Chapter 12.5, Problem 10PPS

a)

To determine

To find : How many batteries will last between $90,000$ and $10,000$ miles.

Answer to Problem 10PPS

  $9500$ Batteries

Explanation of Solution

Given Information :

A car battery costs god, normally distributed, mean is $1,00,000$ miles and standard deviation $10,000$ miles. Company makes $2,00,000$ batteries.

Calculation:

Using Empirical Rule,

  $\begin{array}{l}\text{90k is 1}\sigma \text{ away, 34\%}\\ \text{110k is 1}\sigma \text{ away, 18}\text{.5\%}\\ P(90k<X<110k)=(34+13.5)\\ =47.5\%\end{array}$

Since, $\text{20000 × 47}\text{.5\% = 9500}$ batteries will last $\text{90k}$ to $\text{110k}$ miles.

  

b)

To determine

To Find:- $\text{How many batteries will last more than 120000 miles ?}$

Answer to Problem 10PPS

  $\text{500 batteries}$

Explanation of Solution

Calculation :

  $\begin{array}{l}\text{120k miles is 2}\sigma \text{ away}\text{.}\\ \text{Using Empirical Rule,}\\ \text{P(}x>120k)=2+0.5=2.5\%\\ \text{since, 2}\text{.5\% × 20000}=500\text{ batteries}\\ \text{Will last more than 120k miles}\text{.}\end{array}$

c)

To determine

  $\text{How many batteries will last less than 90000 miles ?}$

Answer to Problem 10PPS

  $\text{7300 batteries}$

Explanation of Solution

Calculation :

  $\begin{array}{l}\text{Using Empirical Rule,}\\ \text{90000 miles is at 1}\sigma \text{, 34\% values}\\ \text{p(x < 90000)}=34+2+0.5=36.5\%\text{ values}\text{.}\\ \text{Since, 36}\text{.5\% × 20000 = 7300 batteries}\text{.}\end{array}$

d)

To determine

  $\begin{array}{l}\text{Probability that a randomly bought car battery will last between}\\ \text{50000 and 110000 miles ?}\end{array}$

Answer to Problem 10PPS

  $\text{95\%}$

Explanation of Solution

Calculation :

  $\begin{array}{l}\text{P(80k}<x<110k\text{)=}\\ \text{Using Empirical Rule}\\ 80kis\text{ at 2}\sigma \text{; }13.5\%\\ \text{90k is at 1}\sigma \text{; 34\%}\\ \text{100k is mean, 34\%}\\ \text{110k at 1}\sigma \text{, 13}\text{.5\%}\\ \text{Probability is = (13}\text{.5 + 13}\text{.5 + 34 + 34)}\\ \text{= 95\%}\end{array}$