Respuesta :
Answer:
The correct options are;
1) The mean (average) score for each class is the SAME
2) The variance of Class B is 159.14
3) Class B has a greater spread
4) Based on the data A score of 8 is more likely in class B than in class A
Step-by-step explanation:
The given data are;
For the 1st period (Class A), we have
The mean, μ = (10 + 15 + 20 + 25 + 30 + 35 + 40)/7 = 175/7 = 25
The standard deviation σ, is given as follows;
[tex]\sigma =\sqrt{\dfrac{\sum \left (x_i-\mu \right )^{2} }{N}}[/tex]
Where;
N = The number population size = 7
[tex]x_i[/tex] = The individual data values
Solving using Microsoft Excel formula STDEV.P gives, σ = 10
[tex]\sigma _A[/tex] ≈ 10
The variance = σ² = 10² = 100
For the 2nd period (Class B), we have
The mean, μ = (7 + 12 + 17 + 25 + 33 + 38 + 43)/7 = 175/7 = 25
The standard deviation σ, is given as follows;
[tex]\sigma =\sqrt{\dfrac{\sum \left (x_i-\mu \right )^{2} }{N}}[/tex]
Where;
N = The number population size = 7
[tex]x_i[/tex] = The individual data values
Solving using Microsoft Excel formula STDEV.P gives, [tex]\sigma _B[/tex] ≈ 12.615
[tex]\sigma _B[/tex] ≈ 12.615 > [tex]\sigma _A[/tex] ≈ 10, class B has greater spread
The variance = σ² = 12.615² ≈ 159.14
A score of 8 is more likely in class B than in class A due to the availability of data before and after the a score of 8
