water carton problem

Apr 2010
160
0
dfsdfdf
Any help offered in solving the below question will be much appreciated.
Thanks
Kingman

The volume of water in milliliters in cartons is normally distributed with mean 't' and the standard deviation 8. Measurements were taken of the volume in 900 of these cartons and it is found that 225 of them contained more then 1002 milliliters.
1) Calculate the value of 't' (0.674)
2) Three of them 900 cartons are chosen at random. Calculate the probability that exactly 2 of them contain more than 1002 milliliters. (0.140)
 
Oct 2009
303
33
1. Okay, to find \(\displaystyle \mu\):

Let X= the number of mL of water in a carton
\(\displaystyle pr(X\geq 1,002)=\frac{225}{900}=0.25\)

This is given in the problem.

Take note that \(\displaystyle pr(X\geq 1,002)=0.25\) is the same as \(\displaystyle pr(X\leq 1,002)=0.75\)

Now standardize the distribution.

\(\displaystyle pr(X\leq 1,002)=0.75\)=\(\displaystyle pr(Z\leq \frac{1,002-\mu}{8})=0.675\) (Taken from z-table)

Now, solve for \(\displaystyle \mu\) and you should get \(\displaystyle \mu = 996.6\)

2. This is a simple binomial distribution:

X~binomial(n=3,p=0.25)

X= the number of cartons containing more than 1,002 mL of water.

\(\displaystyle {n\choose x}(p)^{x}(1-p)^{(n-x)}\)

\(\displaystyle pr(X=2)= {3\choose 2}(0.25)^{2}(0.75)\approx 0.1406\)

Hmm.. I now see that you gave the answers in parentheses. I don't understand why (1) is 0.674. It's approximately the same number I took from the z-table but the problem states that we're trying to find \(\displaystyle \mu\) not \(\displaystyle Z\). Pretty weird. Maybe someone can correct me on that.
 
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Apr 2010
160
0
dfsdfdf
Thanks

Yes there is a typo error in the question.you are right .