Question

where the mean is 845.2 h and standard deviation of 94.2 h, what fraction of the bulbs has the capacity to function for ≤1100 hours?

Answer 1

By definition, cumulative distribution function is

$F(x) = P(X \leq x)$.

So to find what fraction of the bulbs has the capacity to function for ≤1100 hours we need to find F(1100). Let's assume that lifetime of the bulbs are normally distributed. Then

$F(x) = \frac{1}{2}[1 + erf(\frac{x-\mu}{\sigma \sqrt{2}})]$

$F(1100) = \frac{1}{2}[1 + erf(\frac{1100-845.2}{94.2\sqrt{2}})] = 0.9966 \approx 0.997$

Answer: 99.7%