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Exercises: The normal distribution

Practice★☆☆
Using the standard normal distribution table $N(0, 1)$, obtain the following direct probabilities: a) $P(Z < 1,45)$ b) $P(Z \le 0,38)$ c) $P(Z < 2,01)$

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

  1. a) $P(Z < 1,45)$
  2. b) $P(Z \le 0,38)$
  3. c) $P(Z < 2,01)$
0 z P(Z < z) N(0, 1)
Solve
Practice★☆☆
In a standard normal distribution $Z \sim N(0, 1)$, use the symmetry properties of the Gaussian bell curve to calculate the following probabilities: a) $P(Z > 0,82)$ b)…

In a standard normal distribution $Z \sim N(0, 1)$, use the symmetry properties of the Gaussian bell curve to calculate the following probabilities:

  1. a) $P(Z > 0,82)$
  2. b) $P(Z \ge 1,96)$
  3. c) $P(Z > 2,5)$

Remember that $P(Z > k) = 1 - P(Z \le k)$.

Solve
Practice★☆☆
Find the probability that a standard normal variable $Z$ lies in the following intervals: a) $P(0,5 < Z < 1,5)$ b) $P(1 \le Z \le 2,33)$ Illustrate part a) by drawing a…

Find the probability that a standard normal variable $Z$ lies in the following intervals:

  1. a) $P(0,5 < Z < 1,5)$
  2. b) $P(1 \le Z \le 2,33)$

Illustrate part a) by drawing a sketch of the normal curve and shading the corresponding area.

−2 −1 0 1 2 0,5 1,5 Z N(0, 1) P(0,5 < Z < 1,5)
Solve
Practice★☆☆
Let $X$ be a random variable following a normal distribution with mean $\mu = 50$ and standard deviation $\sigma = 10$. Perform the standardisation process to calculate…

Let $X$ be a random variable following a normal distribution with mean $\mu = 50$ and standard deviation $\sigma = 10$. Perform the standardisation process to calculate the probability $P(X < 65)$.

State clearly the value of the transformed variable $Z = \frac{X - \mu}{\sigma}$ before looking it up in the table.

Solve
Practice★☆☆
The marks in an Applied Mathematics exam follow a normal distribution with a mean of $6,2$ marks and a standard deviation of $1,5$ marks. If a student is chosen at…

The marks in an Applied Mathematics exam follow a normal distribution with a mean of $6,2$ marks and a standard deviation of $1,5$ marks. If a student is chosen at random, calculate the probability that their mark is greater than $7$.

Solve
Practice★☆☆
Given a standard normal distribution $Z \sim N(0, 1)$, use the table in reverse to find the value of $k$ in the following cases: a) $P(Z < k) = 0,9452$ b)…

Given a standard normal distribution $Z \sim N(0, 1)$, use the table in reverse to find the value of $k$ in the following cases:

  1. a) $P(Z < k) = 0,9452$
  2. b) $P(Z \le k) = 0,9750$
  3. c) $P(Z < k) = 0,5000$
Solve
Practice★☆☆
The weight of apples from a harvest follows a normal distribution with a mean of $150$ grams and a standard deviation of $15$ grams. If "Type A apples" are defined as…

The weight of apples from a harvest follows a normal distribution with a mean of $150$ grams and a standard deviation of $15$ grams. If "Type A apples" are defined as those weighing between $135$ and $165$ grams, calculate the probability that an apple chosen at random is Type A.

σ=15 σ=15 Type A 135 150 165 μ-σ μ μ+σ
Solve
Practice★☆☆
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 $0,025$, deduce the…

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 $0,025$, deduce the cumulative probability up to that point, $P(Z \leq 1,96)$.

0 1,96 0,025 P(Z ≤ 1,96) = ?
Solve
Practice★☆☆
In quality control for electronic components, the measurement error is assumed to follow a $N(0, 1)$ distribution. Find the probability that the standardised error lies…

In quality control for electronic components, the measurement error is assumed to follow a $N(0, 1)$ distribution. Find the probability that the standardised error lies in the symmetric interval $[-1,5, \, 1,5]$.

