2019-2021 – S3 – Mathematics – TEST 2 « partiel » SOLUTIONS page 1 / 3
IUT of Saint-Etienne – Sales and Marketin department
Mr Ferraris Prom 2019-2021 09/12/2020
MATHEMATIC – 3rd Semester, TEST 2 length: 2 h – coefficient 1/2
SOLUTIONS
Exercise 1 : sampling (5 points)
Assu-Fizan, a small insurance company, wishes to establish its provisional budget for the reimbursement of type S damage for the coming year. It has 400 clients (households) to its credit.
The overall data to which it has access is as follows: last year, 0.5% of French households suffered a type S damage, following which they received an average compensation of €2000.
P is the random variable associated with the proportion of Assu-Fizan clients who will suffer a type S damage next year.
1) a. Give the probability distribution of the variable P. 1 pt
The company's client base is a random SRS sample (a household may have more than one S damage in the coming year) of French households, for which the proportion of S damages is π = 0.005. The variable P is then distributed by the law
( )
, 1
n
π − π
π
N
=N (0.005 0.003527, )
.
b. What is the probability that more than 1% of its clients would be to reimburse next year? 1.5 pt p(P > 0.01) = 0.07813
Casio : NormCd(0.01,1000,0.003527,0.005) TI : NormalFRep(0.01,1000,0.005,0.003527)
2) a. What proportion of future clients to reimburse has only 2% risk to be exceeded? 2 pts
* Answer using the
N
(0 ; 1) table:The value of U that has only 2% risk to be exceeded is 2.055.
However, P
U µ
σ
= − , hence the value of P : 0.005 + 2.055×0.003527 = 0.01225.
There is a 98% chance that less than 1.225% of clients will have to be compensated next year.
* Answer using the calculator directly:
Casio : InvNorm(0.98,0.003527,0.005) TI : FracNormale(0.98,0.005,0.003527) p(P < 1.224) = 98%
* Answer using a function:
Casio : Y1=NormCd(X,1000,0.003527,0.005) TI : Y1=NormalFRep(X,1000,0.005,0.003527) Variable X of the function: start = 0.01, step = 0.00001
It can be seen that the probability calculated by the function becomes less than 0.02 when X is at least 0.01225 = 1.225%.
b. If the latter value is considered to be a plausible maximum, what budget should the insurance company
forecast as a maximum compensation amount? 0.5 pt
1.225%×400 = 4.9. A maximum of 5 clients must be reimbursed, i.e. €10,000.
2019-2021 – S3 – Mathematics – TEST 2 « partiel » SOLUTIONS page 2 / 3 Exercise 2 : (6 points)
The Médiamétrie institute's mission is to estimate the television audience of the public in France.
On a given evening, on a sample of 2000 French households, two independent pieces of information are collected: (a): 360 households watched France 2; (b): the 2000 households watched television on average for 2 hours, with a standard deviation of 30 minutes (thus: 0.5 hours).
1) Processing information (a)
a. Determine the 99% confidence interval of the proportion of households who watched France 2 that
evening. 2 pts
The proportion of households in the sample is as follows: p = 360 / 2000 = 0.18.
The 99% confidence interval is then (u = 2.58) :
( ) ( )
[ ]
1 1 0.18 0.82 0.18 0.82
; 0.18 2.58 ; 0.18 2.58
2000 2000
0.158 ; 0.202
p p p p
I p u p u
n n
α
− − × ×
= − + = − × − ×
=
b. What is the probability that less than 15.8% of French households have watched France 2? 0.5 pt 0.5% (probability that π may be below our interval).
2) Processing information (b)
a. Determine the 95% confidence interval of the average time spent by French households in front of the
television that evening. 2 pts
Since the standard deviation of the population is unknown, the following formula will be used:
;
1 1
s s
I x t x t
n n
α
= − − + − .
where t = 1.96 (Student’s table, 1 – p = 0.05, dof : 1999, hence the row « ∞ »).
The 95% confidence interval is then: 2 1.96 0.5 ; 2 1.96 0.5
[
1.978 ; 2.022]
1999 1999
α
= − + =
I
b. For what sample size would an interval of 0.1 amplitude be obtained (the standard deviation of the population is considered to be known and is equal to that of the sample quoted at the beginning of the
statement). 1.5 pt
1.96 0.5
0.05 19.6 384.16
u n 0.05 n
n
σ = ⇔ = × = ⇔ = . A sample of 384 people is required.
2019-2021 – S3 – Mathematics – TEST 2 « partiel » SOLUTIONS page 3 / 3 Exercise 3 : (9 points)
The Salade-Minute factory specialises in packaging salads in bags. Every day, this company receives batches of a large number of salads on its docks. The specifications specify that the batches must be delivered at a
temperature of 3 degrees Celsius. The worker responsible for checking whether a batch is in conformity has taken a simple random sample of 50 salads, measured their temperature and recorded the results in the following table:
temperature (°C) 2.9 3 3.1 3.2 3.3
number of salads 5 19 18 5 3
The questions 1 and 2 are independent.
1) Test, at a 1% significance level, the assumption that the average temperature of the batch is less than or
equal to 3 degrees Celsius. 4 pts
Null hypothesis: H0 : µ = 3 ; alternate hypothesis: H1 : µ > 3 (one-sided test) Statistics:
The population’s standard deviation is unknown; therefore, we will use a Student’s distribution.
The sample’s standard deviation is s = 0.09962
According to H0, X is distributed by
St
(3 ; 0.09962/√49) =St
(3 ; 0.01407).Rejection area:
The tlim value beyond which H0 may be rejected is 2.405 (49 ddl, 1 – p = 0.02).
Comparison and decision:
The tobs value corresponding to the mean, 3.064°C, of the sample is : tobs = 3.064 3
4.547 0.01407− ≈
As tobs > tlim, we can reject, at a 1% significance level, the assertion that the average temperature of the batch is less than or equal to 3°C (in other words: it is stated, with a 1% risk of being wrong, that the average temperature of this batch is higher than 3°C.).
2) a. Apply a Chi-2 test to determine whether the temperature distribution obtained in this sample conforms to
the theoretical frequency distribution below: 4 pts
temperature (°C) 2.9 3 3.1 3.2 3.3
theoretical frequencies 20% 40% 20% 10% 10%
Let us place in the same chart the observations opposite the theoretical values (calculated from the theoretical frequencies for a total of 50), and let us calculate the partial Chi-square values as well as their sum, Chi²calc:
obs 5 19 18 5 3
th 10 20 10 5 5
part chi² 2.5 0.05 6.4 0 0.8
H0 : The observed temperatures are compatible with the proposed theoretical distribution.
Chi²calc = 9.75
In the Chi-square table, with 4 dof, we can notice that 9.75 is between the limit Chi² values of the 5% and 2% significance levels. Thus, the adequacy between the observations and the expected theoretical distribution can be rejected at the 5% significance level, but not at the 2% significance level.
b. What does the notion of significance level mean? 1 pt
The significance level is the maximum acceptable risk to be wrong on rejecting the null hypothesis.
Here, if H0 was true, we would only have had between 2% and 5% chance of obtaining such a sample.
__________ TEST END __________
