Note for this In each row 4 decimal places In final answer 2 decimal places P(X) 10 30 Prghabily 30 3 30 30 3 30 11 13 3. Number of violations The following table gives the probabilities that a probation officer will receive 0,1,2,3,4, or 5 reports of probation violations on any given day. Solve for the mean. Probability P(X) 0.15 0.25 Number of Violations X 0.36 0.18 0.04 0.02 4. 5

A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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answer 2 and 3 pls
2.
Note for this
In each row 4 decimal places
In final answer 2 decimal places
P(X)
10
30
Probabily
30
3
30
30
30
11
13
3. Number of violations
The following table gives the probabilities that a probation officer will receive 0,1,2,3,4,
or 5 reports of probation violations on any given day.
Solve for the mean.
Number of Violations X
Probability
P(X)
0.15
0.25
0.36
0.18
0.04
0.02
3
4
5
Transcribed Image Text:2. Note for this In each row 4 decimal places In final answer 2 decimal places P(X) 10 30 Probabily 30 3 30 30 30 11 13 3. Number of violations The following table gives the probabilities that a probation officer will receive 0,1,2,3,4, or 5 reports of probation violations on any given day. Solve for the mean. Number of Violations X Probability P(X) 0.15 0.25 0.36 0.18 0.04 0.02 3 4 5
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