• April 21st 2013, 01:33 PM
tdotodot
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.
• April 21st 2013, 01:39 PM
A) Correct

B) p = 0.27425

C) Correct
• April 22nd 2013, 08:25 AM
tdotodot
Quote:

A) Correct

B) p = 0.27425

C) Correct

Can you tell me how you get 0.27425? On the Z-score table, the Z-score 0.6 has a probability of 0.2258? Or am I reading it wrong?
• April 22nd 2013, 12:30 PM
$z = \frac{x - \mu}{\sigma}$
$z = \frac{180 - 210}{50}$
$z=-0.6$