teenagers follows a normal distribution $X \sim N(100, 15)$. a) Standardize the value $X = 115$ and explain what the…

The score in a popular mental agility test for teenagers follows a normal distribution $X \sim N(100, 15)$.

  1. a) Standardize the value $X = 115$ and explain what the obtained result means (how many standard deviations from the mean it is).
  2. b) Standardize the value $X = 80$. What does the negative sign indicate in this case?
Solve
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Two friends, Lucy and Mark, play on different esports teams. Lucy's score in the game "Galaxy Quest" follows a distribution $N(1200, 15)$, while Mark's in "Pixel Arena"…

Two friends, Lucy and Mark, play on different esports teams. Lucy's score in the game "Galaxy Quest" follows a distribution $N(1200, 15)$, while Mark's in "Pixel Arena" follows a distribution $N(850, 10)$.

In the last match, Lucy scored 1,235 points and Mark scored 875 points. Use standardization of variables to determine which of the two obtained a more meritorious result compared to their group.

Solve

Probabilities for Any Normal Distribution

Now you can solve any "find the probability" problem for $X \sim N(\mu, \sigma)$.

A normal curve with mean 25 and the area between 16 and 34 shaded, which is minus 1.5 to 1.5 standard deviations. 16 25 34 P(16 < X < 34) = area between z = −1.5 and 1.5
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The battery life of mid-range wireless headphones follows a normal distribution with a mean of 6 hours and a standard deviation of 0.5 hours ($X \sim N(6; 0,5)$).…

The battery life of mid-range wireless headphones follows a normal distribution with a mean of 6 hours and a standard deviation of 0.5 hours ($X \sim N(6; 0,5)$).

Calculate the probability that, when choosing headphones at random, the battery lasts less than 7 hours. (Remember to standardize the variable first to be able to use the table).

Solve
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The height of 11th-grade students at a high school follows a normal distribution with a mean of $172$ inches and a standard deviation of $8$ inches. If we choose a…

The height of 11th-grade students at a high school follows a normal distribution with a mean of $172$ inches and a standard deviation of $8$ inches. If we choose a student at random, find the probability that:

  1. a) They are taller than $185$ inches.
  2. b) They are shorter than $160$ inches.
  3. c) Their height is between $165$ and $180$ inches.
172 Normal Distribution of Heights X ~ N(μ = 172, σ = 8) Height μ = 172 in σ = 8 156 164 180 188 Student height (in) Probability Density a) P(X > 185) 185 ? b) P(X < 160) 160 ? c) P(165 < X < 180) 165 180 ?
Solve
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A company that manufactures wireless headphones has verified that the battery life follows a normal distribution with a mean of $24$ hours and a standard deviation of…

A company that manufactures wireless headphones has verified that the battery life follows a normal distribution with a mean of $24$ hours and a standard deviation of $3,5$ hours.

  1. a) What percentage of headphones have a battery that lasts between $20$ and $28$ hours?
  2. b) If the company manufactures a batch of $5000$ headphones, how many are expected to last less than $18$ hours?
Solve

Inverse Problems: Finding a Value from a Probability

Sometimes you know the area and need the boundary: "What score puts you in the top 10%?" This is a percentile question. You work backward through the table.

A standard normal curve where the top 10 percent tail is shaded, leaving an area of 0.90 to its left, with the cutoff z at about 1.28. area 0.90 area 0.10 z ≈ 1.28 top 10% means P(Z < z) = 0.90

Useful critical values to remember: the middle $90\%$ of a normal distribution has $z = \pm 1.645$, the middle $95\%$ has $z = \pm 1.96$, and the middle $99\%$ has $z = \pm 2.576$. When the area falls exactly halfway between two table entries (as $0.95$ does between $0.9495$ and $0.9505$), take the middle $z$, here $1.645$.

What happens to $z$ as the area gets closer to $1$? It grows, but slowly. The tails of the bell are very thin and nearly flat, so you have to travel far to pick up the last tiny bits of area.

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Look up the value of the x-coordinate $k$ in the standard normal distribution table $N(0, 1)$ in the following cases (inverse problem): a) $P(Z < k) = 0,8413$ b)…

Look up the value of the x-coordinate $k$ in the standard normal distribution table $N(0, 1)$ in the following cases (inverse problem):

  1. a) $P(Z < k) = 0,8413$
  2. b) $P(Z < k) = 0,9750$
  3. c) $P(Z < k) = 0,5000$

What happens to the value of $k$ as the probability approaches 1? Justify your answer by observing the shape of the bell curve at its tails.

