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# Fall 2013 (Total Marks 15)

February 07 23:59, 2014

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### Objective(s) of this Assignment:

This assignment will strengthen the basic idea about the concept of the following distributions:

• Poisson distribution
• Binomial distribution
• Hypergeometric distribution

Assignment No: 2 (Lessons 27 – 30)

Question 1:                                                                                             Marks: 4+4=8

a)     A batch of 10 gaskets contains 4 defective gaskets. If we draw samples of size 3 without replacement, from the batch of 10, find the probability that a sample contains 2 defective gaskets.

b)     In a binomial distribution, the mean and the standard deviation were found to be 36 and 4.8 respectively. Find the parameters of binomial distribution.

Question 2:                                                                                             Marks: 4+3=7

1. It is known that the computer disks produced by a company are defective with probability 0.02 independently of each other.  Disks are sold in packs of 10. A money back guarantee is offered if a pack contains more than 1 defective disk. What is the probability of sales result in the customers getting their money back?
2. The average number of accidents occurring in an industrial plant during a day is 3. Assuming Poisson distribution for the number of accidents (X) during a day, compute probability that at most two accidents occur in a day.

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Discuss here Assignment 2

Our main purpose here discussion not just Solution

We are here with you hands in hands to facilitate your learning and do not appreciate the idea of copying or replicating solutions.

Question 1:                                                                                             Marks: 4+4=8

a)     A batch of 10 gaskets contains 4 defective gaskets. If we draw samples of size 3 without replacement, from the batch of 10, find the probability that a sample contains 2 defective gaskets.

Solution

A batch of 10 rocker cover gaskets contains 4 defective gaskets. If we draw samples
of size 3 without replacement, from the batch of 10, find the probability
that a sample contains 2 defective gaskets. It is possible to derive formulae for the mean and variance of the Hypergeometric distribution.
However, the calculations are more difficult than their Binomial counterparts, so we will simple
state the results. Question 2:                                                                                             Marks: 4

1. It is known that the computer disks produced by a company are defective with probability 0.02 independently of each other.  Disks are sold in packs of 10. A money back guarantee is offered if a pack contains more than 1 defective disk. What is the probability of sales result in the customers getting their money back?

Solution:

p of a defective disk is equal to .02
sell disks in a pack of 10.
probability of more than 1 disk being defective is equal to 1 minus the probability that 0 or 1 are defective.

probability that 0 out of the 10 pack are defective is equal to 10c0 * .02^0 * .98^10 which is equal to .817073

probability that 1 out of the 10 pack are defective is equal to 10c1 * .02^1 * .98^9 which is equal to .166749

probability that 0 or 1 out of the 10 pack are defective is equal to .817073 + .166749 which is equal to .983822

the probability that more than 1 are defective is therefore equal to 1 minus .983822 which is equal to .016177.

Looks like they will have to give the money back 1.6% of the time.

the complete table of probabilities is shown below. x is equal to the probability of having a defective disk.
ncx is equal to the number of possible combinations of 10 things taken x at a time.
p(x) is the probability of getting one defective disk.
p(1-x) is the probability of not getting one defective disk.
the complete formula is:
the probability of getting x defective disks out of 10 is equal to ncx * p(x)^x * p(1-x)^(n-x).
for example, the probability of getting 3 defective disks out of 10 is equal to:
10C3 * (.02)^3 * (.98)^7 which is equal to 120 * (.02)^3 * (.98)^7 which is equal to 8.33401 * 10^-4.
that's in scientific notation.
in standard decimal notation it is equal to .000833401.

aslam o alikum!

is this correct?

In 3rd equetion after taking SD

npq =23.04

36q=23.04

q=23.04/36

q=0.64

Rosali chek it again

right but tell is this fine solution or the 1 given by admin??????

question # 2 ka b part kasa karna ha plzzzzzzzzzzzzzzzz helppppppppp     anybody

Muhammad Aslam  ap ke calculation wrong ha

friends 1st question mein jo aek formula

P(X=r) = mcr x N-McN-R / NCN kahan sy lea ha