Question

15. The life of a 9-volt battery is normally distributed with a mean of = 2000 hours and a standarddeviation of = 40 hours. What is the proportionof batteries with an average life of 2100 hours ormore? (enter the answer as a percent rounded tothe nearest hundredth as needed)

Answer 1

To answer this question we will use the following formula for the z-score:

`\begin{gathered} z=(x-\mu)/(\sigma), \\ where\text{ }x\text{ is the observed value, }\mu\text{ is the mean, and }\sigma\text{ is the standard deviation.} \end{gathered}`

Substituting = 2000h, = 40h and x=2100h we get:

`z=(2100h-2000h)/(40h).`

Simplifying the above result we get:

`z=(100h)/(40h)=2.5.`

Using tables we get:

`P(z<2.5)=0.99379`

Then:

`P(z\geq2.5)=1-0.99379=0.00621.`

The above result as a percentage is:

`0.621\%.`

Answer:

`0.62\%.`