Question

The batteries from a certain manufacturer have a mean lifetime of 850 hours, with a standard deviation of 70 hours. Assuming that the lifetimes are normally distributed, complete the following statements. Choose a percentage of batteries (68%, 75%, 95%, 99.7%)

Answer 1

`\begin{gathered} (a)\text{ }95\% \\ \\ (b)780.7hrs\text{ and }919.3hrs \end{gathered}`

Step-by-step explanation

(a) Given that:

Mean, μ = 850 hours

Standard deviation, σ = 70 hours

To find the percentage of batteries that have lifetimes between 710 and 990 hours, we have to find:

Hence, 95% of batteries have lifetimes between 710 hours and 990 hours.

(b) The middle 68% of the score represents:

This implies that:

`z=\pm0.99`

Using the z-score formula, we have to find the scores corresponding to the z-scores of -0.99 and 0.99.

That is:

`x=z*\sigma+\mu`

For z = -0.99:

`\begin{gathered} x=(-0.99*70)+850 \\ \\ x=780.7\text{ }hrs \end{gathered}`

For z = 0.99:

`\begin{gathered} x=(0.99*70)+850 \\ \\ x=919.3\text{ }hrs \end{gathered}`

Therefore, approximately 68% of the batteries have lifetimes between 780.7 hours and 919.3 hours.