Download Assignment 8. Probability2010 (IBE).

Assignment 8. Probability2010 (IBE).
Problem:
Scores X and Y on the Probability and Statistics tests follow a bivariate normal distribution with expected values
2
65 and 55, respectively, variances σX
= 100 and σY2 = 81, and correlation ρ = 0.68.1 You are required to:
1. Obtain the vector of means and the variance-covariance matrix (Σ) of the corresponding bivariate distribution
2. Find the marginal distribution of the score in Statistics
3. Find the distribution of the mean of the two scores, H = (X + Y )/2
4. Obtain the conditional distribution of the scores in statistics for the students that have achieved a score
X = 70 in Probability.
5. Compute the probability that a student that obtained a 70 in probability will fail Statistics.
6. In the previous question, find an interval (a, b) for the score in statistics that has a conditional probability
of 95.5%.
7. Find the probability that the mean of the two scores will be between 60 and 80.
8. Find the probability that the score in Statistics will be grater than the score in Probability.
Solution:
1. The mean vector µ =
65
55
100 61.2
The variance covariance matrix Σ =
61.2 81
We used that σ12 = ρσ1 σ2 = 0.68 ∗ 10 ∗ 9 = 5508.
2. Clearly, the marginal distribution of Y is N (55, 81). From tables of the standard normal, we find Φ(.84) =
0.84 (or with R qnorm(0.8)= 0.8416212 . So the percentile 80th of the distribution of Y is : 62.57459
= 55 + 9* 0.8416212 .
3. The distribution of the mean H = (X+Y )/2 is Normal (as it was assumed that (X, Y ) is a bivariate Normal)
2
with mean µH = E(X + Y )/2 = (EX + EY )/2 = (65 + 55)/2 = 60 and variance σH
= V ((X + Y )/2) =
1
4 (V
(X) + V (Y ) + 2C(X, Y )) = 0.25 ∗ (100 + 81 + 2 ∗ 61.2) = 75.85. So, the probability of a student to fail
is P (H < 50) = P (Z <
50−60
sqrt(75.85) )
= P (Z < −1.148212) = 0.1254405, i.e. approximately a probabiliy
12.5% of failing the whole subject Probability + Statistics.
1 This
is fictitious data
4. Using the formula for the conditional distribution of a bivariant Normal distribution, we find that the
conditional distribution will be normal with mean
E(X2 | (X1 = 70)) = 55 +
and variance
V (X2 | (X1 = 7)) = 81 −
61.2
(70 − 65) = 58.06
100
61.22
= 81 − 37.4544 = 43.5456
100
That Y | (X = 7) is N (58.06, 43.5456).
The percentile 80 will thus be (use again the percentile 80 of the standard normal, qnorm(0.8)= 0.8416212)
to obtain for the percentile 80 of this conditional distribution the value 63.60= 58.06 + sqrt(43.5456)*
0.84. So, 80% of the students with a grade of probability equal to 7 have a grade smaller than 63.60.
5. For the above conditional distribution we need to find the probability to be less than 5. That is P (N (58.06, 43.5456) <
5) = Φ((5 − 58.06)/sqrt(43.5456)) = Φ(−8.040723) ≈ 0, so it is very low probability. It would be an
extremely strange event!
6. For the standard normal distribution the interval [−2, +2] is of 95% probability. Thus, the interval
[µ−2σ, µ+2σ] = [58.06−2∗sqrt(43.5456), 58.06+2∗sqrt(43.5456)] = [58.06−13.19782, 58.06+13.19782]
= [44.86218, 71.25782]
contains also the 95% of the distribution.
7. Finally, we need to compute the probability of Statistics (Y) to exceed Probability. We need to find
P (X2 > X1 ) = P (X2 − X1 > 0). We require to find the distribution of X2 − X1 . We know this
distribution is normal, with mean E(X2 ) − E(X1 ) = 55 − 65 = −10 and variance V (X2 − X1 ) =
V (X2 ) + V (X1 ) − 2C(X2 , X1 ) = 81 + 100 − 2 ∗ 61.2 = 58.6. So X2 − X1 ∼ N (−10, 58.6).
The desired probability is · · · = P (N (−10, 58.6) > 0) = 1 − P (N (−10, 58.6) < 0) = 1 − Φ((0 −
(−10))/sqrt(58.6)) = 1 − Φ(1.306325) = 1 − pnorm(1.306325) = 1 − 0.904279 = 0.095721. It is small,
around 10% , the probability that a student gets a grade in Statistics greater than a grade in Probabilitat
2