Solve
Practice★☆☆
The waiting time, in minutes, at a bus stop on an urban line follows a normal distribution $X \sim N(12, 3)$. We want to determine the probability that a passenger waits…

The waiting time, in minutes, at a bus stop on an urban line follows a normal distribution $X \sim N(12, 3)$. We want to determine the probability that a passenger waits between $9$ and $18$ minutes, that is, $P(9 \le X \le 18)$.

  1. a) Before standardising or looking up tables, make a reasoned estimate of this probability by applying the empirical $68\,\%\text{ - }95\,\%\text{ - }99{,}7\,\%$ rule. Justify how you decompose the interval $[9, 18]$ from the mean $\mu = 12$ and the standard deviation $\sigma = 3$.
  2. b) Standardise the endpoints and calculate the exact probability knowing that in the standard normal distribution $P(Z \le 1) = 0{,}8413$ and $P(Z \le 2) = 0{,}9772$.
  3. c) Compare your previous estimate with the exact value obtained and argue why the empirical rule is a useful consistency check tool in probability problems.
Solve
Practice★☆☆
In an exam, to calculate the cumulative probability $P(Z \le -1{,}40)$ of a variable $Z \sim N(0, 1)$, Sergio writes the following working: "Since the curve is symmetric…

In an exam, to calculate the cumulative probability $P(Z \le -1{,}40)$ of a variable $Z \sim N(0, 1)$, Sergio writes the following working:

"Since the curve is symmetric with respect to the origin, the left tail is equal to the right tail but with a negative sign: $P(Z \le -1{,}40) = -P(Z \le 1{,}40)$. Looking up in the table I get $P(Z \le 1{,}40) = 0{,}9192$, so the provisional result is $-0{,}9192$. Given that a probability can never be negative, I change the sign of the final result: $P(Z \le -1{,}40) = 0{,}9192$".

  1. a) Identify and explain with reasoned arguments the two conceptual errors made by Sergio.
  2. b) Relying on the properties of the density function (the total area under the curve is $1$ and the curve is symmetric with respect to $0$), deduce the correct relation to calculate $P(Z \le -z)$ from $P(Z \le z)$ with $z > 0$.
  3. c) Correct Sergio's calculation and find the exact value of $P(Z \le -1{,}40)$.
Solve
Practice★★☆
Given a random variable $Z$ following a standard normal distribution $N(0, 1)$, use the probability table to calculate: a) $P(Z < 1,45)$ b) $P(Z > 0,82)$ c)…

Given a random variable $Z$ following a standard normal distribution $N(0, 1)$, use the probability table to calculate:

  1. a) $P(Z < 1,45)$
  2. b) $P(Z > 0,82)$
  3. c) $P(Z \le -1,24)$
  4. d) $P(-0,5 < Z < 1,72)$
−3 −2 −1 0 1 2 3 −0,5 1,72 z
Solve
Practice★★☆
In the standard normal distribution $Z \sim N(0, 1)$, find the x-coordinate $k$ in each of the following cases: a) $P(Z < k) = 0,9418$ b) $P(Z > k) = 0,1075$ c)…

In the standard normal distribution $Z \sim N(0, 1)$, find the x-coordinate $k$ in each of the following cases:

  1. a) $P(Z < k) = 0,9418$
  2. b) $P(Z > k) = 0,1075$
  3. c) $P(Z < k) = 0,2236$
  4. d) $P(-k < Z < k) = 0,95$
Solve
Practice★★☆
The daily time teenagers at a secondary school spend playing video games follows a normal distribution with a mean of $\mu = 110$ minutes and a standard deviation of…

The daily time teenagers at a secondary school spend playing video games follows a normal distribution with a mean of $\mu = 110$ minutes and a standard deviation of $\sigma = 25$ minutes. If we select a student at random, calculate the probability that they:

  1. a) Play for more than two hours a day.
  2. b) Play between 90 and 130 minutes.
  3. c) Play for less than 60 minutes.
VIDEO GAME TIME DISTRIBUTION Continuous Random Variable: X ~ N(110, 25) Mean (μ) = 110 min | Std. Dev. (σ) = 25 min f(x) = 1 σ√(2π) e -(x-μ)² / 2σ² a) More than 2 hours (X > 120 min) 110 120 P(X > 120) = ? b) Between 90 and 130 min (90 < X < 130) 90 130 110 P(90 < X < 130) = ? c) Less than 60 min (X < 60 min) 110 60 P(X < 60) = ?
Solve
Practice★★☆
It is known that the weight of newborn babies in a hospital follows a normal distribution. If $15,87 \%$ of babies weigh less than $2,8$ kg and the standard deviation of…

It is known that the weight of newborn babies in a hospital follows a normal distribution. If $15,87 \%$ of babies weigh less than $2,8$ kg and the standard deviation of the population is $0,4$ kg, calculate the mean weight of newborn babies in that hospital.

