TSM 1st yearSession P05 Normal distribution

TSM 1st year

Definition 1A random variable X follows a standard normaldistribution N (0;1) ffi X is a continuous randomvariable which probability density function f isdefined for all t ∈R by :

f (t)= 1p2π

e− t22

It will be denoted X vN (0;1).

TSM 1st year

TSM 1st year

Exercise 1

Let X be a random variable following thestandard normal distribution N (0;1). Let’sdenote f the probability density functionassociated to X .On the graph of f below, hatch the areacorresponding to P(−1<X É 2).

TSM 1st year

Property 1

For all real number aÊ 0, we have (because ofthe symmetry in Gauss’ Bell curve) :

1 P(X É−a)=P(X Ê a)2 P(−aÉX É a)= 2P(X É a)−13 P(−1,96<X < 1,96)' 0,95

TSM 1st year

Property 2Let’s consider X vN (0;1). The expected valueof X is :

E(X )= limx→−∞

∫ 0

xtf (t)dt + lim

y→+∞

∫ y

0tf (t)dt

and its variance is :

V (X )=E (X −E(X ))2

We then get E(X )= 0 , and we’ll assume thatV (X )= 1.

TSM 1st year

Exercise 2

Let’s consider X vN (0;1). Compute with yourcalculator :

1 P(0ÉX É 1)2 P(−2ÉX É 0,3)3 P(X Ê 1)4 P(X <−0.5)

TSM 1st year

Theorem 1

Let’s consider X vN (0;1). For all real numberα ∈]0;1[, there existe an unique positive realnumber uα such that :

P(−uα ÉX É uα)= 1−α

TSM 1st year

Property 3

If X vN (0;1), then :u0.05 = 1.96 and u0.01 = 2.58Which means :P(−1.96ÉX É 1.96)= 1−0.05= 0.95 etP(−2.58ÉX É 2.58)= 1−0.01= 0.99

TSM 1st year

Exercise 3

(See property 3). Let’s consider X vN (0;1).1 Give the value of u > 0 such that :

P(−u ÉX É u)= 0.952 Determine the value of v > 0 such that :

P(0ÉX É v)= 0.45. What is the value ofP(X < v)?

3 Find another interval I such thatP(X ∈ I)= 0.95

4 How many intervals I of R are such thatP(X ∈ I)= 0.95 ?

TSM 1st year

Theorem 2

( MOIVRE-LAPLACE)Let’s consider p ∈]0;1[. For all integer n different frpm 0, let’s denote Xn arandom variable following a binomial distribution of parameters n and p. Let’sdenote Zn the random variable defined by :

Zn = Xn −np√np(1−p)

For all real numbers a and b such that a< b, we get :

limn→+∞P(aÉZn É b)=

∫ b

a

1p2π

e− x22 dx

TSM 1st year

In the exercises, we’ll consider that we can usethis approximation if we have simultaneously :

n Ê 30np Ê 5n(1−p)Ê 5

TSM 1st year

Exercise 4

See your calculator’s user manual about normal distributions. Let’s considerX vB(36;0.2) (that’s the notation for a binomial distribution of parametersn = 36 and p = 0.2).

1 For any interger k between 0 and 36, express P(X = k).2 What are the values of E(X ) and σ(X )?3 Compute with a calculator :

1 P(−1É X−7.2

2.4 É 1)

2 P(20<X É 36)3 P(X Ê 7.2)

4 Justify that we can use the approximation : for all real numbers a and b

such that a< b, P(a< X−E(X)

σ(X)< b

)'P(aÉZ É b), where Z ∼N (0;1).

5 Use this approximation to compute :

1 P(−1É X−7.2

2.4 É 1)

2 P(20<X É 36)3 P(X Ê 7.2)

TSM 1st year

Exercise 5

Let’s consider X vB(70;0.7).1 Compute with a calculator the value rounded

to 10−3 of P(45ÉX É 55).2 Justifiy that the conditions for applying

Moivre-Laplace’s theorem are fulfilled.3 Compute the previous probability, using the

approximation given by Moivre-Laplace’stheorem

TSM 1st year

Exercise 6

A company makes sports gear, which can bear two different types ofindependant defects :

D1 : the probability that the fabric bears a defect is 0.02

D2 : the probability of a defect during the sewing is 0.05

1 Compute the probability that an item bears both defects2 Prove that the probability of an item being without defect is 0.931.3 In one week, the company makes 1000 items. Let’s denote X the

random variable equal to the number of items without any defect madein one week.What is the distribution of X ? Compute its expected value and itsstandard deviation.

4 Why can X−9318 be approximated by the standard normal distribution ?

5 Using that approximation, compute : P(927ÉX É 935)6 With the same approximation, for which value of x , rounded to the

nearest integer, do we have P(X É x)= 0.95 ?

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Definition 2

Let’s consider µ ∈R and σ ∈R∗+.

A random variable X follows a normaldistribution of parameters µ and σ2 (generalnormal distribution) if the random variableY = X−µ

σfollows the standard normal distribution

N (0;1).We’ll denote X ∼N (µ;σ2).

µ is the expecetd value (position parameter) andσ the standard deviation (dispersion parameter).

TSM 1st year

Property 4The expected value of X following a normaldistribution of parameters (µ;σ2) is E(X )=µ

TSM 1st year

Probability densities for σ= 1, andµ=−2,−1,0,1,2 :

Each of these curves is symmetric with respectto to the vertical line of equation x =µ.Changing the parameter µ (expected value)therefore "shifts" the curve horizontally.

TSM 1st year

Probability densities for µ= 0 andσ= 0.8;1;1.5;2 :

Changing the parameter σ (standard deviation)therefore changes the magnitude of the "bell".

TSM 1st year

Property 5Let’s consider X ∼N (µ;σ2). We have thefollowing properties :

P(µ−σÉX ɵ+σ)≈ 0.68P(µ−2σÉX ɵ+2σ)≈ 0.95P(µ−3σÉX ɵ+3σ)≈ 0.997

TSM 1st year

Exercise 7

Let’s consider X vN (32;49).1 Compute the expected value, the variance

and the standard deviation.2 Compute (with a calculator) P(30ÉX É 40)3 Compute (with a calculator) P(X < 30)4 Deduce from the previous questions

P(X > 40)5 Check that last result with your calculator

TSM 1st year

Exercise 8

Let’s consider X vN (3;4), and Z following astandard normal distribution.

1 Let’s consider x ∈R. Prove thatP(X É x)=P

(Z É x−3

2

)2 Deduce the value of x for which

P(X É x)= 0.95

TSM 1st year

Exercise 9

Consider X vN (20;4).1 Find x such that P(X É x)= 0.52 Find x such that P(X Ê x)= .,83 Find x > 0 such that P(−x ÉX −20É x)= 0.8

TSM 1st year

exercise 10

We are interested in the requests received by a big company’s web server,from clients scattered all over the internet. If the server receives too manyrequests, it will overload it. We’ll denote X the random variable which valuesrepresent the number of requests received by the server in an interval of 2tenth of seconds.Assuming that X can be approximated by the normal distribution ofparameters µ= 100 and σ= 10, compute :

1 The probability P(X > 120)2 An approximate value of the real number a such that

P(100−aÉX É 100+a)= 0.99

TSM 1st year