∑ SOLUTIONS MATHEMATICS – 3 semester, Test 1 length: 2 hours – coefficient 1/2 IUT of Saint-Etienne – Sales and Marketing department Mr Ferraris Prom 2018-2020 10/2019

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2018-2020 – S2 – Mathematics – TEST 1 SOLUTIONS – page 1 / 3

IUT of Saint-Etienne – Sales and Marketing department

Mr Ferraris Prom 2018-2020 10/2019

MATHEMATICS – 3

^rd

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

SOLUTIONS

Exercise 1: (4.5 points)

In a bag are placed 5 white balls and 10 black ones. You have to draw three of the fifteen, simultaneously and at random. If you get three white balls, you win € 100; if you get only two white ones, you win € 10; only one white makes you win or lose nothing; with three black balls, you lose € 20.

1) Let Y be the random variable of the number of white balls that you will get among the three drawn balls.

Explain the type of this variable’s probability distribution, then give its parameters. 1 pt Each individual ball can be white (success) or black (failure). The three balls are to be drawn

simultaneously (thus without putting back). In the end, Y is the number of successes, among the three.

The probability distribution of Y is then hypergeometric.

n = 3, N = 15, a = 5. Then : ^Y^→^H

(

^{3 5 15}^, ^,

)

^.

2) We name X the random variable « gain in € after one game ».

a. By building a table, give the probability distribution of X (show the detailed calculation of at least one

of the probabilities). 1 pt

( ) ( )

0 3 1 2

5 10 5 10

3 3

15 15

2 1 3 0

5 10 5 10

3 3

15 15

C C C C

p 0 0.26374 ; p 1 0.49451

C C

C C C C

p 2 0.21978 ; p 3 0.021978

C C

Y Y

× ×

= = ≈ = = ≈

× ×

= = ≈ = = ≈

Probability distribution of the gain X:

gains xi 100 10 0 -20

prob. pi 0.021978 0.21978 0.49451 0.26374

b. Give the expected value and the standard deviation of X. Comment them, quickly. 1 pt The gain’s expected value is:

( )

⁴

1

E _i _i 0.87912 €

i

X p x

=

∑

≈ − ^.

You will lose on average 0.87912 € per game if you play a lot of games.

You mustn’t play this game a lot!

The standard deviation is ^σ

( )

^X ^≈^{18 614 €}^. . This is the average variation of the gain, from one game to the following.

c. Considering 10,000 games to be played, what global gain is more likely than any other one? Also, give a confidence interval of the global gain, thanks to the knowledge of X’s standard deviation 1.5 pt In 10,000 games, the real global gain is around 10000× −0.87912 €≈ −8 791 €, .

It’s quite likely (68.3%) that the real average gain would belong to the interval

[ ]

0.87912 ; 0.87912 1.0652 ; 0.69298

10000 10000

σ σ

 

− − − + = − −

 

  , and the that the real global gain

(for this game’s organizer) would be between € 6,930 and € 10,652.

Exercise 2: (4 points)

A telecommunications agency performs a test on the connection status of 20 lines. A line has a 1 in 10 chance of being disconnected, and therefore 9 in 10 chances of not having a problem. The random variable X gives the number of problems encountered, out of the 20 lines tested.

1) Give, and justify, the probability distribution of X. 1 pt

Each tested line leads to two ends: problem (p = 0.1) (success) or no problem (q = 0.9).

Both probabilities are invariable on any tested line: p = 0.1.

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2018-2020 – S2 – Mathematics – TEST 1 SOLUTIONS – page 2 / 3 X, number of problems in 20 tests, is then distributed by B(20 ; 0.1).

2) a. What is the most likely number of problems after 20 tests? Give its probability. 0.5 pt The most likely number of problems is the closest integer to X’s expectation: ^E

( )

^X ⁼^np⁼²^.

( )

p X = ≈2 0.2852.

b. What is the probability that less than two problems would be encountered? 0.5 pt

( )

p X ≤ ≈1 0.3917.

3) 8 groups of tests, 20 lines each, have to be performed. In this question, we name « success » the fact that one group of tests leads to less than two problems. The random variable Y is the number of successes among the 8 groups.

a. What is the most likely value of Y? 0.5 pt

Each group can lead to two opposite outcomes: less than two problems (p = 0.3917) (success), or at least two problems (q = 0.6083). These probabilities are invariable (don’t depend on the group).

The law of Y, number of successes in 8 groups, is then B(8 ; 0.3917).

The most likely value of Y is the closest integer to its expectation: ^E

( )

^Y ⁼^np⁼^3.134. The most likely is that 3 successes would be encountered after 8 tested groups.

b. What is the probability that at least one success would be obtained? 0.5 pt

( )

p Y≥ ≈1 0.9813.

c. Instead of 8, how many groups would have to be tested so that the probability of having at least one

success exceeds 99.9%? 1 pt

( )

^.

( )

13 p 1 0 99844 ; 14 p 1 0.99905 ; at least 14 groups are needed.

n= ⇒ Y ≥ = n= ⇒ Y≥ =

Exercise 3: (3.5 points)

In this exercise, one can either use the calculator directly (sometimes with the support of a transformation formula) or use the standard normal distribution table (most often with the use of transformation formulas and the variable change formula). Below, we have chosen this second option.

