Question

suppose that the lifetimes of light bulbs are approximately normally distributed, with a mean of 56 hours and a standard deviation of 3.2 hours. with this information, answer the following questions
(a) what proportion of light bulbs will last more than 62 hours?
(b) what proportion of light bulbs will last 50 hours or less?
(c) what proportion of light bulbs will last between 58 and 61 hours?
(d) what is the probability that a randomly selected light bulb lasts less than 45 hours?
(a) the proportion of light bulbs that last more than 62 hours is 0.0304
(round to four decimal places as needed.)
(b) the proportion of light bulbs that last 50 hours or less is
(round to four decimal places as needed.)

Explanation:

Step1: Calculate the z - score

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 56\) (mean), \(\sigma=3.2\) (standard deviation), and \(x = 50\) (value).

$$z=\frac{50 - 56}{3.2}=\frac{-6}{3.2}=- 1.875$$

Step2: Find the proportion using the standard normal distribution table

We want to find \(P(X\leq50)\), which is equivalent to \(P(Z\leq - 1.875)\).
Looking up the value of \(z=-1.875\) in the standard - normal table (or using a calculator with a normal - distribution function, e.g., in Excel: NORM.S.DIST(-1.875, TRUE)).
The value of \(P(Z\leq - 1.875)\approx0.0304\)

Answer:

\(0.0304\)