+43 Could you help me with this please.. Its practicing with binomial distribution
A company has determined that 1% of the parts it makes are defective. If the company packs a shipment of 5 parts to a customer, what is the probability that :
a) none are defective
b) more than 2 are defective
I have been searching for help online and found a similar question where i came up with these answers
P(defective) = 1-0.05
a) = 0.95 or is it 4.75
2- 0.95
b) = 1.05
I feel like im atleast in the right direction but honestly still dont understand.. :/
+129850 P(none are defective) = (.99)^5 ≈ .95 ≈ 95%
P( more than 2 are defective) =
P(3 are defective) = C(5,3)(.01)^3(.99)^2
P(4 are defective) = C(5,4)(.01)^4(.99)
P(5 are defective) = C(5,5) (.01)^4
Sum these probabilities ≈ 9.85 * 10^(-6) (almost 0 probability that more than 2 are defective)
+43 Im sorry I still dont understand!! Can you please explain how you got that. I have a assignment due every week in this class and its online. Ive took many online classes but i now know cant teach myself Math :/
+2490 The reason you cannot teach yourself (this level of) math is because you lack the prerequisites.
To understand the solution to the question you posted, requires an understanding of the Binomial Cumulative Distribution Function. Your attempted answer did not include any binomial functions, and this indicates you are at least three prerequisites behind in the skills needed to solve problems of this type.
This video will help you understand the specific skills needed for your presented question.
For the prerequisites, start here: https://www.mathsisfun.com/data/index.html#stats
On this page are links for The Binomial Distribution.
Preceding those links is this one.
https://www.mathsisfun.com/combinatorics/combinations-permutations.html
You will need to understand this (and more) to understand the Binomial Cumulative Distribution Function.
GA
