SOLUTION: Hi there, If you could help me w/ this problem I would be very grateful for your help! Assume the speed of vehicles along a stretch of I-10 has an approximately normal distribut

Question 1048251: Hi there, If you could help me w/ this problem I would be very grateful for your help!
Assume the speed of vehicles along a stretch of I-10 has an approximately normal distribution with a mean of 71 mph and a standard deviation of 8 mph.
a. The current speed limit is 65 mph. What is the proportion of vehicles less than or equal to the speed limit?
b. What proportion of the vehicles would be going less than 50 mph?
Any help is appreciated, thanks!

Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!
mean of 71 mph and a standard deviation of 8 mph.

a. P(x<= 65) = P(z <= -6/8) = P(z <= -3/4)
TI syntax is normalcdf(smaller z, larger z)
normalcdf( ... is a command people using a TI Calculator
use for finding the area under a standard normal curve
between two z=values...-9999 used for the far far left z-value
to give us what we want: area under a standard normal curve for
a particular z-value, in this case z = -.75 (Area under the curve to left of Blue Line)
normalcdf(-9999, -.75) = .2266 0r 22.66%

b . P(x<= 50) = P(z <= -21/8)
normalcdf(-9999, -21/8) = .0043 0r .43%
One can use table as well to find the z-value
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