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.

2. ## Re: Questions Involving Z-Score, Can you please check my answer?

A) Correct

B) p = 0.27425

C) Correct

3. ## Re: Questions Involving Z-Score, Can you please check my answer?

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?

4. ## Re: Questions Involving Z-Score, Can you please check my answer?

I used stattrek.com to get it that far, but my Standard normal distribution table gives me the same number to fewer decimal places.

$\displaystyle z = \frac{x - \mu}{\sigma}$

$\displaystyle z = \frac{180 - 210}{50}$

$\displaystyle z=-0.6$

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]