The Standard Normal Distribution N(0, 1)
Emma and Jayden are in different AP classes and took different tests. Emma scored 1.5 standard deviations above her class mean. Jayden scored 1.1 standard deviations above his. Whose score is more impressive, and what percent of their classmates did each one beat? To answer, you need one single "ruler" for every bell-shaped set of data. That ruler is the standard normal distribution, $N(0,1)$.
The standard normal curve
Test scores, heights, and the weights of a batch of apples all pile up in the same bell shape. The simplest bell has mean $0$ and standard deviation $1$. Its values are called z-scores, and a z-score tells you how many standard deviations a value sits above ($z>0$) or below ($z<0$) the mean.
Observe the following representation of the standard normal distribution. If we know that the area of the shaded region to the right of $z = 1,96$ is $0,025$, reason…
Observe the following representation of the standard normal distribution. If we know that the area of the shaded region to the right of $z = 1,96$ is $0,025$, reason what the cumulative probability up to that point will be, $P(Z \leq 1,96)$.
Reading the z-table: area to the left
You cannot find the area under a bell curve with a simple formula, so we use a z-table (or a calculator). The table lists $\Phi(z)=P(Z\le z)$. The row gives $z$ to one decimal place and the column gives the second decimal place.
| 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 |
To find $\Phi(0.73)$: go to the row $0.7$, then across to the column $0.03$. You get $0.7673$. For a right tail, subtract from $1$.
In a reflex training app, users' results have been standardized following a standard normal distribution $Z \sim N(0, 1)$. If a player obtains a standardized score of…
In a reflex training app, users' results have been standardized following a standard normal distribution $Z \sim N(0, 1)$. If a player obtains a standardized score of $z = 1,45$, calculate the probability that a randomly chosen user has a lower score than theirs; that is, calculate $P(Z < 1,45)$.
The time it takes students to complete a logic challenge on TikTok follows a standard normal distribution once the data has been standardized. Calculate the probability…
The time it takes students to complete a logic challenge on TikTok follows a standard normal distribution once the data has been standardized. Calculate the probability that a student takes more time than the mean corresponding to a value of $Z = 0,82$. That is, find $P(Z > 0,82)$.
Given a random variable $Z$ that follows a standard normal distribution $N(0,1)$, use the probability table to calculate the following areas under the curve: a)…
Given a random variable $Z$ that follows a standard normal distribution $N(0,1)$, use the probability table to calculate the following areas under the curve:
- a) $P(Z \leq 1,48)$
- b) $P(Z < 0,35)$
- c) $P(Z \geq 2,12)$
- d) $P(Z > 1,96)$
Graphically represent the area corresponding to part c) on a bell curve.
Negative z-values and symmetry
The table above stops at positive $z$. You do not need a second table for negative values, because the bell is a mirror image of itself.
A study has been carried out on the monthly savings of young people in a city. After standardizing the variable, it is observed that it follows a $N(0, 1)$ distribution.…
A study has been carried out on the monthly savings of young people in a city. After standardizing the variable, it is observed that it follows a $N(0, 1)$ distribution. Using the symmetry properties of the normal distribution, calculate the probability that a young person has a savings level lower than a standardized value of $z = -1,26$.
In an entrance exam for a vocational training program, the scores fit a standard normal distribution. Knowing that the curve is symmetrical with respect to the y-axis,…
In an entrance exam for a vocational training program, the scores fit a standard normal distribution. Knowing that the curve is symmetrical with respect to the y-axis, determine the probability that a candidate obtains a standardized score greater than $-0,54$.
Making use of the symmetry properties of the standard normal distribution $N(0,1)$, determine the following probabilities for negative values of the variable: a)…
Making use of the symmetry properties of the standard normal distribution $N(0,1)$, determine the following probabilities for negative values of the variable:
- a) $P(Z \leq -0,74)$
- b) $P(Z < -1,25)$
- c) $P(Z \geq -0,58)$
- d) $P(Z > -2,03)$
Areas between two values
To get the probability that $Z$ falls between $a$ and $b$, take the area to the left of $b$ and cut off the part to the left of $a$.
Calculate the probability that a random variable $Z \sim N(0, 1)$ takes values within the interval $[0,45, \, 1,88]$. Remember that to calculate the area between two…
Calculate the probability that a random variable $Z \sim N(0, 1)$ takes values within the interval $[0,45, \, 1,88]$. Remember that to calculate the area between two positive values $a$ and $b$, you must subtract their cumulative probabilities: $P(a < Z < b) = P(Z < b) - P(Z < a)$.
