An automotive guide indicates the Porsche 911 as the car that best retains value. It is expected that a new car of US$ 87,500 has a value of US$48,125 after 3 years of use. Assuming that the price distribution of all Porsches 911 at three years old follows a Normal distribution with a mean of US$48.125 and a standard deviation of US$1600. Find the following probabilities:

• One vehicle chosen at random has a selling price of between US$46,000 and US$49,000,

• One vehicle chosen at random has a selling price higher than US$49,000.

Using the computational commands to obtain the normal distribution cumulative probabilities with parameters μ = 87, and σ = 500:

• Probability of price between US$46,000 and US$49,000:

P(46, 000 ≤ X ≤ 49, 000) = P(X ≤ 49, 000) − P(X ≤ 46, 000) = 0.7078 − 0.0921 = 0.6157

• Probability of price higher than US$49,000:

P(X ≤ 49, 000) + P(X ≥ 49, 000) = 1 → P(X ≥ 49, 000) = 1 − P(X ≤ 49, 000) → P(X ≥ 49, 000) = 1 − 0.7078 → P(X ≥ 49, 000) = 0.2922

from scipy.stats import norm

import matplotlib.pyplot as plt

import numpy as np

#creating an array of values between

#40 to 120 with a difference of 2

x = np.arange(40, 120, 2)

mean = 80

std = 10

y = norm.pdf(x, loc = mean, scale = std)

plt.plot(x,y,'r-',x, y,'bo')

plt.show()

from scipy.stats import norm

import matplotlib.pyplot as plt

import numpy as np

x = [46000, 49000]

mean = 48125

std = 1600

y = norm.cdf(x, loc = mean, scale = std)

print(y)

[0.09206841 0.70776769] 

pinterval = y[1] - y[0]

print('P(46000 <= X <= 49000) = ',pinterval)

P(46000 <= X <= 49000) = 0.6156992859925272 

plower = 1 - y[1]

print('P(X >= 49000) = 1 - P(X <= 49000) = ',plower)

P(X >= 49000) = 1 - P(X <= 49000) = 0.29223230610840834