Suppose that the peak demand for electricity on one summer day in Ontario is nor
ID: 3042780 • Letter: S
Question
Suppose that the peak demand for electricity on one summer day in Ontario is normally distributed with a mean of 1000 units and standard deviation of 60. Ontario’s maximum electricity generation is 1100 units. If the demand exceeds capacity, there will be a blackout.
a.What is the approximate probability that there will be a blackout on that day?
b.If the standard deviation of demand were to fall to 50, what would this do to the probability of a blackout? Draw a picture first, then do the math.
c.If standard deviation were to remain 60, but Ontario purchased an extra 100 units of electricity, what would this do to a probability of a blackout? Draw a picture first, then do the math.
d.If the original conditions were repeated each day for 20 days during one hot month, what is the probability of at least one blackout?
Explanation / Answer
a) P(X > 1100) = P((X - mean)/sd > (1100 - mean)/sd)
= P(Z > (1100 - 1000)/60)
= P(Z > 1.67)
= 1 - P(Z < 1.67)
= 1 - 0.9525 = 0.0475
b) P(X > 1100) = P((X - mean)/sd > (1100 - mean)/sd)
= P(Z > (1100 - 1000)/50)
= P(Z > 2)
= 1 - P(Z < 2)
= 1 - 0.9772 = 0.0228
c) P(X > 1200) = P((X - mean)/sd > (1200 - mean)/sd)
= P(Z > (1200 - 1000)/60)
= 1 - P(Z < 3.33)
= 1 - 0.9996 = 0.0004
d) P(one blackout in one day) = 0.0475
n = 20
P(X = x) = nCx * Px * (1 - P)n - x
P(X > 1) = 1 - P(X < 1)
= 1 - P(X = 0)
= 1 - (20C0 * (0.0475)^0 * (0.9525)^20)
= 1 - 0.3778
= 0.6222
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