MATH SOLVE

4 months ago

Q:
# A manufacturer packages bolts in boxes containing 100 each. each box of 100 bolts contains on average four defective bolts. the quality control staff randomly selects a box at the end of the day from an entire production run. what is the probability that the box will contain less than three defective bolts? use the poisson distribution table.

Accepted Solution

A:

The probability of a Poisson distribution is given by:

[tex]P(k)=e^{-\lambda} \frac{\lambda^k}{k!} [/tex]

where: [tex]\lambda[/tex] is the average number of events and k is the required probability.

Given that a manufacturer packages bolts in boxes containing 100 each and that each box of 100 bolts contains on average four defective bolts. Thus, [tex]\lambda=4[/tex].

The probability that the box will contain less than three defective bolts is given by:

[tex]P(0)+P(1)+P(2)=e^{-4}\cdot \frac{4^0}{0!} +e^{-4}\cdot \frac{4^1}{1!} +e^{-4}\cdot \frac{4^2}{2!} \\ \\ =e^{-4}(1+4+8)=13(0.0183)=0.2381[/tex]

[tex]P(k)=e^{-\lambda} \frac{\lambda^k}{k!} [/tex]

where: [tex]\lambda[/tex] is the average number of events and k is the required probability.

Given that a manufacturer packages bolts in boxes containing 100 each and that each box of 100 bolts contains on average four defective bolts. Thus, [tex]\lambda=4[/tex].

The probability that the box will contain less than three defective bolts is given by:

[tex]P(0)+P(1)+P(2)=e^{-4}\cdot \frac{4^0}{0!} +e^{-4}\cdot \frac{4^1}{1!} +e^{-4}\cdot \frac{4^2}{2!} \\ \\ =e^{-4}(1+4+8)=13(0.0183)=0.2381[/tex]