Question 328285: A painter has learned that the time it takes her to paint a random room in a house is normally distributed with a mean if 4 hours and a standard deviation of 45 minutes. If she is just starting work on a random room and has to leave in five hour, find
a. The probablilty she can have lunch before she has to leave if she needs 30min. for lunch.
b. the probability she does not finish before she leaves.
c. The probability she works for more that 3 hours (don't forget she will stop working to leave on time.)
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! A painter has learned that the time it takes her to paint a random room in a house is normally distributed with a mean of 4 hours and a standard deviation of 45 minutes.
If she is just starting work on a random room and has to leave in five hour, find
a. The probablilty she can have lunch before she has to leave if she needs 30min. for lunch.
She will be home for 4 1/2 hrs
P(the job and lunch will be <= 4.5 hrs) = ?
z(4.5) = (4.5-4)/0.75 = 2.3
So, P(she can do job and lunch) = P(z< 2/3) = 0.7475
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b. the probability she does not finish before she leaves.
P(x > 5) = ?
z(5) = (5-4)/0.75 = 4/3
So, P(x > 5) = P(z > 4/3) = 0.0912
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c. The probability she works for more that 3 hours (don't forget she will stop working to leave on time.)
P(3< x < 5) = ?
z(3) = (3-4)/0.75 = -4/3
So, P(3 < x < 5) = P(-4/3 < z < 4/3) = 0.8176
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Cheers,
Stan H.
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