The batteries from a certain manufacturer have a mean lifetime of 840 hours, with a standard deviation of 70 hours. Assuming that the lifetimes are normally distributed, complete the following statements. (a) Approximately $$ 99.7 \% $$ of the batteries have lifetimes between $$ \square $$ hours and $$ \square $$ hours. (b) Approximately 3 B of the batteries have lifetimes between 700 hours and 980 hours.

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**(a)**  
The empirical rule (68–95–99.7 rule) for normal distributions tells us that approximately $$99.7\%$$ of values lie within 3 standard deviations of the mean. Here, the mean is $$\mu = 840$$ hours and the standard deviation is $$\sigma = 70$$ hours.

Calculate the lower bound:  
$$
840 - 3(70) = 840 - 210 = 630 \text{ hours}
$$

Calculate the upper bound:  
$$
840 + 3(70) = 840 + 210 = 1050 \text{ hours}
$$

Thus, approximately $$99.7\%$$ of the batteries have lifetimes between $$630$$ hours and $$1050$$ hours.

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**(b)**  
The interval from $$700$$ hours to $$980$$ hours can be expressed in terms of standard deviations. Compute the standardized $$z$$-scores:

For $$700$$ hours:
$$
z = \frac{700 - 840}{70} = \frac{-140}{70} = -2
$$

For $$980$$ hours:
$$
z = \frac{980 - 840}{70} = \frac{140}{70} = 2
$$

The empirical rule tells us that approximately $$95\%$$ of the values lie within 2 standard deviations of the mean.

Thus, approximately $$95\%$$ of the batteries have lifetimes between $$700$$ hours and $$980$$ hours.
**(a)** Approximately $$99.7\%$$ of the batteries have lifetimes between $$630$$ hours and $$1050$$ hours. **(b)** Approximately $$95\%$$ of the batteries have lifetimes between $$700$$ hours and $$980$$ hours.

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