IUT of Saint-Etienne – Sales and Marketing department Mr Ferraris Prom 2017-2019 08/11/2018 MATHEMATICS – 3

2017-2019 – S3 – Mathematics – TEST 1 SOLUTIONS Page 1 / 3

IUT of Saint-Etienne – Sales and Marketing department

Mr Ferraris Prom 2017-2019 08/11/2018

MATHEMATICS – 3

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

SOLUTIONS

Exercise 1 (4 points)

In a group of 20 students, 14 had at least ten points in mathematics ("T" students).

You have to choose 10 students at random from this group. The random variable X is the number of "T"

students among these 10.

1) What is your expectation, in terms of "T" students among the 10 chosen? 1 pt 7 "T" students are expected. (population: 14 in 20, hence 7 in 10 for our sample)

2) Give, after having justified, the probability distribution of X. 1.5 pt The population of 20 students is divided into successes ("T") and failures (the others);

a draw of 10 students, without putting back, has to be performed;

the random variable X gives the number of successes among the 10.

X is then distributed by

H

3) What is the probability to get 7 "T" students among the 10? 1.5 pt

( )

C C

p 7 0.3715

X = = C× ≈

Exercise 2 (5.5 points)

A company manufactures bolts (fr.: boulons). At the production line's exit, a sample is taken at random and verified, showing that 1.5 % of the production is defective.

1) Bolts are packed in boxes: 20 bolts in each box. Given that the whole produced quantity is very big compared to a 20-bolt sample, we are allowed to consider the box as made by a draw with putting back. Let be X the random variable "number of defective bolts in a box".

a. What is the probability distribution of X? Justify. 1 pt

The whole bolts production is divided into defective ones (success) and others;

to fill a box, 20 of them have to be chosen without repetition, but we are allowed to consider repetition (with a very low distortion of the following results): the probability of success (1.5%) is considered as invariable;

the random variable X gives the number of successes among the 20.

X is then distributed by

B

b. A box can be sent back by a client, and will have to be reimbursed, if it contains at least two defective

bolts. What is the probability this situation would happen? 2 pts

( ) ( )

p X ≥ = −2 1 p X ≤ = −1 1 0,9643≈0.0357

c. In a total number of 1000 sold boxes, how many could the seller expect to reimburse? 0.5 pt 0.0357 1000× ≈36 boxes

2) According to these former results, let’s think about the number of boxes that would have to be reimbursed, in a batch containing 100 boxes. This number is variable and will be denoted Y. Let's fix its expectation at 4: on average, 4 boxes have to be reimbursed every 100 sold ones.

2017-2019 – S3 – Mathematics – TEST 1 SOLUTIONS Page 2 / 3 a. Justify that Y can be considered as distributed by a Poisson's law, which parameter is then 4. 1 pt The whole production of boxes is divided into "to be reimbursed" (success) and others;

a draw of 100 boxes is made at random, without putting back, but the big size of the production allows us to consider a draw with putting back (with a very low distortion of the following results): the probability of success (4%) is considered as invariable;

the random variable Y gives the number of successes among the 100.

X is then distributed by

B

As n > 30, p < 0.1 and np = 4 < 10, we are indeed allowed to use

P

b. The situation will be considered as problematic for the company as soon as at least 8 boxes have to be reimbursed. What is the probability that this event would occur? 1 pt

( ) ( )

p Y≥8 = −1 p Y≤ = −7 1 0.9489≈0.0511

Exercise 3 (7.5 points)

A company that sells office items decides to send its catalogue to 600 societies. On average, considering all the French societies (much more than 20 times 600), 10% of them place an order after having received such a catalogue. The variable X describes the number of societies that will place an order, among the 600.

1) By creating and analysing the probability distribution of X, show that it can be described by this normal one:

N

The whole set of French society is divided in successes ("will place an order") and failures (other);

a draw of 600 societies is made at random, without putting back, but the big size of the population allows us to consider a draw with putting back (with a very low distortion of the following results): the probability of success (10%) is considered as invariable;

the random variable X gives the number of successes among the 600.

X is then distributed by

B

As n > 30 and npq = 54 > 5, we are indeed allowed to use

N

2) Thanks to this normal distribution, determine:

a. the probability that more than 70 societies place an order. 1.5 pt

( )

p X >70.5 ≈0.07652

b. the probability that at least 55 societies place an order. 1.5 pt

( )

p X >54.5 ≈0.7729

3) Instead of 600, how many catalogues would have to be sent, at least, so that the probability of the event in

question 2b would reach 95%? 2.5 pts

Here, X is distributed by

B

As n > 600 > 30 and npq > 54 > 5, we can use the law

N

Let's use the "function" mode of the calculator, where its variable X will contain the possible values for n.

Let's then test some values of X, from 600 to 700 for instance, and define a function f as a calculation of

( )

p X >54.5 with the "normal law" tool. Finally, we get the following results:

2017-2019 – S3 – Mathematics – TEST 1 SOLUTIONS Page 3 / 3 n ^p

(

)

670 0.94627091 671 0.9475332 672 0.94877011 673 0.949982 674 0.95116927 675 0.95233229 676 0.95347143 677 0.95458708 678 0.95567959

We can see that our probability exceeds 95% as soon as n reaches (and is higher than) 674.

Exercise 4 (3 points)

Only using the form below (standard normal law table), determine with

N

a. p(U < 1.64). 1 pt

( )

p U <1.64 ≈0.9495

b. p(U < – 0.77). 1 pt

( ) ( )

p U < −0.77 = −1 p U<0.77 ≈ −1 0.7794≈0.2206

c. p(-2 < U < 2). 1 pt

( ) ( ) ( ) ( ) ( ( ) ) ^{( )}

p − < <2 U 2 =p U< −2 p U< − =2 p U< − −2 1 p U<2 ≈0.9772− −1 0.9772 ≈0.9544

____________________ TEST END ____________________