SOLUTION: Light Bulbs have an advertised rating of 1750 lumens. If measurements are normally distributed with mean 1784 and standard deviation 38, what percentage of bulbs exceed the adverti

Algebra ->  Probability-and-statistics -> SOLUTION: Light Bulbs have an advertised rating of 1750 lumens. If measurements are normally distributed with mean 1784 and standard deviation 38, what percentage of bulbs exceed the adverti      Log On


   

Question 870306: Light Bulbs have an advertised rating of 1750 lumens. If measurements are normally distributed with mean 1784 and standard deviation 38, what percentage of bulbs exceed the advertised rating?
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!


Hi

z = (1750-1784)/38 = -34/38 = -.8974

P(x > 1750) = P(z > -.8974) = 81.52 %

use z-table or Calculator to find P(z ≥ -.8974) 

TI syntax is normalcdf(smaller z, larger z).

P(z > -.8974) = normalcdf(-.8974, 10)   | z= 10 being a placeholder