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

i think it would be 43 movies that would have to be watched.

step-by-step explanation:

think about it. for the second plan, every month is 8 dollars. 8 x 12= 96. so thats the number for the second plan. now for the first plan, its annual meaning you have to pay 10 dollars every year. so now you have to get an equivalent amount of 86 since 96-10=86. 86 divided by 2 since every movie costs 2$ then you would have to watch 43 movies a year to get the same as the second plan. i really hoped this . have a wonderful day as well!