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Question

ScanSolve AI Tutor · Accepted Answer

Here are the solutions to the exercise:

1) Calculer la moyenne des notes des élèves de cette classe à cette évaluation.
To calculate the mean (average) grade, we sum the product of each grade and its corresponding number of students, then divide by the total number of students.

Calculate the total number of students.
$N = 7 + 4 + 6 + 2 + 4 = 23$
Calculate the sum of (grade $\times$ number of students).
$\sum (\text{note} \times \text{nombre d'élèves}) = (8 \times 7) + (9 \times 4) + (10 \times 6) + (12 \times 2) + (15 \times 4)$ $= 56 + 36 + 60 + 24 + 60$ $= 236$
Calculate the mean grade.
$\text{Moyenne} = \frac{\sum (\text{note} \times \text{nombre d'élèves})}{N} = \frac{236}{23} \approx 10.26$
The mean grade is $\boxed{\frac{236}{23} \approx 10.26}$ .

2.a) Recopier et compléter ce tableau par ligne des effectifs cumulés croissants et par celle des effectifs cumulés décroissants.
Here is the completed table:

Note sur 20 8 9 10 12 15
Nombre d'élèves 7 4 6 2 4
Effectifs cumulés croissants (ECC) 7 11 17 19 23
Effectifs cumulés décroissants (ECD) 23 16 12 6 4
- Effectifs cumulés croissants (ECC):
  
  For note 8: 7
  
  For note 9: $7 + 4 = 11$
  
  For note 10: $11 + 6 = 17$
  
  For note 12: $17 + 2 = 19$
  
  For note 15: $19 + 4 = 23$
- Effectifs cumulés décroissants (ECD):
  
  For note 15: 4
  
  For note 12: $4 + 2 = 6$
  
  For note 10: $6 + 6 = 12$
  
  For note 9: $12 + 4 = 16$
  
  For note 8: $16 + 7 = 23$
b) En déduire la note médiane.
The total number of students is $N = 23$ . Since $N$ is odd, the median is the value of the $\frac{N+1}{2}$ -th observation.
Calculate the position of the median.
$\text{Position de la médiane} = \frac{23+1}{2} = \frac{24}{2} = 12$
The median is the 12th grade when the grades are ordered.
Find the grade corresponding to the 12th position using the cumulative increasing frequencies.
- 7 students have a grade of 8 or less.
- 11 students have a grade of 9 or less.
- 17 students have a grade of 10 or less. The 12th student falls into the category of students who scored 10. The median grade is $\boxed{10}$ .
3) On choisit au hasard et simultanément 5 élèves de la classe.

a) Déterminer le nombre de choix possibles.
This is a combination problem, as the order of selection does not matter. We are choosing 5 students from a total of 23 students.
Use the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ .
$\binom{23}{5} = \frac{23!}{5!(23-5)!} = \frac{23!}{5!18!}$ $= \frac{23 \times 22 \times 21 \times 20 \times 19}{5 \times 4 \times 3 \times 2 \times 1}$ $= \frac{23 \times 22 \times 21 \times 20 \times 19}{120}$ $= 23 \times 11 \times 7 \times 1 \times 19$ $= 33649$
The number of possible choices is $\boxed{33649}$ .

b) Déterminer le nombre de choix possibles contenant exactement trois élèves ayant au moins 10 à cette évaluation.
First, we need to identify the number of students who scored at least 10 and the number of students who scored less than 10.
Count students with a grade of at least 10.
Grades $\ge 10$ : 10, 12, 15
Number of students: $6 + 2 + 4 = 12$ students.
Count students with a grade less than 10.
Grades $< 10$ : 8, 9
Number of students: $7 + 4 = 11$ students.
(Total students: $12 + 11 = 23$ , which is correct).
Calculate the number of ways to choose exactly 3 students from the 12 students with a grade of at least 10.
$\binom{12}{3} = \frac{12!}{3!(12-3)!} = \frac{12 \times 11 \times 10}{3 \times 2 \times 1} = 2 \times 11 \times 10 = 220$
Calculate the number of ways to choose the remaining $5-3=2$ students from the 11 students with a grade less than 10.
$\binom{11}{2} = \frac{11!}{2!(11-2)!} = \frac{11 \times 10}{2 \times 1} = 11 \times 5 = 55$
Multiply the results from Step 3 and Step 4 to find the total number of choices.
$\text{Nombre de choix} = \binom{12}{3} \times \binom{11}{2} = 220 \times 55 = 12100$
The number of possible choices is $\boxed{12100}$ .

Hey Marc-king, good to see you again.

Step-by-step solution

Calculate the total number of students.

Calculate the sum of (grade $\times$ number of students).

Calculate the mean grade.

Calculate the position of the median.

Find the grade corresponding to the 12th position using the cumulative increasing frequencies.

Use the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ .

Count students with a grade of at least 10.

Count students with a grade less than 10.

Calculate the number of ways to choose exactly 3 students from the 12 students with a grade of at least 10.

Calculate the number of ways to choose the remaining $5-3=2$ students from the 11 students with a grade less than 10.

Multiply the results from Step 3 and Step 4 to find the total number of choices.

Still stuck on this one?

More History Questions

Note sur 20	8	9	10	12	15
Nombre d'élèves	7	4	6	2	4
Effectifs cumulés croissants (ECC)	7	11	17	19	23
Effectifs cumulés décroissants (ECD)	23	16	12	6	4

Hey Marc-king, good to see you again.

Step-by-step solution

Calculate the total number of students.

Calculate the sum of (grade ×\times× number of students).

Calculate the mean grade.

Calculate the position of the median.

Find the grade corresponding to the 12th position using the cumulative increasing frequencies.

Use the combination formula (nk)=n!k!(n−k)!\binom{n}{k} = \frac{n!}{k!(n-k)!}(kn​)=k!(n−k)!n!​.

Count students with a grade of at least 10.

Count students with a grade less than 10.

Calculate the number of ways to choose exactly 3 students from the 12 students with a grade of at least 10.

Calculate the number of ways to choose the remaining 5−3=25-3=25−3=2 students from the 11 students with a grade less than 10.

Multiply the results from Step 3 and Step 4 to find the total number of choices.

Still stuck on this one?

More History Questions

Calculate the sum of (grade $\times$ number of students).

Use the combination formula $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ .

Calculate the number of ways to choose the remaining $5-3=2$ students from the 11 students with a grade less than 10.