A) Correct
B) p = 0.27425
C) Correct
1. The lifespan of a certain brand of light bulbs in a photographic machine is normally distributed with a mean of 210 hours and a standard deviation on 50 hours.
a) Determine the z-score of a light bulb that has a lifespan of exactly 124 hours?
z= x - mean / standard deviation
z= 124-210 / 50
z= -86 / 50
z= -1.72
Z-Score is -1.72
b) What is the probability that a randomly chosen light bulb would have a lifespan of less than 180 hours? (Not sure if correct)
z= x- mean / Standard Deviation
z= 180 - 210 / 50
z= -0.60
P(x<180) = P (z<-0.60) = 0.2258
So there is a 0.2258 (22.58%) chance the lifespan is less than 180 hours.
c) What is the probability that a randomly chosen light bulb would have a lifespan of between 200 and 250 hours? (Not sure if right, italic/bold part I'm a little iffy about)
z= x - mean / standard deviation
z= 200-210 / 50
z= -0.2
z= x - mean / standard deviation
z= 250-210 / 50
z= 0.8
P(200<z<250) = p(-0.2<z<-0.8) = P(z<0.8) - p(z<-0.2) = 0.7881 - 0.4207 = 0.3674
There's a 0.3674 (36.74%) chance the lifespan is between 200 and 250.
I used stattrek.com to get it that far, but my Standard normal distribution table gives me the same number to fewer decimal places.
Here's a pic of a standard normal distribution table.
http://classes.engr.oregonstate.edu/...rmal_table.gif
At '-0.6' you go to the '00' column as we are finding '-0.60'
And it gives us a value for p [0.2743]