Question

The vehicle speeds at a particular interstate location can be described by a normal curve. the mean speed is 64 mph, and the standard deviation is 7 mph. what proportion, p of vehicle speeds at this location are faster than 71 mph? (round the answer to four decimal places.)

Answer 1

The speed is normally distributed with mean μ=64 and standard deviation σ =7. Use the transformation
`(X-\mu)/(\sigma)` for calculating the probability. This value is called Z-score.

The Z score is
`Z=(71-64)/(7) =1`.

Refer to the standard normal distribution table.

The probability

`P(X>71)=P(Z>1)=1-P(Z<1)=0.1587</p><p>`

Thus, the proportion of the vehicles that are travelling faster than 71 mph is
`p=0.1587` .