Solve
Practice★★☆
A study on the battery life of a new smartphone model indicates that its useful life follows a normal distribution with a mean of $32$ hours. If it has been found that…

A study on the battery life of a new smartphone model indicates that its useful life follows a normal distribution with a mean of $32$ hours. If it has been found that only $2,28 \%$ of the phones have a battery that lasts less than $26$ hours:

  1. a) Find the standard deviation of this distribution.
  1. b) What is the probability that a randomly chosen phone lasts more than $35$ hours?
Bell curve of the normal distribution of battery life, with mean 32 hours, the area below 26 hours shaded in orange and the area above 35 hours shaded in green μ = 32 h 26 h 35 h 2,28 % P(X > 35) = ?
Solve
Practice★★☆
To gain admission to a highly oversubscribed vocational course, an entrance test is taken whose scores follow a distribution $N(72, 12)$. If there are places for only…

To gain admission to a highly oversubscribed vocational course, an entrance test is taken whose scores follow a distribution $N(72, 12)$. If there are places for only the top $25 \%$ of candidates with the highest marks:

  1. a) What is the minimum score needed to obtain a place?
  2. b) If a student scored $80$ points, are they in the selected group? Justify your answer by calculating their percentile.
Solve
Practice★★☆
In a soft drink packaging plant, the content of cans follows a normal distribution with a mean of $33$ cl and a standard deviation of $0,5$ cl. a) Calculate the…

In a soft drink packaging plant, the content of cans follows a normal distribution with a mean of $33$ cl and a standard deviation of $0,5$ cl.

  1. a) Calculate the probability that a can contains less than $32,2$ cl.
  2. b) Determine the interval of values, symmetrical about the mean, containing $90 \%$ of the contents of the cans.
Distribution of Can Content Modelo Normal: X ~ N(33; 0,5) a) Probability of containing < 32,2 cl 33 32,2 P = ? b) Central 90% interval 33 ? ? 90%
Solve
Practice★★☆
Two students from different secondary schools want to compare their results in an English exam. As it was not the same exam, they use the standardised score $Z$ (which…

Two students from different secondary schools want to compare their results in an English exam. As it was not the same exam, they use the standardised score $Z$ (which follows a $N(0,1)$) to see who is in a better position relative to their group.

  • Student A has a standardised score of $z_A = 1,2$.

  • Student B has a standardised score of $z_B = 0,8$.

Calculate what percentage of students each of them has below them in their respective school, and interpret who has achieved a more outstanding result.

Solve
Practice★★☆
In advanced statistical studies, symmetrical intervals about the origin enclosing a specified probability are often sought. Find the positive value of $k$ such that: a)…

In advanced statistical studies, symmetrical intervals about the origin enclosing a specified probability are often sought. Find the positive value of $k$ such that:

  1. a) $P(-k \leq Z \leq k) = 0,90$
  2. b) $P(-k < Z < k) = 0,95$
  3. 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.

Solve
Practice★★☆
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:

  1. a) $P(0,45 \leq Z \leq 1,82)$
  2. b) $P(-1,05 < Z < 2,15)$
  3. c) $P(-2,30 \leq Z \leq -0,90)$

Explain using a diagram why in case c) the result must be equal to the probability of the corresponding positive interval.

