Let X_1, \ldots , X_5 be i.i.d. random variables. You obtain the sample X_1=5, \, X_2=1.5, \, X_3=-3,\, X_4=0.0,\, X_5=7\,.
Let F(t) be the empirical cdf of this sample. Find
Find F(-4).
F(-4)=\quad
unanswered
Find F(-3).
F(-3)=\quad
unanswered
Find F(10).
F(10)=\quad
unanswered
Find the largest interval of t for which \displaystyle F(t)=3/5. Answer by entering A and B in the equation below:
F(t)=3/5\, for A\leq t< B where
A=\quad
unanswered
B=\quad
To find F(-4), we need to count the number of observations that are less than or equal to -4. In this case, there are 2 observations (-3 and 0) that are less than or equal to -4. The empirical CDF assigns a probability of 0.4 to each observation, so F(-4) = 0.4.
Similarly, to find F(-3), we count the number of observations that are less than or equal to -3. In this case, there is only 1 observation (-3) that is less than or equal to -3. The empirical CDF assigns a probability of 0.2 to this observation, so F(-3) = 0.2.
To find F(10), we count the number of observations that are less than or equal to 10. In this case, all 5 observations are less than or equal to 10. The empirical CDF assigns a probability of 1 to each observation, so F(10) = 1.
To find the largest interval of t for which F(t) = 3/5, we need to find the values of t for which F(t) is equal to or less than 3/5, and F(t) + Δt is greater than 3/5. In this case, since there are no ties in the observations, the values of t for which F(t) = 3/5 are (-3, 0]. Therefore, the largest interval for which F(t) = 3/5 is -3 ≤ t < 0.
To summarize:
F(-4) = 0.4
F(-3) = 0.2
F(10) = 1
The largest interval for which F(t) = 3/5 is -3 ≤ t < 0.
To find the empirical cumulative distribution function (CDF) and the values of F(t) for the given sample, follow these steps:
1. Arrange the sample data in ascending order:
X = {-3, 0.0, 1.5, 5, 7}
2. Assign ranks to each data point:
Rank(X_1) = 1
Rank(X_2) = 2
Rank(X_3) = 3
Rank(X_4) = 4
Rank(X_5) = 5
3. Calculate the proportion of values less than or equal to t:
F(t) = Rank(t) / n
where Rank(t) is the rank of t within the ordered sample and n is the total number of observations in the sample.
Now, let's calculate the values of F(t) for the given sample:
For F(-4):
Since -4 is smaller than the smallest value in the sample (-3), F(-4) = 0.
For F(-3):
Since -3 is the smallest value in the sample, F(-3) = Rank(X_1) / n = 1 / 5 = 1/5.
For F(10):
Since 10 is greater than the largest value in the sample (7), F(10) = 1.
For the largest interval of t where F(t) = 3/5:
We need to find the smallest and largest values of t that satisfy F(t) = 3/5.
To find A:
3/5 = Rank(A) / n
3/5 = Rank(A) / 5
Rank(A) = 3
Since Rank(A) is an integer, A must be the third smallest value in the sample, which is 1.5.
To find B:
3/5 = Rank(B) / n
3/5 = Rank(B) / 5
Rank(B) = 3
Since Rank(B) is an integer, B must be the third largest value in the sample, which is 5.
Therefore, the largest interval of t where F(t) = 3/5 is given by:
F(t) = 3/5 for 1.5 ≤ t < 5
So, A = 1.5 and B = 5.
Summary:
- F(-4) = 0
- F(-3) = 1/5
- F(10) = 1
- F(t) = 3/5 for 1.5 ≤ t < 5