Standardizing and z-Scores
Aaliyah scored 1350 on the SAT. Her friend Ben scored 29 on the ACT. The two tests use different scales, so who actually did better? Almost no real data has mean $0$ and standard deviation $1$: test scores, heights, phone battery life, commute times. Standardizing is the bridge. It turns any normal distribution into the standard normal $N(0,1)$ without losing information, so you can always use the same table (Table A) or the same calculator commands.
The z-Score
Why does this work? Subtracting $\mu$ slides the center of the bell to $0$. Dividing by $\sigma$ stretches or squeezes the bell so its spread becomes $1$. The shape and the areas under the curve do not change.
Some problems call $z$ the standardized variable, and they call going back from $z$ to $x$ unstandardizing. These are the same ideas as the z-score and converting back.
Drag the sliders below. The bell moves and spreads out, but its shape stays the same. Standardizing undoes exactly that shift and stretch.
In a community of content creators, the number of daily video views follows a normal distribution with a mean of $\mu = 1200$ and a standard deviation of $\sigma = 300$.…
In a community of content creators, the number of daily video views follows a normal distribution with a mean of $\mu = 1200$ and a standard deviation of $\sigma = 300$. If an "influencer" has achieved $1650$ views today, calculate the value of the standardized variable $Z$ corresponding to this data.
The battery life of a smartphone model under heavy use follows a normal distribution $N(8, 1.5)$, where the mean is $8$ hours. A user notices that their phone only lasts…
The battery life of a smartphone model under heavy use follows a normal distribution $N(8, 1.5)$, where the mean is $8$ hours. A user notices that their phone only lasts $5$ hours after the latest update. Standardize this value and state whether the result is positive or negative. What is the meaning of the sign in this context?
In an e-sports competition, the reaction time of the players follows a normal distribution with a mean of $210$ milliseconds and a standard deviation of $15$…
In an e-sports competition, the reaction time of the players follows a normal distribution with a mean of $210$ milliseconds and a standard deviation of $15$ milliseconds. Standardize the following reaction times and interpret the results:
- a) $195$ milliseconds.
- b) $210$ milliseconds.
- c) $240$ milliseconds.
Going Back: From z to x
Sometimes you know the z-score and need the original value. Just undo the two steps of the formula in reverse order: multiply by $\sigma$, then add $\mu$.
Given a continuous random variable $X$ that follows a normal distribution with a mean of $\mu = 50$ and a standard deviation of $\sigma = 8$, find the value of the…
Given a continuous random variable $X$ that follows a normal distribution with a mean of $\mu = 50$ and a standard deviation of $\sigma = 8$, find the value of the original variable $X$ knowing that its standardized value is $Z = 2.25$.
The weight of popcorn bags at a movie theater follows a normal distribution $N(150, 12)$ in ounces. If we select a bag at random and we know that its standardized value…
The weight of popcorn bags at a movie theater follows a normal distribution $N(150, 12)$ in ounces. If we select a bag at random and we know that its standardized value is $Z = -1.5$, does this bag weigh more or less than the mean? Calculate its exact weight in ounces.
Comparing Values from Different Distributions
Why? Subtracting the mean puts both groups on the same center. Dividing by $\sigma$ removes the units and the spread. After that, a $z$ means the same thing everywhere: "this many standard deviations above average."
Back to the start. Use the simplified values SAT $\sim N(1050, 200)$ and ACT $\sim N(21, 5)$. Aaliyah has $z = \dfrac{1350 - 1050}{200} = 1.5$. Ben has $z = \dfrac{29 - 21}{5} = 1.6$. Ben did slightly better relative to the other test takers.
Two students from different high schools want to compare their scores on an Economics exam. Emma scored $8.2$ at her school, where the mean was $\mu = 6.5$ and the…
Two students from different high schools want to compare their scores on an Economics exam.
-
Emma scored $8.2$ at her school, where the mean was $\mu = 6.5$ and the standard deviation was $\sigma = 1.1$.
-
Ethan scored $8.5$ at his school, where the mean was $\mu = 7.2$ and the standard deviation was $\sigma = 1.4$.
Calculate the standardized variable for both and determine which of the two obtained a better score relative to their own group.
On an online gaming platform, the players' scores in the "Legend" level follow a normal distribution with a mean of $\mu = 4500$ points and a standard deviation of…
On an online gaming platform, the players' scores in the "Legend" level follow a normal distribution with a mean of $\mu = 4500$ points and a standard deviation of $\sigma = 600$ points. On the other hand, in the "Master" level, the scores follow a normal distribution with $\mu = 3800$ points and $\sigma = 450$ points.
A player has obtained $5200$ points in the "Legend" level and their friend has obtained $4400$ points in the "Master" level. Use the standardized variable to determine which of the two friends has obtained a better score in relation to their gaming group.
Three students from different high schools want to compare their results in a linguistic aptitude test. Because the exams were different, the scores follow different…
Three students from different high schools want to compare their results in a linguistic aptitude test. Because the exams were different, the scores follow different normal distributions:
-
Ashley obtained $72$ points in an exam with $N(60, 8)$.
