The lifespan of lightbulbs in a photographic machine is normally distributed with a mean of 210 hours and a standard deviation of 50 hours.

1) Determine the z-score of a light bulb with a lifespan of exactly 124hours.

z= x-mean/standard deviation

z= (124-210)/50

x= -1.72

Z-Score = -1.72

2) What is the probability that a randomly chosen light bulb would have a lifespan of less than 180 hours?z= x-mean/standard deviation

z= (180-210)/50

z= -0.60

p(x<180) = p(z<-0.60) = 0.2258

Probability= 0.2258 / 22.58%

3) What is the probability that a randomly chosen light bulb would have a lifespan of between 200 and 250 hours?

z= x-mean/standard deviation FOR BOTH 200 and 250:

For200:

z= x-mean/standard deviation

z= (200-210)/50

z= -0.2

For250:

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**Not sure if those italic/underlined numbers are correct on this line**0.3674

Probability = 0.3674 / 36.74%