# Thread: Can any one pls help with this Probibality question with formulas?

1. ## Can any one pls help with this Probibality question with formulas?

XYZ Manufacturing has 6 machines that perform a particular task. Breakdowns occur frequently for this machine. Past records indicate that the number of breakdowns that occur each day is described by the following probability distribution:

Number of
Breakdowns------------- Probability
0------------------------------- 0.3
1 -------------------------------0.4
2------------------------------- 0.1
3------------------------------- 0.2
More than 3----------------- 0.0

(a) What is the expected number of breakdowns in any given day?
(b) What is the variance for this distribution?
(c) What is the probability that there will be at least 2 breakdowns in a day?

2. $\displaystyle E(X)=(0)(.3)+(1)(.4)+(2)(.1)+(3)(.2)$

Likewise $\displaystyle E(X^2)=(0)^2(.3)+(1)^2(.4)+(2)^2(.1)+(3)^2(.2)$

and $\displaystyle V(X)=E(X^2)-[E(X)]^2$

$\displaystyle P(X\ge 2)=.1+.2$

3. ## Can you solve the whole question pls?

Will be highly appreciated

4. Originally Posted by stoorrey
Will be highly appreciated
Please show some effort. What have you done? Where are you still stuck? What part of matheagle's reply do you not understand?

5. XYZ Manufacturing has 6 machines that perform a particular task. Breakdowns occur frequently for this machine. Past records indicate that the number of breakdowns that occur each day is described by the following probability distribution:

Number of
Breakdowns------------- Probability
0------------------------------- 0.3
1 -------------------------------0.4
2------------------------------- 0.1
3------------------------------- 0.2
More than 3----------------- 0.0

(a) What is the expected number of breakdowns in any given day?
(b) What is the variance for this distribution?
(c) What is the probability that there will be at least 2 breakdowns in a day?

a) probability of 1 breakdown + probability of 2 breakdowns + probability of 3 breakdowns = 0.4+0.1+0.2 = 0.7 (70%)
c) Atleast two break downs means 2 and above. ie. probability of 2 breakdowns + probability = 0.1+0.2 = 0.3 (30%)

6. Originally Posted by malaynsamani
XYZ Manufacturing has 6 machines that perform a particular task. Breakdowns occur frequently for this machine. Past records indicate that the number of breakdowns that occur each day is described by the following probability distribution:

Number of
Breakdowns------------- Probability
0------------------------------- 0.3
1 -------------------------------0.4
2------------------------------- 0.1
3------------------------------- 0.2
More than 3----------------- 0.0

(a) What is the expected number of breakdowns in any given day?
(b) What is the variance for this distribution?
(c) What is the probability that there will be at least 2 breakdowns in a day?

a) probability of 1 breakdown + probability of 2 breakdowns + probability of 3 breakdowns = 0.4+0.1+0.2 = 0.7 (70%)
c) Atleast two break downs means 2 and above. ie. probability of 2 breakdowns + probability = 0.1+0.2 = 0.3 (30%)
a) is incorrect, that's not E(X)

7. Originally Posted by malaynsamani
a) probability of 1 breakdown + probability of 2 breakdowns + probability of 3 breakdowns = 0.4+0.1+0.2 = 0.7 (70%)
c) Atleast two break downs means 2 and above. ie. probability of 2 breakdowns + probability = 0.1+0.2 = 0.3 (30%)
A. You should know by definition why this isn't right. Expected value isn't (necessarily) a percentage, unless the random value is a percentage (but then it still wouldn't mean what it normally means. . .). The expected value is just that the value you would expect to get. In this case, since our random variable is a discreet number, referring to number of breakdowns, then we should get a specific number (and not necessarily an integer). To calculate expected value we take the sum of all random variables, multiplied by their probabilities: 0*P(X=0)+1P(X=1)+2P(X=2)+3P(X=3). Why does this work? Remember what the expected value is - the mean. Imagine you have a perfect sample of 100 days. Among this sample I would expect to have 30% have 0 breakdowns, 40% have 1 breakdown, 10% have 2 breakdowns, 20% have 3 breakdowns, and 0% to have more than 3 breakdowns. That means I would EXPECT to have, among a perfect sample of 100: 30 0's, 40 1's, 10 2's, 20 3's and 0' "More than 3's". Now how would I go about averaging those numbers? And that is your expected value.

8. and I thought I answered this a week ago

9. Oh man you are right. And the OP isn't the fella who I responded to. I'm all mixed up here.

10. ## Hello Matheagle

Originally Posted by matheagle
a) is incorrect, that's not E(X)

If (a) is not E(X) the would you mind tell me the correct answer or formula if you know it. thnaks

11. Originally Posted by stoorrey
If (a) is not E(X) the would you mind tell me the correct answer or formula if you know it. thnaks
He was responding to someone else. Question A is asking for the expected value of X, and was answered.

12. so answer this, who's on first?

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### a production machine has a likelihood of 0.2 to breakdown any day

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