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.
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%