Solve
Practice★★☆
Using the symmetry properties of the standard normal distribution $N(0,1)$, determine the following probabilities for negative values of the variable: a)…

Using the symmetry properties of the standard normal distribution $N(0,1)$, determine the following probabilities for negative values of the variable:

  1. a) $P(Z \leq -0,74)$
  2. b) $P(Z < -1,25)$
  3. c) $P(Z \geq -0,58)$
  4. d) $P(Z > -2,03)$
Solve
Practice★★☆
An olive oil cooperative in Jaén bottles jugs of oil whose net volume, in litres, follows a normal distribution $X \sim N(\mu, \sigma)$. Quality control classifies as…

An olive oil cooperative in Jaén bottles jugs of oil whose net volume, in litres, follows a normal distribution $X \sim N(\mu, \sigma)$. Quality control classifies as "anomalous" those jugs containing less than $4{,}92\text{ l}$ or more than $5{,}14\text{ l}$. Historical records show that $2{,}5\,\%$ of the jugs contain less than $4{,}92\text{ l}$ and that $16\,\%$ contain more than $5{,}14\text{ l}$.

Data from the $N(0, 1)$ distribution: if $P(Z \le z_1) = 0{,}025$, then $z_1 = -1{,}96$; if $P(Z > z_2) = 0{,}16$, then $z_2 = 1{,}00$ (since $P(Z \le 1{,}00) \approx 0{,}84$).

  1. a) Set up the system of two linear equations relating $\mu$ and $\sigma$ by standardising the given conditions.
  2. b) Solve the system to calculate the mean volume $\mu$ and the standard deviation $\sigma$ of the bottling process.
  3. c) Justify conceptually why the mean $\mu$ obtained does not coincide with the midpoint of the cut-off values ($5{,}03\text{ l}$), relating this fact to the asymmetry in the rejection percentages of both tails.
Solve
Practice★★☆
Based on statistics from the Instituto Nacional de Estadística (INE) on wages in Spain, the net monthly salary of recent graduates hired in the technology sector is…

Based on statistics from the Instituto Nacional de Estadística (INE) on wages in Spain, the net monthly salary of recent graduates hired in the technology sector is modelled using a normal variable $X \sim N(1500, 250)$ euros.

A professional association wishes to classify these workers into three categories for its annual report:

  1. "Entry-level support salary": the $10\,\%$ of professionals with the lowest compensation.

  2. "Standard benchmark salary": those who earn between $1300$ and $1800$ euros.

  3. "Outstanding compensation": those who exceed the $95$th percentile.

Data from the $N(0, 1)$ table: $P(Z \le 0{,}80) = 0{,}7881$; $P(Z \le 1{,}20) = 0{,}8849$; $P(Z \le 1{,}28) = 0{,}8997 \approx 0{,}90$; $P(Z \le 1{,}645) = 0{,}9500$.

  1. a) Model condition 1 using a probability inequality and determine the maximum salary to belong to the "support" group.
  2. b) Calculate the percentage of graduates who fall in the "standard benchmark salary" bracket (between $1300$ and $1800$ euros).
  3. c) Find the minimum compensation threshold from which a professional is included in the "outstanding compensation" group.
Solve
Practice★★★
At a prestigious music academy, it has been observed that the scores obtained by applicants in the sight-singing test follow a normal distribution $N(\mu, \sigma)$. It…

At a prestigious music academy, it has been observed that the scores obtained by applicants in the sight-singing test follow a normal distribution $N(\mu, \sigma)$. It is known that $15\%$ of applicants obtained a score above $82$ points, whilst only $10\%$ of students failed to achieve the minimum score of $45$ points.

Calculate the value of the mean ($\mu$) and the standard deviation ($\sigma$) of this score distribution.

σ 45 μ 82 10% 15%
Solve
Practice★★★
The reaction time of players in a professional e-sports tournament follows a normal distribution $N(180, 25)$, where time is measured in milliseconds. The tournament…

The reaction time of players in a professional e-sports tournament follows a normal distribution $N(180, 25)$, where time is measured in milliseconds. The tournament organisers want to classify players into three categories according to their speed:

  • Elite Category: the fastest $5\%$ of players.

  • Amateur Category: the $75\%$ of players with intermediate times.

  • Beginner Category: the remaining players (the slowest ones).

