Question

An electrical firm manufactures light bulbs that have a length of life that is normally distributed with mean equal to 778 hours and a standard deviation of 43 hours. Find the probability that a bulb burns between 755 and 892 hours.

Answer 1

Let $X=$ a length of life of the light bulb: $X\sim N(\mu, \sigma^2).$

Then $Z=\dfrac{X-\mu}{\sigma}\sim N(0, 1)$

Given $\mu=778\ hr, \sigma=43\ hr$

$$P(755<X<892)=P(X<892)-P(X\leq755)$$

$$=P(Z<\dfrac{892-778}{43})-P(Z\leq\dfrac{755-778}{43})$$

$$\approx P(Z<2.651163)-P(Z\leq-0.534884)$$

$$\approx0.9959892-0.2963651\approx0.699624$$

The probability that a bulb burns between 755 and 892 hours is 0.699624.