Q. 8.11 - A First Course in Probability

Question

Each of the batteries in a collection of $40$batteries is equally likely to be either a type A or a type B battery. Type A batteries last for an amount of time that has a mean of $50$and a standard deviation of $15$; type B batteries last for a mean of $30$ and a standard deviation of 6.

(a) Approximate the probability that the total life of all $40$ batteries exceeds $1700$ 

(b) Suppose it is known that $20$ of the batteries are type A and $20$are type B. Now approximate the probability that the total life of all $40$ batteries exceeds $1700.$

Step-by-Step Solution

Answer

a) The approximate probability of all $40$ batteries exceeding $1700$ is $\approx .1492$

b) If $20$ batteries are type A and $20$ are type B, The approximate probability of all $40$ batteries exceeding $1700$ is$\approx .0838.$

Part (a) Step 1: Given information

Total batteries $=40$

Type A batteries have a mean of $50$ and a standard deviation of $15$ type B batteries last for a mean of $30$ and a standard deviation of $6$.

a) Find the approximate probability if it exceeds $1700$

b) Find the approximate probability if type A is $20$ and type B is $20.$

Part (a) Step 2: Explanation

Assume that there are $n=40$ batteries in a collection. Let $X_A$ represents the lifetime of a type A battery and $X_B$ represents the lifetime of a type B battery. It is given that:

${\mu }_A=E\left[X_A\right]=50,{\sigma }_A=15,{\mu }_B=E\left[X_B\right]=30,{\sigma }_B=6$

Now see the next two events

$A=\text{'}\text{'}$the battery is of type $A\text{'}\text{'}$

$B=\text{'}\text{'}$the battery is of type $B\text{'}\text{'}$

Then, because each battery in a collection has an equal chance of being either a type A or a type B battery, we have that

$p=P\left\{A\right\}=P\left\{B\right\}=\frac{1}{2}$

Let $N_A$ denote the number of type A batteries in a collection, and $N_B$denote the number of type B batteries in a collection of $n$ batteries. These random variables are then binomial random variables with parameters $n$ and p, and they are plainly not independent. Take note of this:

$N_A=40-N_B$

Therefore $X$ denotes the total life of all batteries

$X=\sum _{i=1}^{N_A}{X}_A+\sum _{i=1}^{N_B}{X}_B=\sum _{i=1}^{N_A}{X}_A+\sum _{i=1}^{40-N_A}{X}_B$

Now we can find the mean and variance of $X$

$E\left[X\right]=E\left[E\left[X\mid N_A\right]\right]\stackrel{\left(1\right)}{=}E\left[20N_A+1200\right]=20E\left[N_A\right]+1200\stackrel{\left(3\right)}{=}1600.$

Next use proposition $5.2$

$$\begin{gathered} Var\left(X\right)=E\left[Var\left(X\mid N_A\right)\right]+Var\left(E\left[X\mid N_A\right]\right) \\ \stackrel{ \left(1\right)}{ =}E\left[Var\left(X\mid N_A\right)\right]+Var\left(20N_A+1200\right) \\ \stackrel{\left(5\right)}{=}E\left[189N_A+1440\right]+Var\left(20N_A+1200\right) \\ =189E\left[N_A\right]+1440+400Var\left(N_A\right)\stackrel{\left(3\right)}{=}9220. \end{gathered}$$

Part (a) Step 3: Explanation

The probability of all $40$ batteries that the total life exceeds $1700$ is

$P\{X>1700\}$

Then $N_A=n_0$, the central limit theorem sums,$\sum _{i=1}^{n_0}{X}_A$ and $\sum _{i=1}^{40-n_0}{X}_B$ are normally distributed.

$X\mid N_A=n_0$ is also normally distributed, so we can tell that $X$ is a normally distributed variable

$$\begin{gathered} P\{X>1700\}=1-P\{X\le 1700\}=1-P\left\{\frac{X-1600}{\sqrt{9220}}\le \frac{1700-1600}{\sqrt{9220}}\right\}\approx 1-\Phi \left(\frac{1700-1600}{\sqrt{9220}}\right) \\ =1-\Phi (1.04)\stackrel{\text{ Table }5.1\text{ (textbook, Ch 5) }}{=}1-.8508=\mathbf{.}\mathbf{1492}. \end{gathered}$$

Part (b) Step 1: Given information

Total batteries $=40$

Type A batteries have a mean of $50$ and a standard deviation of $15$ type B batteries last for a mean of $30$ and a standard deviation of $6$.

a) Find the approximate probability if it exceeds $1700$

b) Find the approximate probability if type A is $20$ and type B is $20$.

Part (b) Step 2: Explanation

Let $X$ denote the total life of all $40$ batteries,

$X=\sum _{i=1}^{20}{X}_A+\sum _{i=1}^{20}{X}_B$

Assume $20$ are type A and $20$ are type B

Sums above are nearly regularly distributed, according to the central limit theorem. As a result, $X$ is also a variable with a roughly normal distribution. Since

$E\left[X\right]=20{\mu }_A+20{\mu }_B=1600$ and

$Var\left(X\right)=20{\sigma }_A^2+{\sigma }_B^2=5220$

We conclude

$$\begin{gathered} P\{X>1700\}=1-P\{X\le 1700\}=1-P\left\{\frac{X-1600}{\sqrt{5220}}\le \frac{1700-1600}{\sqrt{5220}}\right\} \\ \approx 1-\Phi \left(\frac{1700-1600}{\sqrt{5220}}\right)=1-\Phi (1.38)\stackrel{\text{ Table }5.1\left(\text{ textbook, Ch }5\right)}{=} \\ =1-.9162=\mathbf{.}\mathbf{0838}\mathbf{.} \end{gathered}$$