# Normal Distribution Help Needed

1. May 7, 2014

### vanceEE

1. The problem statement, all variables and given/known data
"Cans of lemon juice are supposed to contain 440 ml of juice. It is found that the actual volume of
juice in a can is normally distributed with mean 445 ml and standard deviation 3.6 ml."

It is found that 94% of the cans contain between 445−c ml and 445+c ml of juice.
(ii) Find the value of c

2. Relevant equations
X~N(445,3.6)
$$p(445-c ≤ x ≤ 445+c) = 0.94$$

3. The attempt at a solution
$$p(445-c ≤ x ≤ 445+c) = 0.94$$
$$p(x≤ 445+c)-p(x≥445-c)= 0.94$$
$$p(x≤ 445+c)-p(x≥445-c)= 0.94$$
$$p(x≤ 445+c)-[1-p(x≤445-c)]= 0.94$$
$$p(x≤ 445+c)+p(x≤445-c)= 1.94$$

..whenever I standardize this normal distribution, the terms cancel out. How would I go about solving this problem?

2. May 7, 2014

### vanceEE

Please disregard, I came up with a solution!

$$P(445-c≤x≤445+c) = 0.94$$
$$P(\frac{-c}{3.6}≤z≤\frac{c}{3.6}) = 0.94$$
$$\phi(\frac{c}{3.6})-\phi(\frac{-c}{3.6}) = 0.94$$
$$\phi(\frac{c}{3.6})-[1-\phi(\frac{c}{3.6})] = 0.94$$
$$2\phi(\frac{c}{3.6}) = 1.94$$
$$\phi(\frac{c}{3.6}) = 0.97$$
$$\frac{c}{3.6} = \phi^{-1}(0.97)$$
$$c = 3.6(1.88079...) = 6.77$$