Question

4. Assume that the mean life of a particular brand of car battery is normally distributed with a mean of 28 months and a standard variation of 4 months.

(a) for a randomly selected battery of this make, what is the probability that it will last between 30 and 34 months?.

(b) what is the probability that a randomly selected battery of this make will fail within years of date of purchase?.

(c) after what time period will 60% of all batteries of this makes fail?.

Answer 1

We denote the respective random variable (life of battery) by $X$. We remind that for the normal distribution with parameters $\mu=28$ and $\sigma=4$ the probability density function is $p(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}$

(a) $P(30\leq X\leq34)=\int_{30}^{34}p(x)dx=\frac{1}{\sigma\sqrt{2\pi}}\int_{30}^{34}e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}dx\approx0.2417$

We computed the integral using the following code in Anaconda(https://www.anaconda.com/):

_{from scipy import integrate}

_{import numpy as np}

_import math

_{func = lambda x:(1/(4*math.sqrt(2)*math.sqrt(math.pi)))*math.exp(-1/2*((x-28)/4)*((x-28)/4))}

_{e = integrate.quad(func, 30, 34)}

_print(e)

(b) We assume that the time is 1 year. We get: $P(-\infty\leq X\leq12)=\int_{-\infty}^{12}p(x)dx=\frac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^{12}e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}dx\approx0.00003167$

(c) The aim is to find such $\alpha$ that: $P(-\infty\leq X\leq\alpha)=\frac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^{\alpha}e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}dx=0.4$ It means that the battery will fail with the probability 60%. We get that $\alpha\approx26.99$. We received $\alpha$ by substituting different values in the code (Anaconda):

_{from scipy import integrate}

_{import numpy as np}

_import math

_{func = lambda x:(1/(4*math.sqrt(2)*math.sqrt(math.pi)))*math.exp(-1/2*((x-28)/4)*((x-28)/4))}

_{e = integrate.quad(func,  -np.inf, 26.99)}

_print(e)