1) The random variable U is distributed by the standard normal law. Determine:

a. p(U < 2.28) ; b. p(U > 1.44) ; c. p(–1.5 < U < 1.5) 1 pt a. ^p

(

^U ^<^2.28

)

^≈^0.9887

b. ^p

(

^U^>^1.44

)

^{= −}^{1 p}

(

^U ^<^1.44

)

^{≈ −}^{1 0.9251}^≈^0.0749

c. ^p

(

⁻^1.5^{< <}^U ^1.5

) (

⁼^p ^U ^<^1.5

) (

⁻^p ^U ^{< −}^1.5

) (

⁼^p ^U^<^1.5

)

^{− −}

(

^{1 p}

(

^U^<^1.5

) )

^≈^0.9332^{− −}

(

^{1 0.9332}

)

^≈^0.8664

d. the value u0 such that p(U > u0) = 0.8 1 pt

We can read on the table: ^p

(

^U^<^{0 84}^.

)

^≈^0.8^{. So,}^p

(

^U^{> −}^0.84

)

^≈^0.8^andu0= −0.84.

2) The random variable X is distributed by the normal law with a mean of 126 and a standard deviation that values 6. Determine:

a. p(X > 135) 0.5 pt

(

^X ^>

)

^{≈ −} ⁼

p 135 1 0.9332 0.0668

b. the value x0 such that p(X > x0) = 0.01 1 pt

We can read on the table: ^p

(

^U^<^2.33

)

^≈^0.99^{. So,}^p

(

^U^>^2.33

)

^≈^0.01^.

The variable change formula helps us to calculate values of X:

0 126 2.33 6 139.98 140 X mean

U X mean U st deviation x

st deviation

= − ⇔ = + × ⇔ ≈ + × = ≈ .

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2018-2020 – S2 – Mathematics – TEST 1 SOLUTIONS – page 3 / 3 Exercise 4: (4.5 points)

An internal study conducted in a large company revealed that 75% of employees are willing to work overtime.

A new agency is about to be opened, in which 80 employees of the company will have to work. Of these 80 employees, we are interested in the number of those who will agree to work overtime, and we note X the random variable associated with this number.

1) a. Show that X is distributed by a binomial law and give it’s parameters. 0.5 pt An employee will accept overtime (success, p = 0.75) or not (failure, q = 0.25). These probabilities are proportions observed in a large firm, whose size is considered very large compared to 80 (more than twenty times 80): p and q are considered constant

In 80 employees, X is the number of those who will accept overtime.

The probability distribution of X is then B(80 ; 0.75).

b. Show that this law can be approximated by a normal law and give its parameters. 0.5 pt n = 80 > 30 ; np = 60 > 5 ; nq = 20 > 5. X can be approximated by N(60 ; 3.873).

2) a. Calculate with this normal law the probability that less than 55 people will accept overtime. 1 pt

( )

p X <54.5 ≈0.07779

c. Similarly, calculate the probability that this number of people will be between 55 and 65. 1 pt

( )

p 54.5< <X 65.5 ≈0.8444

3) What is the minimum number of employees who would be likely to accept overtime, which can be given

with a 99% confidence level? 1,5 pt

We are looking for the number x0 such that p(X ≥ x0) ≥ 0.99.

We can read on the table that ^p

(

^U ^<^2.33

)

^≈^0.99. Therefore, ^p

(

^U^{> −}^2.33

)

^≈^0.99^.

The variable change formula allows us to go back to X:

0 60 2.33 3.873 50.97 51 X mean

U X mean U st deviation x

st deviation

= − ⇔ = + × ⇔ ≈ − × ≈ ≈ .

There is a 99% chance that at least 51 out of 80 people will agree to work these hours.

Exercise 5: (3.5 points)

A device in a production line fills vials (fr : flacons), which are sold as containing 100 mg of product. But the machine is not perfect: the real quantity X introduced into a vial is a random variable of normal distribution N(m , σ = 1.1 mg), m being adjustable by the operator.

1) The machine is set to m = 101.2 mg. Out of 1000 vials, how many will actually contain less than 100 mg of

product? 1.5 pt

With the law N(101.2 ; 1.1): ^p

(

^X ^<¹⁰⁰

)

^≈^0.1377, or 138 vials out of 1000!

2) On what mean should the machine be set so that less than 1% of the vials contain less than 100 mg of

product? 2 pts

We are looking for the mean µ such that p(X < 100) < 0.01 with the law N( µ^{; 1.1).}

We can read on the table that ^p

(

^U ^<^2.33

)

^≈^0.99^{. So,}^p

(

^U ^{< −}^2.33

)

^≈^0.01^.

The variable change formula allows us to find out this mean:

100 2.33 1.1 102.563 X mean

U mean X U st deviation

st deviation⁻ µ

= ⇔ = − × ⇔ ≈ + × ≈ .

The machine must be set to an average quantity of 102.56 mg.