In the quality control of some electronic components, it is accepted that the measurement error follows a $N(0, 1)$ distribution. Find the probability that the…
In the quality control of some electronic components, it is accepted that the measurement error follows a $N(0, 1)$ distribution. Find the probability that the standardized error lies in the symmetrical interval $[-1,5, \, 1,5]$.
Calculate the probability that the random variable $Z \sim N(0,1)$ takes values within the following intervals: a) $P(0,45 \leq Z \leq 1,82)$ b) $P(-1,05 < Z < 2,15)$ c)…
Calculate the probability that the random variable $Z \sim N(0,1)$ takes values within the following intervals:
- a) $P(0,45 \leq Z \leq 1,82)$
- b) $P(-1,05 < Z < 2,15)$
- c) $P(-2,30 \leq Z \leq -0,90)$
Explain by means of a drawing why in case c) the result must be equal to the probability of the corresponding positive interval.
Symmetric intervals and two tails
Intervals like $[-k,k]$ come up constantly, and the tails outside them too. Symmetry gives you a quick formula.
Calculate in a reasoned way, using the properties of the standard normal distribution $Z \sim N(0, 1)$, the value of the following probabilities without directly using…
Calculate in a reasoned way, using the properties of the standard normal distribution $Z \sim N(0, 1)$, the value of the following probabilities without directly using the table for negative values (only use the table for positive values):
- a) $P(|Z| \ge 1,25)$
- b) $P(-0,5 \le Z \le 0,5)$
- c) $P(Z \le -1,96 \cup Z \ge 1,96)$
A study has been carried out on the reaction time of teenagers to a visual stimulus. After standardizing the data to a $N(0, 1)$, it is observed that $15,87\%$ of the…
A study has been carried out on the reaction time of teenagers to a visual stimulus. After standardizing the data to a $N(0, 1)$, it is observed that $15,87\%$ of the subjects have a very slow reaction time (high values of $Z$).
If we choose two teenagers at random, what is the probability that both have a value of $Z$ between $-1$ and $1$? Assume that the reactions of both are independent.
Prove algebraically that for any variable $Z \sim N(0, 1)$ and any real value $k > 0$, it holds that $P(-k \le Z \le k) = 2 \cdot \Phi(k) - 1$. Subsequently, use this…
Prove algebraically that for any variable $Z \sim N(0, 1)$ and any real value $k > 0$, it holds that $P(-k \le Z \le k) = 2 \cdot \Phi(k) - 1$.
Subsequently, use this formula to find the probability that $Z$ is at a distance from the mean greater than $1,645$ standard deviations.
Reverse lookup: from area to z
Sometimes you know the probability and need the z-value. Now you search the inside of the table for the area, then read the row and column to build $z$.
Given a standard normal distribution $Z \sim N(0, 1)$, use the table in reverse to find the value of the constant $k$ such that the probability of obtaining a value less…
Given a standard normal distribution $Z \sim N(0, 1)$, use the table in reverse to find the value of the constant $k$ such that the probability of obtaining a value less than it is exactly $0,9750$. That is, look for $k$ in: $P(Z < k) = 0,9750$.
In the standard normal distribution $N(0,1)$, it is common to have to take the "reverse path": knowing the probability and looking for the value of the abscissa. Find…
In the standard normal distribution $N(0,1)$, it is common to have to take the "reverse path": knowing the probability and looking for the value of the abscissa. Find the value of the constant $k$ in each of the following cases:
- a) $P(Z \leq k) = 0,8749$
- b) $P(Z < k) = 0,9913$
- c) $P(Z \leq k) = 0,0516$
- d) $P(Z < k) = 0,3336$
Find the value of the abscissa $k$ in a distribution $N(0,1)$ knowing that: a) The area to the right of $k$ is $0,1515$. b) The area to the right of $k$ is $0,9418$. c)…
Find the value of the abscissa $k$ in a distribution $N(0,1)$ knowing that:
- a) The area to the right of $k$ is $0,1515$.
- b) The area to the right of $k$ is $0,9418$.
- c) $P(Z > k) = 0,0228$
Symmetric intervals in reverse: critical values
A critical value is the $k$ for which the middle interval $[-k,k]$ captures a given probability. You will use these constantly for confidence intervals later.
