A certain inspection process has been designed in a manner that its time took a normal distribution with a mean equal to 40 minutes and a standard deviation equal to 8 minutes. Find the following quantities:

• Compute the six-sigma efficiency of the process, i.e., the probability that the inspection time spent by a randomly chosen product is in the range given by the mean minus three standard deviations and the mean plus three standard deviations.

• Probability of time between the six-sigma interval is much easier to describe using standard normal Z(0,1) since μ = 0, and σ = 1: 

P(μ − 3σ ≤ z ≤ μ + 3σ ) = P(−3 ≤ z ≤ 3) = P(z ≤ 3) − P(z ≤ −3) = 0.9986 − 0.0013 = 0.9973

• Lean agile principles will be applied to improve the process and reduce its time. How much should the standard deviation decrease so that the range of six standard deviations is in the interval [37, 43]?

• Considering μ = 40 and σ = 8 to find the new upper bound of the interval: μ + 3 · σ = 43 → σ = (43−μ)/3 → σ = (43−40)/3 → σ = 1

So, the reduction should be from 8 minutes to 1 minute standard deviation.

from scipy.stats import norm

import matplotlib.pyplot as plt

import numpy as np

#creating an array of values between

#-3 to 3 with a difference of 0.2

x = np.arange(-3, 3, 0.2)

mean = 0

std = 1

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 = [-3, 3]

mean = 0

std = 1

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

print(y)

[0.0013499 0.9986501] 

pinterval = y[1] - y[0]

print('P(-3 <= X <= 3) = ',pinterval)

P(-3 <= X <= 3) = 0.9973002039367398