Question

the life of light bulbs is distributed normally. the variance of the lifetime is 900 and the mean lifetime of a bulb is 570 hours. find the probability of a bulb lasting for at least 626 hours. round your answer to four decimal places.

Explanation:

Step1: Calculate the standard deviation

The standard deviation $\sigma=\sqrt{900}=30$.

Step2: Calculate the z - score

The z - score formula is $z=\frac{x-\mu}{\sigma}$. Here, $x = 626$, $\mu=570$, $\sigma = 30$.
So, $z=\frac{626 - 570}{30}=\frac{56}{30}\approx1.87$.

Step3: Find the probability

We want $P(X\geq626)$. Using the standard normal distribution, $P(X\geq626)=1 - P(X < 626)$.
From the standard normal table, $P(Z<1.87)=0.9693$.
So, $P(X\geq626)=1 - 0.9693 = 0.0307$.

Answer:

$0.0307$