Calculate the reaction times that mark the boundaries of each of these categories.

t₁ μ = 180 ms t₂ Elite 5% Amateur 75% Beginner 20% N(180, 25)
Solve
Practice★★★
A factory produces packets of coffee whose theoretical weight is $500$ grams. Due to the machinery, the actual weight follows a normal distribution with a standard…

A factory produces packets of coffee whose theoretical weight is $500$ grams. Due to the machinery, the actual weight follows a normal distribution with a standard deviation of $12$ grams. Under consumer regulations, any packet weighing less than $485$ grams or more than $525$ grams is considered "out of specification".

Calculate the mean weight ($\mu$) to which the filling machine must be set so that the probability of a packet being "out of specification" due to overweight (more than $525$ g) is exactly $2,5\%$. With this new mean, what percentage of packets will weigh less than the legal minimum of $485$ g?

Solve
Practice★★★
The volume of soft drink poured by a vending machine into $330$ ml cups follows a normal distribution $N(\mu, \sigma)$. It has been verified that $95\%$ of the cups…

The volume of soft drink poured by a vending machine into $330$ ml cups follows a normal distribution $N(\mu, \sigma)$. It has been verified that $95\%$ of the cups contain an amount of soft drink between $318$ ml and $338$ ml.

Determine the value of the mean ($\mu$) and the standard deviation ($\sigma$) of the volume of soft drink poured by the machine, assuming that the given interval is symmetrical about the mean.

μ 318 338 95% d d
Solve
Practice★★★
It is known that the weight of newborn babies in a region follows a normal distribution with a mean of $3,25$ kg and a standard deviation of $0,6$ kg. A private clinic…

It is known that the weight of newborn babies in a region follows a normal distribution with a mean of $3,25$ kg and a standard deviation of $0,6$ kg. A private clinic considers a newborn baby to be of "low weight" if their weight is less than $2,5$ kg.

  1. a) If $150$ babies are born in that region in a month, what is the probability that more than $35$ of them are of low weight? (Use the normal approximation).
  2. b) What should be the maximum weight so that only the heaviest $5\%$ of babies in the region are placed in a "special monitoring" category?
Solve
Practice★★★
Let $Z$ be a random variable following a standard normal distribution $N(0, 1)$. Calculate the value of the positive constant $a$ such that the following equality holds:…

Let $Z$ be a random variable following a standard normal distribution $N(0, 1)$. Calculate the value of the positive constant $a$ such that the following equality holds:

$$P(|Z| \le a) = 3 \cdot P(Z > a)$$

Justify your answer graphically using the symmetry properties of the Gaussian bell curve.

0 -a a 3A A A N(0, 1) P(Z < -a) = A P(|Z| ≤ a) = 3A P(Z > a) = A
Solve
Practice★★★
It is known that the time it takes high school students to commute to school from home follows a normal distribution $N(\mu, \sigma)$. After conducting a survey, it is…

It is known that the time it takes high school students to commute to school from home follows a normal distribution $N(\mu, \sigma)$. After conducting a survey, it is observed that 15% of students take less than 25 minutes to arrive, while only 20% need more than 55 minutes.

Normal distribution of school commute time with known areas and target interval15 %?20 %25μ553040μX ~ N(μ, σ)X (min)

Calculate the value of the mean $\mu$ and the standard deviation $\sigma$ of this distribution. Then, determine the probability that a randomly chosen student takes between 30 and 40 minutes to arrive at school.

Solve
Practice★★★
Prove algebraically that for any variable $Z \sim N(0, 1)$ and any real value $k > 0$, $P(-k \le Z \le k) = 2 \cdot \Phi(k) - 1$ holds. Subsequently, use this formula to…

Prove algebraically that for any variable $Z \sim N(0, 1)$ and any real value $k > 0$, $P(-k \le Z \le k) = 2 \cdot \Phi(k) - 1$ holds.

Subsequently, use this formula to find the probability that $Z$ lies at a distance from the mean greater than $1,645$ standard deviations.

Solve
Practice★★★
A study has been conducted on the reaction time of teenagers to a visual stimulus. After standardising the data to a $N(0, 1)$, it is observed that $15,87\%$ of subjects…

A study has been conducted on the reaction time of teenagers to a visual stimulus. After standardising the data to a $N(0, 1)$, it is observed that $15,87\%$ of subjects have a very slow reaction time (high values of $Z$).

