Probability

Probability theory and statistics

Probability

Postby jzotter » Sat Aug 03, 2019 2:45 pm

Dear Math Forum,

please could you help me to understand how to solve these two following tasks:
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Re: Probability

Postby Guest » Mon Aug 05, 2019 3:25 pm

If 10% of goods are defective then 90% are not and the probability that none are defective in a sample of 15 is [tex]0.9^{15}= 0.206[/tex]. The probability that there is exactly one defective part in the sample of 15 is [tex]15(0.9)^{14}(0.1)= 0.343[/tex]. So the probability of "no more than one defective part" is 0.343 and the probability of "more than one defective part" is 1- 0.343= 0.657. The probability that a shipment with 10% defectives will be rejected is 0.657. Using this method, the probability that a shipment with 10% defectives will be rejected is 0.657.

If 10% of goods are defective then 90% are not and the probability that none are defective in a sample of 30 is [tex]0.9^{30}= 0.0424[/tex]. The probability that there is exactly one defective part in the sample of 30 is [tex]30(0.9)^{29}(0.1)= 0.141. The probability that there are exactly two defective parts in the sample of 30 is [tex]\begin{pmatrix}30 \\ 2\end{pmatrix}(0.9)^{28}(0.1)^{2}= 29(15)(0.9)^{28}(0.1)^2= 0.228[/tex]. So the probability of "no more than 2 defective parts" is 0.0424+ 0.141+ 0.228= 0.4114 and the probability of "no more than two defective parts is 1- 0.4114=
0.5886. Using this method, the probability that a shipment with 10% defectives will be rejected is 0.5886.

For (4), do you know what a "Poisson distribution" is?

Here is a link to a "poisson distribution calculator": https://stattrek.com/online-calculator/poisson.aspx
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