Solve
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Using the standard normal distribution table $N(0, 1)$, calculate the following probabilities: a) $P(Z < 1,48)$ b) $P(Z > 0,72)$ c) $P(Z < -1,15)$ d) $P(-0,5 < Z < 1,2)$…

Using the standard normal distribution table $N(0, 1)$, calculate the following probabilities:

  1. a) $P(Z < 1,48)$
  2. b) $P(Z > 0,72)$
  3. c) $P(Z < -1,15)$
  4. d) $P(-0,5 < Z < 1,2)$

Furthermore, find the value of the positive constant $a$ such that $P(-a < Z < a) = 0,90$.

Solve
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On a music streaming platform, the time users spend listening to a "Top 50" playlist follows a distribution $N(45, 12)$ minutes. The platform decides to send a…

On a music streaming platform, the time users spend listening to a "Top 50" playlist follows a distribution $N(45, 12)$ minutes. The platform decides to send a satisfaction survey to the $15\%$ of users who spend the most time listening to this playlist. After how many minutes of listening will a user receive the survey?

Solve

Finding an Unknown Mean or Standard Deviation

Sometimes the problem gives a probability and a boundary, and you must find $\mu$ or $\sigma$. The plan: use the probability to get the z-score, then solve $z = \dfrac{x - \mu}{\sigma}$ for the unknown.

A normal curve with the lowest 10 percent shaded to the left of the value 7, with the mean marked with a question mark. 7 μ = ? 10% known: P(X < 7) = 0.10
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The weight of the chocolate bars produced by a factory follows a normal distribution with a mean of $200$ ounces. Quality control rejects those bars that weigh less than…

The weight of the chocolate bars produced by a factory follows a normal distribution with a mean of $200$ ounces. Quality control rejects those bars that weigh less than $195$ ounces. Knowing that $3\%$ of the bars are rejected, calculate the value of the standard deviation of this production.

Solve
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In a prestigious design school, the final grade of the applicants follows a normal distribution. It is known that the mean grade is $6,8$ and that $33\%$ of the students…

In a prestigious design school, the final grade of the applicants follows a normal distribution. It is known that the mean grade is $6,8$ and that $33\%$ of the students obtain a grade higher than $7,5$.

  1. a) Find the standard deviation of the grades.
  2. b) If earning a merit scholarship requires being in the top $5\%$ of grades, what is the minimum grade needed to get this scholarship?
Solve
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A logistics company analyzes the delivery time of its local packages. The delivery time follows a normal distribution with a standard deviation of $12$ minutes. It is…

A logistics company analyzes the delivery time of its local packages. The delivery time follows a normal distribution with a standard deviation of $12$ minutes. It is known that $75\%$ of the packages are delivered in less than $50$ minutes.

  1. a) Calculate the mean delivery time.
  2. b) The company guarantees a shipping refund if the package takes more than an hour ($60$ minutes) to arrive. If $5000$ packages are sent in a month, how many refunds is the company estimated to have to make?
  3. c) If the company wants to halve the number of refunds by improving logistics but keeping the same mean, what should the new standard deviation be?
Solve

Harder Problems: Two Unknowns, Symmetric Cutoffs, and Comparisons

Two unknowns: a system

If both $\mu$ and $\sigma$ are unknown you need two probabilities. Each one gives an equation $z = \dfrac{x - \mu}{\sigma}$, and you solve the pair by elimination.

Symmetric cutoffs and a conditional probability

A normal curve centered at 0 with both tails beyond minus k and k shaded, and the middle accepted region unshaded. −k 0 k rejected rejected accepted

Remember: $P(A \mid B) = \dfrac{P(A \text{ and } B)}{P(B)}$. And when the distribution is centered at $0$, half of the accepted balls have positive error, because the accepted interval $(-k, k)$ is symmetric.

Comparing two normal models

To compare models, standardize against the same boundary. A higher z-score means the boundary is farther above that model's mean.

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The latency (ping) in a popular online video game follows a normal distribution $N(\mu, \sigma)$. After taking thousands of measurements, the developers have observed…

The latency (ping) in a popular online video game follows a normal distribution $N(\mu, \sigma)$. After taking thousands of measurements, the developers have observed that $15,87\%$ of the games have a latency greater than $110$ ms, while $6,68\%$ of the games have a latency of less than $75$ ms.

  1. a) Calculate the parameters $\mu$ (mean) and $\sigma$ (standard deviation) of this distribution.
  2. b) If a player considers a game to be "unplayable" when the latency exceeds $125$ ms, what percentage of games will fall into this category?
Solve
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A technology company manufactures precision sensors for drones. The margin of error in the height measurement of these sensors follows a normal distribution with a mean…

A technology company manufactures precision sensors for drones. The margin of error in the height measurement of these sensors follows a normal distribution with a mean of $0$ inches and a standard deviation of $1,5$ inches. For safety reasons, the company discards all sensors whose absolute error is greater than a critical value $k$.

