Assignment No.4 (Course STA301)
Spring 2010 (Total Marks 30)
Deadline
Your Assignment must be uploaded/ submitted before or on
23:59,1st July, 2010
Assignment.4 (Lessons 23-27)
Question 1: Marks: 3+7=10
a)
Find mean from the following probability distribution.
Solution:
|No of petals |P(X) |X.P(X) |
|X | | |
|3 |0.05 |0.15 |
|4 |0.10 |0.4 |
|5 |0.20 |1 |
|6 |0.30 |1.8 |
|7 |0.25 |1.75 |
|8 |0.075 |0.6 |
|9 |0.025 |0.225 |
|Total |1 |∑X.P(X) = 5.925 |
b) A random variable X has the following probability distribution:
|X |P(X) |
|-2 |0.1 |
|-1 |k |
|0 |0.2 |
|1 |2k |
|2 |0.3 |
|3 |3k |
Find
i) K
|X |P(X) |
|-2 |0.1 |
|-1 |K |
|0 |0.2 |
|1 |2k |
|2 |0.3 |
|3 |3k |
|Total |∑P(X) = 0.6+6k |
The sum of the probabilities is equal to 1
[pic]
Substituting values of k,
|X...