Mrs. yan's drive to work is found to range between 12 and 18 minutes.
a) sketch the graph of this distribution
b) determine the probability that
her drive on a randomly selected day will be less than 14 minutes.
c) determine the probability that her drive on a randomly selected day will be more than 19 minutes.
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We have the range of values is 12 to 18.
Assume that the mean is 16. We can approximate the standard deviation using the range rule -- `s~~"range"/4` so `s~~1.5`
(1) If there is no other information given we would assume that the drive times were approximately normally distributed. So draw a bell curve with the peak at x=16 and standard deviation 1.5
(2) Find P(X<14)
Converting to a z-score we get `z=(14-16)/1.5=-1.bar(3)`
Then P(x<14)=P(z<-4/3). Consulting the standard normal table or a calculator, we find the area of the curve to the left of z=-4/3 (and thus the probability of a random z being in this area) is approximately .0912 .
The probability of a trip taking less than 14 minutes is about 9%.
(3) Find P(x>19)
Convert 19 to a z-score: `z=(19-16)/1.5=2`
Then P(x>19)=P(z>2) Again consulting the standard normal table we find P(z>2) is approximately .0228
The probability of a trip lasting more than 19 minutes is about 2%
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