1. ## Help, and I have no statistical background.

I am taking a course in safety and now have one class left, but ran into a big problem. This class has statistics and I don't have a clue. The text is of no help and doesn't explain anything but theory. I am over my head and need some directions with a number of problems.

1. A system has 5 sensors. The probability of any sensor failing is equal and 3 sensors are in a failed state. As an inspector if 1 sensor is selected randomly what is the probability that it wil be in a failed state?

******** If 2 senors are selected randomly what is the probablity of at least one in a failed state?

******** If 3 sensors are selected what is the probability that the system will be found in a failed state.

2. Consider 4 books on a shelf with 4 spaces. how many unordered ways may they be arranged.

3. Consider the 4 books, but only 3 spaces. how many unordered ways may these books be arranged.

4. A bin contains 7 parts. The probability that any part in the bin is defective is 3.5E-3. If 1 part is elected at random from the in, what is the probability that it is defective.

******* If 3 parts are selcted at random, what is the probability that exactly 2 are defetive?

5. A component has a failure rate per hour of 4.5E-3. The failures are governed by the exponential distribution. What is the probability of this component failing exactly once in 100 hours?

******** what is the probability of this component failing at least once in 100 hours?

6. A fire detection sensor is tested for its failure arte and found to have a mean failure rate of 3.6e-2 per hour of exposure with a standard deviation of .4E-2 per hour. The distribution of the failure rate is normal. If a sensor is selected at random from a large lot for installation in a sysyem, what is the probability that it will have a failure rate greater than 4.5e2 per hour.

I do not even know where to begin with these sample problems. Any assistance is appreciated.

2. 1.
part 1 --> p(1) = 3/5 = 0.6
part 2 --> p(at least 1) = 1-p(none) = 1- [ (2/5)*(1/4) ]
part 3 --> assuming that the system will be viewed as in a failed state if all 3 sampled sensors are defective --> p(3) = (3/5)*(2/4)*(1/3)

2. and 3.
use the combinations formula: nCr = n! / (r! (n-r)!)
n = number of books
r = number of spaces

4.
uses the binomial probability distribution
p(r) = nCr * p^r * q^(n-r)

part 1: you can assume that the 1 sampled part was taken directly from the process that made the parts
p = 3.5E -3 = .0035
q = 1-p = .9965
n = 1
r = 1

part 2: same assumption in part 1 is valid here as well
p = 3.5E -3 = .0035
q = 1-p = .9965
n = 3
r = 2

5.
working on it

6.
uses the normal distribution
mean = .0036
std dev = .004
x = .0045

z = (x - mean) / std dev
compute z, then use the normal distribution table to find the proportion of the area under the curve that is above z

3. ## need assistance

This is all I have and can figure out so far, I need some help, I have absolutely no stat background, I do not know what most of the symbols mean, or how to actually work out these problems. I have asked my instructor for help and he said "buy the cliff notes". I even had a grad student/educator try to help me and she said she didn't have a clue. Any assistance is appreciated.

1. A system has 5 sensors. The probability of any sensor failing is equal and 3 sensors are in a failed state. As an inspector if 1 sensor is selected randomly what is the probability that it wil be in a failed state?

p(1)=3/5=0.6
p=0.6

******** If 2 senors are selected randomly what is the probablity of at least one in a failed state?

p(at least)= 1-p(none)
= 1-[(2/5)*(1/4)]
p=0.9

******** If 3 sensors are selected what is the probability that the system will be found in a failed state.

p(3)=(3/5)*(2/40*(1/3)
1-p=(3/5)*(2/4)*(1/3)
p=.93

2. Consider 4 books on a shelf with 4 spaces. how many unordered ways may they be arranged.

This number is 1, because order is of no consequences.

3. Consider the 4 books, but only 3 spaces. how many unordered ways may these books be arranged.

4*3*2/3*2*1=24/6
=4

4. A bin contains 7 parts. The probability that any part in the bin is defective is 3.5E-3. If 1 part is elected at random from the in, what is the probability that it is defective.

p(r)=nCr*p^r*q^(n-r)
1*(.0035*.9965)
p=.0035

******* If 3 parts are selcted at random, what is the probability that exactly 2 are defetive?

p(r)=nCr*p^r*q^(n-r)
p(r)=6*(.0035*.0035)*.9965(1)
p(r)=6*(.000012)*.9965
P=.00072

5. A component has a failure rate per hour of 4.5E-3. The failures are governed by the exponential distribution. What is the probability of this component failing exactly once in 100 hours?

.0045*100

******** what is the probability of this component failing at least once in 100 hours?

?

6. A fire detection sensor is tested for its failure arte and found to have a mean failure rate of 3.6e-2 per hour of exposure with a standard deviation of .4E-2 per hour. The distribution of the failure rate is normal. If a sensor is selected at random from a large lot for installation in a sysyem, what is the probability that it will have a failure rate greater than 4.5e2 per hour.

m=.0036
Std. dev=.004
x=.0045
z=(.0045-.0036/.004)
z=.225

4. Check your math on the third part of question number 1, and the second part of question number 4.

You computed z correctly on the last question - now you need to use the normal distribution table to find the probability of z >= .225.

In defense of your instructor, this course sounds like it has a stats prerequisite, and he probably feels like it's not his job to get you up to speed...

5. ## still trying

On the third part of question 1-this is the re-work.
3/5*2/4*1/3
.60*.50*.33
P=.099

if I then put in the 1-p
that gives me p=.90

The second part of question 4
p(r)=6*(.0035*.0035)*.9965(1)
p(r)=6*(.000012)*.9965
p=.0000717

I looked at the nominal distribution table to find z
I am not quite sure what to look for, I looked down at 2.2, then went to .05 across the table and found .48776

6. On the third part of question 1, you don't need to subtract the multiplication from 1.0 - it's simply:

P(3 defective parts) = P(part 1 defective AND part 2 defective AND part 3 defective)

= (3/5) * (2/4) * (1/3) = 6/60 = 0.1

In the second part of question 4, you should get 3 for 3C2:

3!/(2!(3-2)!) = 6/2 = 3

For question 7, to find the probability of z >= 0.225, it's 0.411

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