1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Normal Distribution Help Needed

  1. May 7, 2014 #1
    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. jcsd
  3. May 7, 2014 #2
    Please disregard, I came up with a solution!

    $$P(445-c≤x≤445+c) = 0.94$$
    $$P(\frac{-c}{3.6}≤z≤\frac{c}{3.6}) = 0.94$$
    $$\phi(\frac{c}{3.6})-\phi(\frac{-c}{3.6}) = 0.94$$
    $$\phi(\frac{c}{3.6})-[1-\phi(\frac{c}{3.6})] = 0.94$$
    $$2\phi(\frac{c}{3.6}) = 1.94$$
    $$\phi(\frac{c}{3.6}) = 0.97$$
    $$\frac{c}{3.6} = \phi^{-1}(0.97)$$
    $$ c = 3.6(1.88079...) = 6.77 $$
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted