Marketing and Consumer Behavior The city Kingston, Ontario, has approximately 37
ID: 3149431 • Letter: M
Question
Marketing and Consumer Behavior The city Kingston, Ontario, has approximately 3700 parking spaces. Metered parking is available on some streets and in certain parking garages, and the maximum length of stay at a metered spot is between one and three hours. 4 The probability density function for the length of time a car is parked (in hours) at a metered spot in a certain lot is given below. Suppose a car parked at a metered spot in this lot is selected at random. What is the probability that the car is parked for less than 2 hours? What is the probability that the 1.4 hours? What is the probability that the car is parked for more than 2.6 hours? W hat is the probability that the car is parked for between 1.4 and 2.6 hours?Explanation / Answer
AS PER THE GRAPH OR THE DISTRIBUTION THE MEAN = 2 (HIGHEST PEAK)
THE STANDARD DEVIATION =1 (AS 3-2=1, AND 2-1=1)
IT WILL BE A NORMAL DISTRIBUTION
WITH Z = (X-MEAN)/STANDARD DEVIATION
A) P(X<2) =
For x = 2, the z-value z = (2 - 2) / 1 = 0
Hence P(x < 2) = P(z < 0), now from the z table we will take the value of z score = 0
And that value will be the probability required.
= [area to the left of 0] = 0.5
B) P(X<1.4) =
For x = 1.4, the z-value z = (1.4 - 2) /1 =-0.6
Hence P(x < 1.4) = P(z < -0.6), now from the z table we will take the value of z score = -0.6
And that value will be the probability required.
= [area to the left of -0.6] = 0.2743
C) P(X>2.6) =
b) For x = 2.6, z = (2.6 - 2) / 1 = 0.6
Hence P(x > 2.6) = P(z >0.6) = [total area] - [area to the left of 0.6]
1 - [area to the left of 0.6]
now from the z table we will take the value of z score = 0.6
= 1 - 0.7257 = 0.2743
D) P(1.4<X<2.6) =
) For x = 1.4 , z = (1.4 - 2) / 1 = -0.6 and for x = 2.6, z = (2.6 - 2) / 1 = 0.6
Hence P(1.4 < x < 2.6) = P(-0.6 < z < 0.6) = [area to the left of z = 0.6] - [area to the left of -0.6]
= 0.7257 - 0.2743 = 0.4514