# Can you Check Over MY Work Please - Normal Distribution (Z-SCORE) Word Problem

#### tdotodot

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 = 0.3674 **Not sure if those italic/underlined numbers are correct on this line**
Probability = 0.3674 / 36.74%

Last edited:

#### chiro

MHF Helper
Hey tdotodot.

1) is correct
2) is not. Using R we get
> pnorm(-0.6,0,1)
[1] 0.2742531
3) Looks good. Again using R:
pnorm(0.8,0,1) - pnorm(-0.2,0,1)
[1] 0.3674043

#### tdotodot

Hey tdotodot.

1) is correct
2) is not. Using R we get
> pnorm(-0.6,0,1)
[1] 0.2742531

3) Looks good. Again using R:
pnorm(0.8,0,1) - pnorm(-0.2,0,1)
[1] 0.3674043
Thank you, really appreciate your help. Could you tell me how you got 0.2752531 for question #2? And what is "R"?

I am using the Z-Score chart, row 0.6 / column 0.00 shows the number 0.2258 and column 0.01 shows .2291. Other than that, looks good.

#### chiro

MHF Helper
pnorm(z,0,1) calculates the probability P(Z < z) where Z ~ N(0,1). I used this command to verify your answer.

R is an open source statistical package that is becoming one of the most popular platforms used in statistics.

The R Project for Statistical Computing