Question

A manufacturer of pins knows that 2% of his product is defective. If he sells pins in
boxes of 100 and guarantees that not more than 4 pins will be defective. What is the
probability that a box will fail to meet the guaranteed quality?

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Answer 1

Step-by-step explanation:

Given A manufacturer of pins knows that 2% of his product is defective. If he sells pins inboxes of 100 and guarantees that not more than 4 pins will be defective. What is the  probability that a box will fail to meet the guaranteed quality?

• Given n = 100
• Probability of the defective bulbs p = 2% = 0.02
• Mean number of defective bulbs m = n x p
•                                                            = 100 x 0.02
•                                                            = 2
• So P = e^-m m ^x / x! , x = 0,1,2….
• Probability that the box will fail to meet the guaranteed quality will be
•            P (X > 4)
•             = 1 – P (X < = 4)
•              = 1 - ∑ x = 0 to 4 e^-2 2^x / x!
•              = 1 – e^-2 ∑ x = 0 to 4 4^x / x!

Reference link will be

https://brainly.in/question/16145676

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

GIVEN

A manufacturer of pins knows that 2% of his product is defective. If he sells pins in boxes of 100 and guarantees that not more than 4 pins will be defective.

TO DETERMINE

The probability that a box will fail to meet the guaranteed quality

CONCEPT TO BE IMPLEMENTED

POISSON DISTRIBUTION

The probability density function is given by

$\displaystyle \sf{ \: }P(X = i) = \frac{ {e}^{ - \mu \: } { \mu}^{i} }{ i \:! } \: \:$

$\sf{Where \: \mu \: \: is \: the \: parameter}$

CALCULATION

Let p be the probability that a pin will be defective

$\displaystyle \sf{ \: }Then \: \: p \: = \frac{2}{100} = \frac{1}{50}$

Again n = Total number of pins = 100

This problem may be assumed to follow Possion distribution

$\sf{Where \: \mu \: \: is \: the \: parameter}$

$\displaystyle \sf{ \: }Then \: \: \mu\: = np \: = 100 \times \frac{1}{50} = 2$

Now the manufacturer guarantees that not more than 4 pins will be defective

In order to determine probability that a box will fail to meet the guaranteed quality

So atleast 5 pins will be defective

Hence the required probability

$\sf{ = P(X \geqslant 5)}$

$\sf{ = 1 - P(X < 5)}$

$\sf{ = 1 - P(X = 0) -P(X = 1) - P(X = 2) - \:P(X = 3) - P(X = 4)\: }$

$\displaystyle \sf{ \: } = 1 - \frac{ {e}^{ - 2 \: } { 2}^{0} }{ 0 \:! } - \frac{ {e}^{ - 2 \: } { 2}^{1} }{ 1 \:!}\: - \frac{ {e}^{ - 2 \: } { 2}^{2} }{ 2\:!} - \frac{ {e}^{ - 2 \: } { 2}^{3} }{ 3\:!}\: - \frac{ {e}^{ - 2 \: } { 2}^{4} }{ 4\:!}$

$\displaystyle \sf{ \: } = 1 - {e}^{ - 2} \bigg[ \: \frac{ { 2}^{0} }{ 0 \:! } + \frac{ { 2}^{1} }{ 1 \:!}\: + \frac{ { 2}^{2} }{ 2\:!} + \frac{{ 2}^{3} }{ 3\:!}\: + \frac{ { 2}^{4} }{ 4\:!} \bigg]$

$\displaystyle \sf{ \: } = 1 - {e}^{ - 2} \bigg[ \: 1 + + 2 + \frac{4}{3} + \frac{2}{3} \: \bigg]$

$\displaystyle \sf{ \: } = 1 - 7 \: {e}^{ - 2}$

$\displaystyle \sf{ \: } = 1 - ( 7\times 0.135)$

$\displaystyle \sf{ \: } = 1 - 0.945$

$\displaystyle \sf{ \: } = 0.055$

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p=1/2,q=1/2, n =6.

Then find p(x=2)

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