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 (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 \(x = 62\), \(\mu=56\), and \(\sigma = 3.2\).

$$z=\frac{62 - 56}{3.2}=\frac{6}{3.2}=1.875$$

Step2: Find the proportion

We want \(P(X>62)\), which is equivalent to \(P(Z > 1.875)\). Since \(P(Z>z)=1 - P(Z\leq z)\), and looking up \(P(Z\leq1.875)\) in the standard normal table (or using a calculator with a normal - distribution function), \(P(Z\leq1.875)\approx0.9696\).

$$P(Z>1.875)=1 - 0.9696 = 0.0304$$

Answer:

\(0.0304\)