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

1. 3. can also be expressed as (22/7)

2. -4.5

3. 3.6

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