# I am new..

• Jun 5th 2007, 09:16 AM
robins
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 :eek: :confused:

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! :o

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! ;)
• Jun 6th 2007, 03:52 AM
CaptainBlack
Quote:

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
• Jun 6th 2007, 03:56 AM
CaptainBlack
Quote:

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! :o

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