Question

5. Let x and Y be the scores of a student on ther midterm and final exam. We model these scores as X=S+E_1,Y=S+E², where S,E ₁ and E ₂, are independent random variables distributed as as S∝N(70,49), E ₁ ,E₂ ∝N(0,25). We think of S as the "skill" component of the score and E₁ and E₂ as the "luck" component. (a) What is the joint distribution of (X,Y) ? (b) Assume that a student received a midterm score that is one standard deviation below the midterm mean. What do you expect their final score to be? (c) Students achieving a midterm score of at least two standard deviations above the midterm mean are praised by the instructor. However, when taking the final exam many of these praised students do not achieve a score two or more standard deviations above the mean. Do you think this is due to the praise?

Answer 1

(a) To find the joint distribution of (X, Y), we have:

X = S + E₁

Y = S + E₂

Given that S, E₁, and E₂ are independent random variables with the following distributions:

S ~ N(70, 49)

E₁ ~ N(0, 25)

E₂ ~ N(0, 25)

We need to find the joint distribution of (X, Y). Since S, E₁, and E₂ are independent, their sums X and Y will also be independent.

First, let's find the mean and variance of X and Y:

Mean of X:

E[X] = E[S + E₁] = E[S] + E[E₁] = 70 + 0 = 70

Variance of X:

Var(X) = Var(S + E₁) = Var(S) + Var(E₁) = 49 + 25 = 74

Mean of Y:

E[Y] = E[S + E₂] = E[S] + E[E₂] = 70 + 0 = 70

Variance of Y:

Var(Y) = Var(S + E₂) = Var(S) + Var(E₂) = 49 + 25 = 74

Now, we know that X and Y are normally distributed with means 70 and variances 74. So, the joint distribution of (X, Y) can be represented as a bivariate normal distribution:

(X, Y) ~ N(μ, Σ)

Where:

μ = [70, 70] (mean vector)

Σ = [[74, 0], [0, 74]] (covariance matrix)

(b) If a student received a midterm score that is one standard deviation below the midterm mean, their midterm score would be 70 - 1 * √74 = 70 - √74 ≈ 62.24.

Since the final exam score follows the same distribution as the midterm, and the mean of Y is also 70, we would expect their final score to be around 70 as well.

(c) The fact that some students who achieved a midterm score of at least two standard deviations above the mean do not achieve a score two or more standard deviations above the mean on the final exam is not necessarily due to the praise. This can be explained by the nature of random variables and the "luck" component (E₁ and E₂) in the scores.

Even though students with high midterm scores are more likely to have high final exam scores on average, the "luck" component can introduce variability. Luck can either boost or hinder a student's performance on any given day, and this variability can cause some students to not perform consistently at a very high level.

In other words, the "luck" component introduces randomness, and high scores on the midterm do not guarantee equally high scores on the final exam due to this inherent variability. The praise itself is unlikely to be a direct cause of the variability in final exam scores.

Explanation: