1. ## I am new..

can I post my Probability and Statistics question? or should I say questions.... I believe I can because this is the section of Probability and Statistics! duh! okay anyhoo on to my question! this question has me stumped more than a tree! haha get it! sorry anyhoo here it is, but before I write it there is two parts to this annyoing and confusing question for part A I showed my work but for part B I dont know how to do it! in fact I have no clue

Part A
A box of cookies contains a means of mass of 180g with a standard deviation of 2g. One box had a mass that correspondence to a z score of -1.5 how many grams of cookies did the box contain?

what I have done:
We have
-1.5 = (x - 180) / 2
-3 = x - 180,
x = 180 - 3 = 177 grams

Did I by miracle stumble on the right answer?

Part B:

if the mean mass of theses boxes is assumed to be normally distributed, what is the probability a box of cookies will contain less than 179g?

here the lightbulb in my head is trying so hard to turn on that it is already burnt!

anyhoo any help with theses Probability and Statistics question would be a huge greatfulness on my part! and if I could I would go through the computer and hug you!

2. Originally Posted by robins

Part A
A box of cookies contains a means of mass of 180g with a standard deviation of 2g. One box had a mass that correspondence to a z score of -1.5 how many grams of cookies did the box contain?

what I have done:
We have
-1.5 = (x - 180) / 2
-3 = x - 180,
x = 180 - 3 = 177 grams

Did I by miracle stumble on the right answer?
The z-score of a statistic x from a population with mean m and standard
deviation s, is defined to be:

z =(x-m)/s

In your case m=180g, and s= 2g. So if

z = -1.5 = (x-180)/2

so rearranging:

x = (-1.5)*2+180

so: x=180-3=177 g.

RonL

3. Originally Posted by robins
Part B:

if the mean mass of theses boxes is assumed to be normally distributed, what is the probability a box of cookies will contain less than 179g?

here the lightbulb in my head is trying so hard to turn on that it is already burnt!
Compute the z-score for 179g, then look the z-score up in a table of the
cumulative standard normal distribution, and that will tell you the required
probability.

z = (179 -180)/2 = -0.5.

P(z<=-0.5) = 0.3085

RonL