The Question
The average lifetime of a light bulb is 3000 hours with a standard deviation of 696 hours. A random sample of 35 bulbs is taken.
(i) Calculate the probability that the average lifetime in a sample will be between 2670.56 and 2809.76 hours?
(ii) Calculate the probability that the average lifetime in a sample will last at least 3200 hours?
(iii) How large of a sample needs to be taken to provide a 0.01 probability that the average lifetime in the sample will be equal to or greater than 3219.24 hours?
My Attempt
(i) $p(2670.56 < x < 2809.76) = p(z < -1.62) - p(z < -2.8) = 0.5$
This is correct.
(ii) $p(x > 3200) = 1 - p(x < 3200) = 1 - p(z < 1.7) = 1 - 0.9554 = 0.0446$
This too is correct.
(iii) I do not understand how would I go about this section. Tips will be appreciated.
If the sample is of size $n$, then for the sample mean $\bar x$ you have $$\frac{\bar x - \mu}{\frac{\sigma}{\sqrt{n}}} \sim N(0, 1)$$
$$P(\bar x \geq 3219.24) = P(z \geq \frac{3219.24 - 3000}{696}\sqrt{n}) \stackrel{!}{=} 0.01$$
So, you need to find $n$ such that
$$\frac{3219.24 - 3000}{696}\sqrt{n} = z_{0.01} = 2.33$$
I leave the calculations up to you :-).