  1. a) If it is known that $10\%$ of the total production is discarded, calculate the value of this margin of error $k$.
  2. b) What is the probability that, when choosing an accepted (not discarded) sensor, its error is positive?
σ = 1,5 0 -k k 90% Accepted 5% Discarded 5% Discarded N(0; 1.5)
Solve
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Two models of batteries for electric scooters are being compared. Model A has a mean range of $40$ miles with a standard deviation of $4$ miles. Model B has a mean range…

Two models of batteries for electric scooters are being compared. Model A has a mean range of $40$ miles with a standard deviation of $4$ miles. Model B has a mean range of $42$ miles with a standard deviation of $8$ miles. Both durations follow normal distributions.

  1. a) If a user needs to make a trip of $48$ miles without recharging, which battery model offers them greater reliability (higher probability of success)? Justify the calculations.
  2. b) For what distance is the probability of both models failing the same?
48 miles (Target) Distance "x" = ? Left areas Probability Density 20 30 40 42 48 50 60 70 Distance (miles) Scooter Battery Comparison Normal Distributions of Range Model A: A ~ N(40, 4) Model B: B ~ N(42, 8) a) Greater reliability for 48 miles? Area to the right = P(Success) P(A > 48) = ? P(B > 48) = ? b) Equal probability of failure Area to the left = P(Failure) P(A < x) = P(B < x) = ?
Solve

Normal Approximation to the Binomial

A binomial variable $X \sim B(n, p)$ counts successes in $n$ independent trials, each with success probability $p$ (and $q = 1 - p$). Its exact probabilities are

$$P(X = k) = \binom{n}{k} p^k q^{n-k}, \qquad \mu = np, \qquad \sigma = \sqrt{npq}$$

For small $n$ the formula is fine. For example, with $n = 5$ and $p = 0.3$: $P(X = 2) = \binom{5}{2}(0.3)^2(0.7)^3 = 10(0.09)(0.343) \approx 0.3087$. For large $n$, adding hundreds of terms is hopeless, but the histogram starts to look like a bell curve.

A histogram of a binomial distribution with a normal bell curve drawn over the tops of the bars. number of successes bars: binomial. curve: N(np, √(npq))

Because the binomial takes only whole numbers but the normal is continuous, you need a continuity correction: each whole number $k$ becomes a bar of width $1$, covering $[k - 0.5,\ k + 0.5]$. (Some problems call this Yates's continuity correction.)

Three adjacent histogram bars with the middle bar for k highlighted, spanning from k minus 0.5 to k plus 0.5. k − 0.5 k k + 0.5 the bar for k covers [k − 0.5, k + 0.5]
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A survey on social media use has been carried out and it is known that $30\%$ of American teenagers prefer to use TikTok over other platforms. If we select a random…

A survey on social media use has been carried out and it is known that $30\%$ of American teenagers prefer to use TikTok over other platforms. 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 prefer TikTok" using a normal distribution.
  2. b) Using this approximation (with Yates's continuity correction), calculate the probability that more than $70$ teenagers prefer TikTok.
μ = 60 x' = 70.5 40 50 60 70 80 Teenagers (X) B(200, 0.3) ≈ N(60, 6.48) P(X > 70) = P(X ≥ 71) Yates: P(X' ≥ 70.5)
Solve
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A music streaming app has launched a new recommendation algorithm. It is estimated that the probability of a user liking a recommended song is $65\%$. If in one day the…

A music streaming app has launched a new recommendation algorithm. It is estimated that the probability of a user liking a recommended song is $65\%$. If in one day the algorithm recommends songs to a group of $400$ users:

  1. a) Justify whether it is possible to approximate the distribution of the number of users who like the recommendation using a normal distribution.
  2. b) Using the appropriate approximation (with continuity correction), calculate the probability that the recommendation is successful for a number of users between $250$ and $275$ (inclusive).
Solve
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The weight of the recycled plastic bottles collected at a sustainability center follows a distribution $N(120, 15)$ ounces. The center classifies the bottles into three…

The weight of the recycled plastic bottles collected at a sustainability center follows a distribution $N(120, 15)$ ounces. The center classifies the bottles into three categories: "Light" (the lightest $20\%$), "Standard" (the middle $60\%$), and "Heavy" (the heaviest $20\%$).

  1. a) Determine the weights that delimit each of the three categories.
  2. b) If we choose $5$ bottles at random, what is the probability that exactly $3$ of them are classified as "Standard"?
Solve

Back to the start: a phone's charge time is continuous, so $P(X = 87) = 0$ exactly. We can only ask about intervals. If charge time follows $N(87, 5)$ minutes, then $P(85 \le X \le 90)$ has $z_1 = -0.4$ and $z_2 = 0.6$, and the probability is $0.7257 - (1 - 0.6554) = 0.7257 - 0.3446 = 0.3811$, about $38\%$.

Summary

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