Question

2. A random test score X is obtained from a class with students who fall into two groups. For the first group X is conditionally Gaussian with mean 80 and variance 20, while for the second group X is conditionally Gaussian with mean 20 and variance 10. The probability that a student is in the first group is 0.6. (a) Find E [X] and Var [X] . (b) Assuming the grades are assigned as in the notes, find the probability that a randomly selected exam gets an A, B, C,D or F grade (in terms of the Φ function

Answer 1

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:

$X_{1}\sim N(80, 20)\\X_{2}\sim N(20, 10)$

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)

       $=(0.60\times 80)+(0.40\times 20)\\=48 +8\\=56$

Thus, the expected test score obtained is E (X) = 56.

Compute the value of E (X₁²) as follows:

$V(X_{1})=E(X_{1}^{2})-(E(X_{1}))^{2}\\20=E(X_{1}^{2})-80^{2}\\E(X_{1}^{2})=20+6400\\E(X_{1}^{2})=6420$

Compute the value of E (X₂²) as follows:

$V(X_{2})=E(X_{2}^{2})-(E(X_{2}))^{2}\\10=E(X_{2}^{2})-20^{2}\\E(X_{2}^{2})=10+400\\E(X_{2}^{2})=410$

Compute the value of E (X²) as follows:

$E(X^{2})=p\times E(X_{1}^{2})+q\times E(X_{2}^{2})\\=(0.60\times 6420)+(0.40\times 410)\\=3852+164\\=4016$

Compute the variance of the test scores obtained as follows:

$V(X)=E(X^{2})-(E(X))^{2}\\=4016-56^{2}\\=880$

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.