Batteries of each type have lifetimes before recharge that can be assumed independent and Normally distributed. The mean and standard deviation of the lifetimes of the Duxcell batteries are 10 and 2 minutes respectively, the mean and standard deviation for the Infinitycell batteries are 14 and 3 minutes respectively.
Part a) What is the probability that a Duxcell battery will last longer than an Infinitycell battery?,

Solution:
X = N(10, 2^2)
This means that:
mean is 10
variance is 2^2

Y = N(14, 3^2).

Asked:P(X > Y)
= P(X - Y > 0).
mean of Z = X - Y
= 10 - 14
variance = 2^2 + 3^2
Z = N(-4, 2^2 + 3^2).

Answer Link

mathjohndough mathjohndough · Answer 2 · 2018-12-09T07:03:09+01:00

Answer:

0.1562

Step-by-step explanation:

Step 1

In this tep we define the normal distributions in this problem. . The lifetime of the Duracell battery is a normal random variable X with parameters (10,4). The lifetime of the Infinitycell battery is a normal random variable with parameters (14.9).

Step 2

Define the new normal distribution formed by the combination of these two distributions. We define a random variable Z such that

,[tex]Z=X-Y.[/tex]

This will be a random variable with mean of [tex]\mu_x-\mu_y=10-14=-4[/tex]. This random variable will have a variance of [tex]\mu_z^2 =\mu_x^2+\mu_y^2=2^2+3^2=13[/tex]. The mean for this distribution is [tex]\mu_z=\sqrt{13}[/tex] . The random variable Z is normally distributed with parameters (-4,13)

Step 3

The next step is to calculate the probability that [tex]Z=X-Y>0[/tex]. For that we perform the calculation shown below. Along the way , we will use the method of transforming any normal distribution to the standard normal distribution.This transformation is [tex]z=\frac{x-u}{\sigma}[/tex].

[tex]P(Z>0)=P(X-Y>0)=P(Z>0)\\P(Z>0)=1-P(Z<0)\\P(Z>0)=1-P(\tfrac{0-(-4)}{\sqrt{13}})=1-P(\tfrac{4}{\sqrt{13}})\\P(Z>0)=1-0.8438\\P(Z>0)=0.1562[/tex]