-
Ben obtained $80$ points in an exam with $N(70, 12)$.
-
Chloe obtained $65$ points in an exam with $N(55, 5)$.
Determine, by standardizing the variables, which of the three students is in a better relative position compared to their reference group. Justify your answer by comparing the values of the standardized variable $Z$.
Why Standardizing Works
The key fact is that standardizing does not change the event, only the way it is written. For $\sigma > 0$ you may subtract the same number from all parts of an inequality and divide all parts by the same positive number, and the inequality keeps its direction.
The standard normal curve is symmetric about $0$. So the area to the left of $-a$ equals the area to the right of $+a$. This is the most useful shortcut for reading the table with negative $z$.
If a standardized variable $Z$ follows a standard normal distribution $N(0, 1)$, use the symmetry of the Gaussian bell curve to explain why the probability $P(Z < -1)$…
If a standardized variable $Z$ follows a standard normal distribution $N(0, 1)$, use the symmetry of the Gaussian bell curve to explain why the probability $P(Z < -1)$ is the same as $P(Z > 1)$. Draw a diagram of the situation.
Let $X$ be a continuous random variable that follows a normal distribution $N(\mu, \sigma)$. If we define the standardized variable as $Z = \frac{X - \mu}{\sigma}$,…
Let $X$ be a continuous random variable that follows a normal distribution $N(\mu, \sigma)$. If we define the standardized variable as $Z = \frac{X - \mu}{\sigma}$, prove analytically, using the properties of standardization and the symmetry of the Gaussian bell curve, that the following probabilities are equivalent:
-
$P(\mu - \sigma < X < \mu + 2\sigma)$
-
$P(-1 < Z < 2)$
Calculate this value using the table for the distribution $N(0, 1)$.
From a z-Score to a Probability
Table A gives the area to the left of $z$: $P(Z < z)$. A graphing calculator does the same without rounding $z$: normalcdf(lower, upper, μ, σ) works directly on $X$.
The time taken by young people in a city to get to high school has been measured. The data follows a normal distribution with $\mu = 25$ minutes and $\sigma = 5$…
The time taken by young people in a city to get to high school has been measured. The data follows a normal distribution with $\mu = 25$ minutes and $\sigma = 5$ minutes. Standardize the value $X = 35$ minutes. If the standard normal table tells us that $P(Z < 2) = 0.9772$, what percentage of students takes less than $35$ minutes to arrive?
A study has been carried out on the daily time that 17-year-olds spend watching content on streaming* platforms. The results show that this time follows a normal…
A study has been carried out on the daily time that 17-year-olds spend watching content on streaming platforms. The results show that this time follows a normal distribution $X \sim N(110, 25)$, where time is measured in minutes.
- a) Calculate the probability that a randomly chosen young person spends more than $135$ minutes a day on these platforms.
- b) What is the probability that the time spent is between $90$ and $120$ minutes?
The weight of the backpacks of high school seniors follows a normal distribution with a mean of $6,5$ pounds and a standard deviation of $1,4$ pounds. a) Determine the…
The weight of the backpacks of high school seniors follows a normal distribution with a mean of $6,5$ pounds and a standard deviation of $1,4$ pounds.
- a) Determine the value of the standardized variable $Z$ corresponding to a backpack of $8$ pounds.
- b) The school doctor recommends that the weight of the backpack should not exceed $9$ pounds. What percentage of students is failing to comply with this recommendation?
Conditional Probability
Some questions say "given that". Then you need the definition of conditional probability:
$$P(A \mid B) = \frac{P(A \cap B)}{P(B)}.$$For intervals of a normal variable, $A \cap B$ is just the overlap of the two intervals.
The time that teenagers in a city spend daily on social media follows a normal distribution $X \sim N(120, 20)$, where time is measured in minutes. Calculate the…
The time that teenagers in a city spend daily on social media follows a normal distribution $X \sim N(120, 20)$, where time is measured in minutes.
Calculate the probability that a randomly chosen teenager spends between $100$ and $150$ minutes on social media, knowing beforehand that they spend more than $110$ minutes. (Note: use the definition of conditional probability along with the standardization of the variable).
Percentiles and Cutoffs
Now the question runs backward. You know the area and want the value of $x$. Find the area in the body of Table A, read the matching $z$ from the margins, then convert back with $x = \mu + z\sigma$. On a calculator, invNorm(area, μ, σ) does it in one step.
The final scores on a college entrance exam in a certain state follow a normal distribution with a mean of $6,8$ and a standard deviation of $1,2$. The university has…
The final scores on a college entrance exam in a certain state follow a normal distribution with a mean of $6,8$ and a standard deviation of $1,2$.
The university has decided to award a special excellence scholarship to the top $5\%$ of students who have obtained the best scores. Using the standard normal distribution table $N(0,1)$, calculate the minimum score required to obtain this scholarship.
We want to compare the mental agility of two groups of students using a logic test. Group A obtains results that follow a $N(75, 8)$ and Group B a $N(70, 5)$. a) If a…
We want to compare the mental agility of two groups of students using a logic test. Group A obtains results that follow a $N(75, 8)$ and Group B a $N(70, 5)$.
