Answer to Question #158579 in Statistics and Probability for Zwavhudi Mangenge

Question

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

Solution:

a) To solve this task we need to find the following probability:

$P\{30 \leq \xi \leq 34\}$

To perform this we will use the Laplace integral table, but to do this we will need the following transformation:

$P \{x_1 \leq \xi \leq x_2\} = P \{ (x_1-a)/\sigma \leq (\xi - a)/ \sigma \leq (x_2 - a)/\sigma\} = \Phi_0((x_2 - a)/ \sigma) - \Phi_0((x_1 - a)/\sigma).$

Where a - expected value (mean), a = 28;

$\sigma = std dev = 4$

$x_1 = 30, x_2=34$

From it, we get:

$\Phi_0((34-28)/4) - \Phi_0((30-28)/4) = \Phi_0(6/4) - \Phi_0(2/4) =$

$= \Phi_0(1.5) - \Phi_0(0.5) =$ by the table of Laplace Integrals $=$

$= 0.4332 - 0.1915 = 0.2417$

b) In this task, we need to find if battery fails within a year. It is the same as the probability of a car to live less than or equal to a year:

$P \{ \xi \leq12\}$ , because 1 year = 12 months

Actually, a car battery can not live for less 0 months, so we get lower boundary for our random variable:

$P \{0 \le \xi \le 12 \}$

Using the same logic, as in part a) we get $x_2 = 12, x_1 = 0$ , and:

$P \{0 \le \xi \le 12 \} = \Phi_0((12 - 28)/4) - \Phi_0((0-28)/4) =$

$= \Phi_0((-16)/4) - \Phi_0((-28)/4) = \Phi_0((-4)) -\Phi(-7) =$

Then, we will use the fact that $\Phi_0(-x) = -\Phi_0(x)$

$= -\Phi_0(4) - (-\Phi_0(7)) = \Phi_0(7) - \Phi_0(4) =$ from the table $=$

$= 0.49999 - 0.49996=0.00003$

since all the values of $\Phi_0(x), x>5$ are considered to be 0.49999

c) This problem can be restated as follows: probability that car battery lives for x months is 0.4. Therefore, it fails with 0.6 chance

Find x

It gives us this expression:

$P \{0 \le \xi \le x\} =0.4$

Rewriting the left part:

$P \{0 \le \xi \le x\} = \Phi_0((x-28)/4) - \Phi_0((0-28)/4) =$

from previous task, for 0 months we get:

$= \Phi_0((x-28)/4) + \Phi_0(7) = 0.4$

$\Phi_0((x-28)/4) + 0.49999 = 0.4$

$\Phi_0((x-28)/4) = -0,09999$

$-\Phi_0((x-28)/4) = 0.09999$

Then, remembering the formula:

$-\Phi_0(x) = \Phi_0(-x)$

We get:

$\Phi_0((28-x)/4) = 0.09999$

Using the Laplace table, getting slightly bigger value $\Phi_0(0.26) = 1.026$ :

$(28-x)/4 = 0.26$

$28-x = 1.04$

$x = 26.94$

Answer:

a) 0.2417

b) 0.00003

c) 26.94