Answer:
Step-by-step explanation:
Forecast for period 1 is 5
Demand For Period 1 is 7
Demand for Period 2 is 9
Forecast can be given by
where
Forecast for Period 3
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We are looking for the probability :
Transform the law to standard normal like this:
The above formula is equivalent to this one:
From normal law table, we read the value of
.
Solving the above equation for the score n:
, it is the score we are looking for.
Transform the law to standard normal like this:
The above formula is equivalent to this one:
From normal law table, we read the value of
Solving the above equation for the score n:
Answer:
The probability that no flaws occur in a certain portion of wire of length 5 millimeters = 1.1156 occur / millimeters
Step-by-step explanation:
<u>Step 1</u>:-
Given data A copper wire, it is known that, on the average, 1.5 flaws occur per millimeter.
by Poisson random variable given that λ = 1.5 flaws/millimeter
Poisson distribution
<u>Step 2:</u>-
The probability that no flaws occur in a certain portion of wire
On simplification we get
P(x=0) = 0.223 flaws occur / millimeters
<u>Conclusion</u>:-
The probability that no flaws occur in a certain portion of wire of length 5 millimeters = 5 X P(X=0) = 5X 0.223 = 1.1156 occur / millimeters