## anonymous one year ago The life expectancy of a typical lightbulb is normally distributed with a mean of 2,000 hours and a standard deviation of 27 hours. What is the probability that a lightbulb will last between 1,975 and 2,050 hours? I figure that since one standard deviation is 27, we are looking at 3 standard deviations away from the mean(1 less than the mean and 2 more than the mean). Is this right? Where do I go from here?

1. anonymous

The answer options are as follows: A 0.17619 B 0.32381 C 0.79165 D 0.96784

2. anonymous

I have no idea how to do this, will fan and medal

3. MrNood

no - it's not 3sd from the mean it's from -1 to +2sd from mean |dw:1433548569142:dw| So you know tharea between mean and -1 you know th earea between mean and +2 so add them up (use 68, 95, 99.7 rule)

4. MrNood

(however - the answers are more accurate than the approximations you have used - so choose the answer nearest to your approximation

5. IrishBoy123

first you need to calc the z scores/ ie normalise. you want: $$z_{lower} = \frac{1,975 - \mu}{\sigma}; \ z_{upper} = \frac{2,050- \mu}{\sigma}$$ $$\mu = 2,000; \ \sigma = 27$$ then you need to look $$z_{lower}$$ and $$z_{upper}$$ up in your tables. or on your calculator/ spread sheet. or here is a fun visual interactive: https://www.mathsisfun.com/data/standard-normal-distribution-table.html