B H SOLUTIONS MATHEMATICS – 3 semester, Test 1 length: 2 hours – coefficient 1/2 IUT of Saint-Etienne – Sales and Marketing department Mr Ferraris Prom 2019-2021 12/11/2020

2019-2021 – S3 – Mathematics – TEST 1 Solutions page 1 / 4

IUT of Saint-Etienne – Sales and Marketing department

Mr Ferraris Prom 2019-2021 12/11/2020

MATHEMATICS – 3

semester, Test 1 length: 2 hours – coefficient 1/2

SOLUTIONS

Exercise 1 : (4.5 points)

Among the managers of the 120 shops in a shopping mall, 6 of them said they wanted to open an additional outlet in the town centre. Ms Shaw Ping, who regularly visits the gallery but lives in the city centre, asks every shop she visits whether it would be possible to open an additional outlet near her home. After asking the question in 12 shops, she finds that she still has not received a positive answer.

1) Build the probability distribution of the variable X = « number of yes after 12 trials ». 1.5 pt Each answer can be a « yes » (success) or a « no » (failure). The question was asked to twelve different shop managers (draw without putting back) and N < 20n since 120 < 240. The law that distributes X is then hypergeometric and can’t be correctly modeled by a binomial one.

Correct answer : ^{X →}

H (

)

2) Is it so unlikely that Ms. Shaw Ping did not receive a positive answer? 1.5 pt

( )

C C

p 0 0.5239

X = = C× ≈ . That was rather likely!

3) What was the likelihood that Ms. Shaw Ping had at least two positive answers? 1.5 pt

( ) ( ) ( )

120

C C

p X 2 1 p X 0 p X 1 1 0.5239 1 0.5239 0.3662 0.1099

C

≥ = − = − = = − − × = − − = .

Exercice 2 : (4 points)

In a seaside resort, at the end of the summer, the weather is not very nice. Meteorological services have estimated that, for each day of the coming week, there is a 30% chance of rain the next day. The municipal team, whose municipality is quite rich, has decided to organise activities on days when it would rain: each rainy day will be given a budget of €5,000 for this.

1) Build the probability distribution of the variable X = « number of rainy days int the next seven days ». 1 pt For each day, the next one will be rainy (success) or not (failure). The probability of success is invariable:

p = 30%. X is then distributed by the law

B (

)

2) Give the expectation and the standard deviation of the number of rainy days to be expected for the whole of these seven days. Interpret these parameters by discussing their usefulness for the resort. 1.5 pt Expectation : np = 2.1. Standard deviation = npq = 1.47≈1.212.

On a large number of weeks placed in similar conditions (30% chances that it rains the next day), we can expect an average of 2.1 rainy days per week, and the average variation of the number of weekly rainy days is 1.2 day from one week to an other.

These results are completely useless for the seaside resort, since its situation is only true for one week.

2019-2021 – S3 – Mathematics – TEST 1 Solutions page 2 / 4

3) What is the overall budget that has a maximum 1% chance of being exceeded? 1.5 pt Let's first look for the number of rainy days which has a 99% chance of not being exceeded. For that purpose, the tools BCd (Casio) and binomFRep (TI) will have to be used, because they will give probabilities that X be less than or equal to a given number.

Let’s generate the function Y1=BinomialCD(X,7,0.3) on a Casio or Y1=binomFRep(7,0.3,X) on a TI, for X going from 0 to 7. The results table is:

X Y1

0 0.0823543 1 0.3294172 2 0.6470695

3 0.873964

4 0.9712045 5 0.9962092 6 0.9997813

7 1

We can see that the probability that the week has no more than 4 rainy days is 97.12% and the probability that the week has no more than 5 rainy days is 99.6%. There is therefore less than 1% risk of exceeding 5 rainy days (and more than 1% risk of exceeding 4 rainy days) and the budget to be planned is therefore 5×5 000 € = 25 000 €.

Exercise 3 : (4.5 points)

1) The variable X is distributed by the law

N

a. Explain why we can assess, without calculations, that p(24 < X < 36) = 68.3 %. 0.75 pt The interval [24 ; 36] is the one of the values less than one standard deviation from the mean. In normal law, whatever its parameters, it is an interval with 68.3% confidence.

b. Give, thanks to the calculator: 0.75 pt

b1. p(X < 39) = 0.9332 b2. p(X > 24) = 0.8413 b3. p(23 < X < 28) = 0.2478

c. Determine the value x0 such that p(X > x0) = 5%. 1 pt

* With the form: p(U > u0) = 5% ⇔ p(U < u⁰) = 0.95. Hence u0 = 1.645.

As u0 = ⁰ 30 6 x −

, we get x0 = 30 + 1.645×6 = 39.87.

* With the tool InvNormale (Casio) or FracNormale (TI): we give a probability of 0.95. (indeed, these tools return a value of X when we give them a probability to be less than this value)

* By generating a function:

Casio : Y1=NormCd(X,1000,6,30) SET : Start 39, End 40, Step 0.01 TABL TI : Y1=NormalFRep(X,1000,30,6) 2^nd-Fenêtre : Début 39, Pas 0.01 Table

2) X is a discrete variable distributed by the law

B

N

a. Determine, with both laws, the probability that X would be at least equal to 22. 1.5 pt binomial: 0.3460. Casio : 1-binomialCd(21,100,0.2) TI : 1-binomFRep(100,0.2,21)

normal: 0.3538. Casio : normCd(21.5,1000,4,20) TI : normalFRep(21.5,1000,20,4)

b. Explain the visible difference between both results. 0.5 pt

The normal law is a continuous one (also called « density distribution ») and gives an imperfect account of the probabilities linked to a discrete variable, even if the pass criteria are verified.

2019-2021 – S3 – Mathematics – TEST 1 Solutions page 3 / 4 Exercise 4 : (7 points)

On a can manufacturing line, 3% of the cans do not comply with the manufacturer's specifications. A sample of 200 cans is taken from this production line and the random variable X is called the "number of non- compliant cans in this sample".

1) Among these laws: hypergeometric, binomial and Poisson, say which of them could be used for X (justify).

1.5 pt Each box drawn is non-compliant (success) or compliant (failure). The draw is carried out without putting back, but the total production, unknown, will be considered higher than twenty times 200; therefore, the probability of success (p = 0.03) is almost invariable. Finally, X refers to the random number of hits out of 200 boxes drawn (n = 200).

The law of X is then hypergeometric, but, assuming that the production is large, we are allowed to use a binomial law: ^{X →}

B (

)

The conditions for the pass to a Poisson law are met: n = 200 > 30, p = 0.03 < 0.1, np = 6 < 10.

We could then use

P ( )

2) In this question, we will use a Poisson distribution.

a. Give the probability that all boxes in the sample are compliant. 0.5 pt p(X = 0) = 0.00248 (by comparison, with a binomial law: C₂₀₀⁰ ×0.03⁰×0.97²⁰⁰≈ 0.0023)

b. Give the probability of the most likely value of X. 0.75 pt

The most likely value is the closest integer to the expectation, hence 6 successes.

p(X = 6) = 0.16062 (by comparison, with a binomial law: C⁶₂₀₀×0.03⁶×0.97¹⁹⁴≈ 0.1631)

3) a. Can we approximate the binomial law of X with a normal law? Which one? 0.5 pt n =200 > 30, np = 6 > 5, nq = 194 > 5. So

B (

)

N (

)

b. Calculate, with this normal distribution, the probability to get six non-compliant boxes. 0.5 pt p(X = 6) = p(5.5 < X < 6.5) = 0.1642. The answer is similar to question 2b.

c. Calculate, with this normal distribution, the probability to get more than ten non-compliant boxes. 0,75 pt p(X > 10.5) = 0.03104.

d. Calculate, with this normal distribution, the probability to get more than 9 and less than 13 non-compliant boxes (we will consider an interval, instead of adding several probabilities together). 1 pt p(9.5 < X < 12.5) = 0.06986.

e. What should be the maximum sample size so that the probability of having more than 10 non-compliant

boxes does not exceed 1%? 1.5 pt

First way to look for an answer:

The tools NCd (Casio) and normalFRep (TI) will be useful because they will give us the probability that X exceeds 10.5, for several sample sizes. It is essential to realise that the parameters of the normal distribution depend on these sizes, in order to make the right application:

The law of X is here

N (

) (

N

)

Let’s generate the function Y1=NormCd(10.5,1000,√(X×0.0291),0.03×X) on Casio

or Y1= normalFRep(10.5,1000,0.03×X,√(X×0.0291),) on TI, for X (sample’s size) between 150 and 200 for example. The values table contains:

X Y1

172 0.00849602 173 0.00897615 174 0.0094764 175 0.00999727 176 0.01053925 177 0.01110282

2019-2021 – S3 – Mathematics – TEST 1 Solutions page 4 / 4

It can then be said that with a sample of up to 175 cans, the probability of having more than 10 non- compliant cans does not exceed 1%.

An other way to look for an answer:

According to the form, the value of U that has only 1% chances to be exceeded is 2.33.

Moreover, the law of X is here

N (

) (

N

)

The variable change formula X

U µ

σ

= − is translated in: 10.5 0.03 2.33 0.0291

n n

= − .

This last fraction can be entered into the calculator as a function of n to obtain a table of values of this fraction, which will show us from which value of n it will fall below 2.33. We then obtain: n = 175.

One can also solve the proposed equation:

10.5 0.03

2.33 10.5 0.03 2.33 0.0291 0.3975 0.03 0.3975 10.5 0

0.0291

n n n n n n

n

= − ⇔ − = = ⇔ + − = .

By defining √n = N : 0.03N²+0.3975N 10.5− =0, quadratic equation.

∆ = 1.418 : two real roots for N : –26.47 (impossible because negative) and 13.22.

A possibility for n (= N²) : 13.22² = 174.8 rounded to 175.

__________ TEST END __________