The life expectancy (in hours) of an electric bulb is normally distributed with a mean of 5000 and a standard deviation of 1000. Find the probability that a bulb lasts for more than 6300 hours. Round answer to four decimal places. (TIP: calculate z-value and use z-probability distribution table.)

P (z>1.3) = 0.9032

Step-by-step explanation:

We are given:

Mean = 5000

Standard deviation = 1000

x = 6300

P (x>6300) = ?

z-score = ?

z-score = x - mean/standard deviation

z-score = 6300 - 5000/1000

z - score = 1300/1000

z-score = 1.3

So, P (x>6300) = P (z>1.3)

Looking at the z-probability distribution table and finding value:

P (z>1.3) = 0.9032

So, P (z>1.3) = 0.9032