If we select two teenagers at random, what is the probability that both have a value of $Z$ between $-1$ and $1$? Assume that both reaction times are independent.

Solve
Practice★★★
Calculate with clear reasoning, using the properties of the standard normal distribution $Z \sim N(0, 1)$, the value of the following probabilities without using the…

Calculate with clear reasoning, using the properties of the standard normal distribution $Z \sim N(0, 1)$, the value of the following probabilities without using the table directly for negative values (use the table for positive values only):

  1. a) $P(|Z| \ge 1,25)$
  2. b) $P(-0,5 \le Z \le 0,5)$
  3. c) $P(Z \le -1,96 \cup Z \ge 1,96)$
Solve
Practice★★★
An automotive parts factory produces pieces whose length follows a distribution $X \sim N(200, 5)$ in millimetres. To calibrate the machinery, a tolerance interval…

An automotive parts factory produces pieces whose length follows a distribution $X \sim N(200, 5)$ in millimetres. To calibrate the machinery, a tolerance interval centred at the mean, $[200 - k,\, 200 + k]$, must be established containing exactly $90\,\%$ of the produced pieces.

Two quality technicians propose different approaches to calculate $k$:

  • Technician Alba: "Since the $N(0, 1)$ table gives the cumulative probability, I look for the value $z$ such that $P(Z \le z) = 0{,}90$, which gives $z = 1{,}28$. Thus, $k = 1{,}28 \cdot 5 = 6{,}4\text{ mm}".

  • Technician Bruno: "Since the interval is centred and leaves out $10\,\%$, it corresponds to $5\,\%$ in each outer tail. To cover the central interval, the left tail up to the upper limit accumulates $0{,}90 + 0{,}05 = 0{,}95$. Therefore, I look for $P(Z \le z) = 0{,}95$, which gives $z = 1{,}645$, resulting in $k = 1{,}645 \cdot 5 = 8{,}225\text{ mm}".

Data from the $N(0, 1)$ table: $P(Z \le 1{,}28) = 0{,}8997$; $P(Z \le 1{,}645) = 0{,}9500$.

  1. a) Analyse both methods rigorously: show which one contains an error in geometric formulation, indicate what percentage of pieces the erroneous method actually keeps inside, and justify the validity of the correct method.
  2. b) Propose a third formal procedure based on the symmetry formula $P(-z_0 \le Z \le z_0) = 2\cdot P(Z \le z_0) - 1 = 0{,}90$ and check whether it confirms the correct solution.
  3. c) Determine the final tolerance interval and reason which of the two valid methods is preferable and safer when working with general confidence levels in Social Sciences.
Solve
Practice★★★
In the university entrance examinations (EBAU) of a certain academic year, the marks of the general phase in two autonomous communities follow normal distributions with…

In the university entrance examinations (EBAU) of a certain academic year, the marks of the general phase in two autonomous communities follow normal distributions with the following parameters:

  • Community A: $X_A \sim N(6{,}8;\, 1{,}2)$.

  • Community B: $X_B \sim N(7{,}2;\, 0{,}8)$.

A degree programme with highly coveted places reserves preferential admission for the top $10\,\%$ of students with the highest marks from each community.

Data from the $N(0, 1)$ distribution:

$P(Z \le 1{,}25) = 0{,}8944$; $P(Z \le 1{,}28) = 0{,}8997 \approx 0{,}90$; $P(Z \le 1{,}83) = 0{,}9664$; $P(Z \le 2{,}25) = 0{,}9878$.

  1. a) Model the preferential admission condition and determine the minimum cut-off mark required in each autonomous community.
  2. b) Two students, Lucía (from Community A) and Mateo (from Community B), both obtain a mark of $8{,}2$. Standardise both marks and interpret who is in a better relative position within their respective community.
  3. c) A newspaper headline states: "Since the average mark in Community B is higher ($7{,}2$ compared to $6{,}8$), it is more likely to obtain an excellent mark ($9$ points or more) in Community B than in Community A". Refute this statement mathematically by calculating and comparing $P(X_A \ge 9)$ and $P(X_B \ge 9)$, and reason how dispersion (standard deviation) can be more decisive than the mean in the probability of extreme values.
Solve