Question 1162961
to answer question A, do the following:
mean = 110
standard deviation = 2
use the following online calculator:
<a href = "https://homepage.divms.uiowa.edu/~mbognar/applets/normal.html" target = "_blank">https://homepage.divms.uiowa.edu/~mbognar/applets/normal.html</a>
your inputs will be:
mean = 110
standard deviation = 2
x = 106
select p(X < x) and hit the return.
the calculator then tells you that the probability of getting a chocolate bar with a weight less than 106 grams is equal to .02275.
here's what it looks like:
<img src = "http://theo.x10hosting.com/2020/073001.jpg" >


to answer question B, do the following.
set the mean equal to 0 and the standard deviation equal to 1
set p(X < x) to .01
hit the return and the calculator tells you that the x is equal to -2.32635.
that's the z-score that will have an area to the left of it equal to .01
here's what it looks like.
<img src = "http://theo.x10hosting.com/2020/073002.jpg" >


use the z-score formula of z = (x - m) / s
set z = -2.32635.
set x = 115
set s = 2
the z-score formula becomes -2.32635 = (115 - m) / 2
multiply both sides of this equation by 2 to get -2.32635 * 2 = 115 - m
subtract 115 from both sides of this equation to get -2.32635 * 2 - 115 = -m
simplify to get -119.6527 = -m
multiply both sides of the equation by -1 to get:
119.6527 = m
that's your mean.
confirm by using the calculator again.
inputs will be:
mean = 119.6527
standard deviation = 2
select p(X < x)
x = 115
hit the return and the calculator tells you that p(X < x) = .01
here's what it looks like.
<img src = "http://theo.x10hosting.com/2020/073003.jpg" >


this means that the probability of getting a chocolate weighing less than 115 grams when the mean is 119.6527 grams and the standard deviation is 2 grams is equal to .01.
.01 is the same as 1/100.


i'll be available to answer anyquestions you might have in regard to this.
you could solve it by using the z-score table, but there's no necessity to do so unless you are told you can't use an online calculcor to solve it.