Question

If 5% of the electric bulbs manufactured by a company are defective use poison distribution to find

the probability in a sample of 100 bulbs none is defective and 5 bulbs will be defective.(Poison

Distribution) ​

Answer 1

You will get the answer if you solve the above.

(i) for none

(ii) 5 will be defective

Here is the answer to your question.

Hope it helps you!!!!

Please mark this answer as the BRAINLIEST.

Answer 2

The probability in a sample of 100 bulbs none is defective  is $e^{-5}$

Also  the probability in a sample of 100 bulbs 5 bulbs will be defective is $= 26 .04 e^{-5}$.

Step-by-step explanation:

Given: If 5% of the electric bulbs manufactured by a company are defective.

To Find: The probability in a sample of 100 bulbs none is defective.

Also  The probability in a sample of 100 bulbs 5 bulbs will be defective.

Formula Used:

P (X = k) =  $\frac{e^{-z} . z^{k} }{k!}$  

Here, P(X=k) = The probability of  getting k electric bulb defective  out of  number of n electric bulb  selected.          

            X= X is a Poisson random variable.

            k = number  of  getting  defective electric bulb

            p = probability of getting  one defective electric bulb

            n = number of  electric bulb  selected

            z = n.p

Solution:

Consider - in sample of 100 electric bulb   none is defective.

As given 5% of the electric bulbs manufactured by a company are defective

k=0

p = probability of getting  one defective  bolt out of 100

= 5/100 = 1/20

n= 100

z = n.p = 100 x (1/20) = 5

Substituting the value of n,k,p and z in the formula

P (X = 0)  $= \frac{e^{-5} . 5^{ 0} }{0 !}$

               $= \frac{e^{-5} .1 }{1}$

              $= e^{-5}$

The probability in a sample of 100 bulbs none is defective  is $e^{-5}$.

Consider - in sample of 100 electric bulb  5  is defective.

As given 5% of the electric bulbs manufactured by a company are defective

k=5

p = probability of getting  one defective  bolt out of 100

= 5/100 = 1/20

n= 100

z = n.p = 100 x (1/20) = 5

Substituting the value of n,k,p and z in the formula

P (X = 5)  $= \frac{e^{-5} . 5^{5} }{5!}$

              $= \frac{e^{-5} . 5^{5} }{5.4!}$

              $= \frac{e^{-5} . 5^{4} }{4!}$

              $= 26 .04 e^{-5}$

The probability in a sample of 100 bulbs 5 bulbs will be defective is $= 26 .04 e^{-5}$.

Question:

If 5% of the electric bulbs manufactured by a company are defective use poison distribution to find the probability in a sample of 100 bulbs none is defective. Also  the probability in a sample of 100 bulbs   5 bulbs will be defective.