Ex:
A die is rolled it n times. Let Xk = the spots recorded on the kth roll and X = X1 + X2 + …+ Xn = sum of spots resulting.
Q1. Roll the die twice (n = 2). Describe the sample space.
Q2. Find the mean of X1 and the variance of X1.
Q3. Are the first roll and the second roll independent? Are X1 and X2 independent?
Q4. Roll the die twice (n = 2). Find the sampling distribution of X = sample total = X1 + X2.
Q5. Roll the die 100 times. P[3.5 ≤ ≤ 3.9] is ______________.
Q1)
Sample space is as follows (given by Totals)
2 with P(1/36), 3 with P(2/26), 4 with P (3/36), 5 with P (4/36), 6 with P (5/36), 7 with P (6/36), 8 with P (5/36), 9 with P (4/36), 10 with P (3/36), 11 with P(2/36), 12 with P (1/36)
Q2)
To find the mean of X1 and the Variance of X1, is it just:
Common Mean = E(X1) = Mew (U)
E(X1) = (1 * P(x1=1))
+ (2 * P(x1 = 2)
+ (3 * P(x1 = 3)
+ (4 * P(x1 = 4)
+ (5 * P(x1 = 5)
+ (6 * P(x1 = 6)
=3.5 = Mew
Common Variance = V(x1) = E{[x1 - E(x1)]^2}
*Where E(x1) = Mew
=
(1-3.5)^2 * 1/6
+ (2-3.5)^2 * 1/6
+ (3-3.5)^2 * 1/6
+ (4-3.5)^2 * 1/6
+ (5-3.5)^2 * 1/6
+ (6-3.5)^2 * 1/6
=2.916 = Sigma Squared
Q3)
Yes indépendant because P(any outcome) = 1/6 for both X1 and X2
Q4) If die rolled twice (N=2) Sampling Distribution of X is = sigma^2/n which is
2.916/2 = 1.458
** I'm going by his wednesday lecture notes here where he stated several times Common Variance of X1 = Sigma^2, and that Variance(Xbar) = Sigma^2/n
EDIT: I'm a retard I read his notes wrong -.- I hate Scott Pai I don't understand why he throws so many equations at us .. I seriously had less equations in physics -.-
It's actually Sqrt(2.916)/Sqrt(2) = 1.207
Q5) Die is rolled 100 times,
P[3.5 ≤ ≤ 3.9]
The distribution is Approx. Normal with N[Mew, Sqrt (Sigma^2)/ Sqrt(n)]
Therefore, N[3.5, 0.17076]
(3.5 - 3.5)/0.17076 [LessthanEqualto] Z [LessthanEqualTo] (3.9-3.5)/0.17076
0 ≤ Z ≤ 2.3424
=0.9904 - 0.5 = 49.04%
i have a feeling I'm way off
If any RS-ers wants to study for stats 201 my email is
dhillon09@gmail.com (I check that way more than RS) so just email me your number and we can meet up!