- a) If a student from Group A scores $85$ points and one from Group B scores $78$ points, which of the two is in a better relative position within their own group?
- b) In Group B, what is the maximum score of the lowest $20\%$ of students?
Central Intervals and Exact Values
A symmetric interval around the mean that holds the central $C\%$ leaves $100\% - C\%$ in the two tails, split equally.
What about the probability of one exact value? In a continuous model, the area above a single point is a line with no width, so
$$P(X = a) = 0.$$Only intervals have positive probability. So $P(X < a)$ and $P(X \le a)$ are the same number.
A content creator on a social network analyzes the number of "likes" their posts receive. They have observed that the number of "likes" follows a normal distribution…
A content creator on a social network analyzes the number of "likes" their posts receive. They have observed that the number of "likes" follows a normal distribution with a mean of $1200$.
- a) If the standard deviation is $\sigma = 150$, calculate the symmetric interval around the mean that contains $90\%$ of the posts.
- b) What is the probability that a randomly chosen post has exactly $1300$ "likes"? Justify your answer based on the properties of continuous variables.
The heights of the high school seniors at a school are distributed according to a normal distribution $X \sim N(172, 9)$. The principal wishes to select the tallest…
The heights of the high school seniors at a school are distributed according to a normal distribution $X \sim N(172, 9)$. The principal wishes to select the tallest $5\%$ of the students for the basketball team and the shortest $5\%$ of the students for an ergometric study.
- a) From what height are students selected for the basketball team?
- b) What is the maximum height to take part in the ergometric study?
- c) What is the probability that a student measures exactly $172$ cm? Justify your answer based on the nature of continuous variables.
Finding an Unknown Mean or Standard Deviation
If you know one probability and one of the two parameters, you can find the other. Get $z$ from the area, then use $x = \mu + z\sigma$ and solve for the unknown.
The battery life of a new smartphone* model under heavy use follows a normal distribution with a mean of $\mu$ hours and a standard deviation of $\sigma = 1,5$ hours. It…
The battery life of a new smartphone model under heavy use follows a normal distribution with a mean of $\mu$ hours and a standard deviation of $\sigma = 1,5$ hours.
It is known that $25,14\%$ of the phones have a battery life of less than $8$ hours. Find the value of the mean $\mu$ of this distribution using the process of unstandardization.
In a track meet, the time in seconds taken by the participants to complete the 400 meter sprint follows a $X \sim N(52, \sigma)$ distribution. If it is known that…
In a track meet, the time in seconds taken by the participants to complete the 400 meter sprint follows a $X \sim N(52, \sigma)$ distribution.
If it is known that $84,13\%$ of the runners take less than $55$ seconds to finish the race, calculate the standard deviation $\sigma$ of the runners' times.
It is known that the weight of coffee packages automatically packed by a machine follows a normal distribution with a standard deviation of $\sigma = 15$ grams. The…
It is known that the weight of coffee packages automatically packed by a machine follows a normal distribution with a standard deviation of $\sigma = 15$ grams. The machine has been adjusted so that $97,5\%$ of the packages have a weight greater than $470,6$ grams.
- a) Calculate the mean weight $\mu$ to which the machine has been programmed.
- b) With this adjustment, what is the probability that a randomly chosen package weighs more than $510$ grams?
Two Conditions: Solve for μ and σ Together
When both $\mu$ and $\sigma$ are unknown, you need two pieces of information. Each one gives an equation $x = \mu + z\sigma$. Then solve the system, for example by elimination.
In a test preparation center for college entrance exams, it has been observed that the scores on a practice AP exam follow a normal distribution $N(\mu, \sigma)$. It is…
In a test preparation center for college entrance exams, it has been observed that the scores on a practice AP exam follow a normal distribution $N(\mu, \sigma)$. It is known that $15,87\%$ of the students obtained a score higher than $8$ points, while $2,28\%$ did not reach $3$ points. Calculate the value of the mean $\mu$ and the standard deviation $\sigma$ of this distribution using the standard normal table $Z \sim N(0, 1)$.
A company that manufactures electronic components guarantees that the lifespan of its batteries follows a normal distribution with a mean of $800$ days. After a quality…
A company that manufactures electronic components guarantees that the lifespan of its batteries follows a normal distribution with a mean of $800$ days. After a quality study, it is determined that the probability of a battery lasting more than $1000$ days is only $0,0062$.
- a) Find the value of the standard deviation $\sigma$ of the production.
- b) If the company decides to replace all batteries that last less than $k$ days, what value must $k$ have so that they only have to replace $5\%$ of the batteries sold?
In a study on the monthly leisure expenditure of families in a county, it has been proven that this follows a normal distribution. It is known that the mean expenditure…
In a study on the monthly leisure expenditure of families in a county, it has been proven that this follows a normal distribution. It is known that the mean expenditure is $\$250$. If $20\%$ of the families spend more than $\$320$:
- a) Calculate the standard deviation of the distribution.
- b) Determine the expenditure interval, symmetric around the mean, in which the central $90\%$ of the families in the county is found.