In advanced statistical studies, symmetrical intervals with respect to the origin that enclose a given probability are often sought. Find the positive value of $k$ such…
In advanced statistical studies, symmetrical intervals with respect to the origin that enclose a given probability are often sought. Find the positive value of $k$ such that:
- a) $P(-k \leq Z \leq k) = 0,90$
- b) $P(-k < Z < k) = 0,95$
- c) $P(-k \leq Z \leq k) = 0,99$
Note: These values of $k$ are known as critical values and are fundamental for calculating confidence intervals.
A streaming platform classifies its users according to their "loyalty index", which follows a distribution $N(0, 1)$. The system considers "VIP Users" to be those who…
A streaming platform classifies its users according to their "loyalty index", which follows a distribution $N(0, 1)$. The system considers "VIP Users" to be those who are in the central interval that concentrates $68,26\%$ of the population. However, for an aggressive marketing campaign, they want to redefine this group.
Find the symmetrical values $-z_{\alpha/2}$ and $z_{\alpha/2}$ that enclose a probability of $0,98$. Explain what percentage of users would be left out of this new group at each end.
Find a positive value $m$ such that the area enclosed under the standard normal curve $N(0, 1)$ between $-m$ and $m$ is three times the total area of the two outer tails…
Find a positive value $m$ such that the area enclosed under the standard normal curve $N(0, 1)$ between $-m$ and $m$ is three times the total area of the two outer tails (the sum of the areas to the left of $-m$ and to the right of $m$).
Percentiles and comparing scores
A percentile tells you what percent of values are below a given value. For $Z\sim N(0,1)$, the percentile of a z-score is $100\cdot\Phi(z)$. This is how you compare scores from different tests: standardize each, then compare their percentiles.
Back to Emma and Jayden: Emma's $z=1.5$ gives $\Phi(1.5)=0.9332$, so she beat about $93.3\%$ of her class. Jayden's $z=1.1$ gives $\Phi(1.1)=0.8643$, so he beat about $86.4\%$ of his. Emma's result is more outstanding, by about $7$ percentage points.
Two students from different high schools want to compare their results on an English exam. As it is not the same exam, they use the standardized score $Z$ (which follows…
Two students from different high schools want to compare their results on an English exam. As it is not the same exam, they use the standardized score $Z$ (which follows a $N(0,1)$) to know who is in a better position with respect to their group.
-
Student A has a standardized score of $z_A = 1,2$.
-
Student B has a standardized score of $z_B = 0,8$.
Calculate the percentage of students below each of them in their respective school and interpret who has achieved a more outstanding result.
In the world of esports, the skill of players of a famous video game is measured with a standardized score that follows a $N(0, 1)$. The developer company has decided to…
In the world of esports, the skill of players of a famous video game is measured with a standardized score that follows a $N(0, 1)$. The developer company has decided to award an exclusive skin only to players who exceed the 99th percentile. On the other hand, players who are below the value $-2,33$ enter a "basic training" program.
What is the probability that a randomly chosen player does not receive the skin but also does not have to enter the training program?
Equations with probabilities
Some problems describe a relationship between areas, like "the right tail is three times the left area". Translate the words into an equation in $\Phi(a)$, solve the linear equation for $\Phi(a)$, then run the table in reverse.
Given a variable $Z \sim N(0,1)$, calculate the value of $a$ knowing that the probability of the variable being greater than $a$ is seven times greater than the…
Given a variable $Z \sim N(0,1)$, calculate the value of $a$ knowing that the probability of the variable being greater than $a$ is seven times greater than the probability of it being less than $a$, where $a$ is a negative value.
Use the relationship $P(Z < a) + P(Z > a) = 1$ and the standard normal tables to help you.
In a standard normal distribution $N(0, 1)$, it is known that there is a positive value $k$ such that the probability that the variable $Z$ is greater than $k$ is…
In a standard normal distribution $N(0, 1)$, it is known that there is a positive value $k$ such that the probability that the variable $Z$ is greater than $k$ is exactly a quarter of the probability that $Z$ is less than $k$. Using the symmetry properties of the bell curve and the distribution tables, calculate the value of this parameter $k$ to two decimal places.
Given a random variable $Z \sim N(0, 1)$, calculate the value of the abscissa $a$ knowing that the following probability relationship holds:…
Given a random variable $Z \sim N(0, 1)$, calculate the value of the abscissa $a$ knowing that the following probability relationship holds:
$P(Z < a) = 3 \cdot P(a < Z < 1,5)$
Note: You will need to set up an equation using the distribution function $\Phi(z)$ and solve by looking it up in the table.