Find the average, standard deviation, and proportion

Question

The speeds of cars on main highway are approximately normal. Data collected at a certain point shows at 95% of the cars travel at the speed less than 140 km/h and 10% travel has been less than 90 km/h. 

a) find the average speed and the standard deviation for the car is traveling on that specific stretch of the highway.

b) find the proportion of cars that travel at speeds exceeding 110 km/h. 

Answer 1

If 95% travel less than 140 km/hour, then z = Φ^-1(.95), where Φ is the standard normal distribution (mean 0, sd 1)

In Excel, we get this using normsinv(.95) = 1.64485362695147

10% travel less than 90 km/hour, so z =Φ^-1(.10) = normsinv(.1) = -1.2815515655446

Then, z = 1.64485362695147 and z = -1.2815515655446 are 140 - 90 = 50 km apart, so

The standard deviation σ = 50/(1.64485362695147 - -1.2815515655446) = 17.085808939996

The average speed µ = 140 - 1.64485362695147 * 17.085808939996 = 111.896345195648

Then, for the fraction exceeding 110 km/hr, z = (x-µ)/σ = (110 - 111.896345195648)/17.085808939996 =

-0.110989488546185, and 1 - Φ(-0.110989488546185) = 1 - normsdist(-0.110989488546185) =

0.544187658922233