Respuesta :
Answer:
E (X) = 56 and V (X) = 880.
Step-by-step explanation:
The random variable X denotes the test scores obtained from a class with students who fall into two groups.
Let's denote the test scores of first group as X₁ and the test scores of second group as X₂.
The information provided is:
[tex]X_{1}\sim N(80, 20)\\X_{2}\sim N(20, 10)[/tex]
The probability of selecting a student from the first group is, p = 0.60.
Then the probability of selecting a student from the second group is,
q = 1 - p = 1 - 0.60 = 0.40.
(a)
Compute the expected test score obtained as follows:
E (X) = p × E (X₁) + q × E (X₂)
[tex]=(0.60\times 80)+(0.40\times 20)\\=48 +8\\=56[/tex]
Thus, the expected test score obtained is E (X) = 56.
Compute the value of E (X₁²) as follows:
[tex]V(X_{1})=E(X_{1}^{2})-(E(X_{1}))^{2}\\20=E(X_{1}^{2})-80^{2}\\E(X_{1}^{2})=20+6400\\E(X_{1}^{2})=6420[/tex]
Compute the value of E (X₂²) as follows:
[tex]V(X_{2})=E(X_{2}^{2})-(E(X_{2}))^{2}\\10=E(X_{2}^{2})-20^{2}\\E(X_{2}^{2})=10+400\\E(X_{2}^{2})=410[/tex]
Compute the value of E (X²) as follows:
[tex]E(X^{2})=p\times E(X_{1}^{2})+q\times E(X_{2}^{2})\\=(0.60\times 6420)+(0.40\times 410)\\=3852+164\\=4016[/tex]
Compute the variance of the test scores obtained as follows:
[tex]V(X)=E(X^{2})-(E(X))^{2}\\=4016-56^{2}\\=880[/tex]
Thus, the variance of the test scores obtained is, V(X) = 880.
(b)
Since the division of grades is not provided, i.e. which score is assigned what grade we cannot compute the probability of randomly selecting an exam with grade A, B, C, D or F.
