Statistics Question

Question

I need help on the following question. I do not know what to do and I really want to know how to solve these types of questions. I have a test coming up soon and when it comes to these types of questions I cannot solve them.

Thank you.

The amount of corn chips dispensed into a 12-ounce bag by the dispensing machine has been identified as possessing a normal distribution with a mean of 12.5 ounces and a standard deviation of 0.2 ounce. What chip amount represents the 67th percentile for the bag weight distribution? Round to the nearest hundredth.

Answer 1

12.5+.2=mean+one standard deviation=12.7 where .32/2=.16 > than 12.7, using the 68-95-99 rule

But that's the 84th percentile at 12.7 chips.

You want 67th percentile. 50th percentile is the mean at 12.5

67 is 17 more than 50, and 17 less than 84. 67 is half the distance from 50 to 84

and about half the way between the two percentiles

half of .2 is .1 so add .1 to 12.5 to get 12.60 approximately.

To get a more accurate reliable answer use z tables.

Best I can do just estimating.