can someone please help me out with this problem i been trying to figure out what the answer is .. ? or at least give me some ti
ps so i can remember for the other problems ??
1 answer:
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"Variability (also called spread or dispersion) refers to how spread out a set of data is. Variability gives you a way to describe how much data sets vary and allows you to use statistics to compare your data to other sets of data. The four main ways to describe variability in a data set are:
<span>Range Interquartile range Variance Standard deviation.it is hard to explain but i hope this helped:/:)</span>
Answer:
Step-by-step explanation:
Hello!
There are 4 3D printers available to print drone parts, then be "Di" the event that the printer i printed the drone part (∀ i= 1,2,3,4), and the probability of a randomly selected par being print by one of them is:
D1 ⇒ P(D1)= 0.15
D2 ⇒ P(D2)= 0.25
D3 ⇒ P(D3)= 0.40
D4 ⇒ P(D4)= 0.20
Additionally, there is a chance that these printers will print defective parts. Be "Ei" represent the error rate of each print (∀ i= 1,2,3,4) then:
P(E1)= 0.01
P(E2)= 0.02
P(E3)= 0.01
P(E4)= 0.02
Ei is then the event that "the piece was printed by Di" and "the piece is defective".
You need to determine the probability of randomly selecting a defective part printed by each one of these printers, i.e. you need to find the probability of the part being printed by the i printer given that is defective, symbolically: P(DiIE)
Where "E" represents the event "the piece is defective" and its probability represents the total error rate of the production line:
P(E)= P(E1)+P(E2)+P(E3)+P(E4)= 0.01+0.02+0.01+0.02= 0.06
This is a conditional probability and you can calculate it as:
To reach the asked probability, first, you need to calculate the probability of the intersection between the two events, that is, the probability of the piece being printed by the Di printer and being defective Ei.
P(D1∩E)= P(E1)= 0.01
P(D2∩E)= P(E2)= 0.02
P(D3∩E)= P(E3)= 0.01
P(D4∩E)= P(E4)= 0.02
Now you can calculate the probability of the piece bein printed by each printer given that it is defective:
P(D2)= 0.25 and P(D2/E)= 0.33 ⇒ The prior probability of D2 is smaller than the posterior probability.
The fact that P(D2) ≠ P(D2/E) means that both events are nor independent and the occurrence of the piece bein defective modifies the probability of it being printed by the second printer (D2)
I hope this helps!