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Where am I going wrong? (Standard Deviation)

  1. May 8, 2007 #1
    Hi, I was trying to solve this question but my answer is different to the one given by my textbook.

    "The weight of a randomly chosen plastic washer is normally distributed with mean 5g. Calculate the standard deviation in grams given that the probability that a randomly chosen washer weighs less than 3g is 0.123."

    I said X is equivalent to a normal distribution with mean 5 and variance A.
    As X is a continuous distribution, the binomial probability of it being less than 3g can be approximated to the normal distribution being less than 2.5g, so:
    P(X<2.5) = 0.123
    let sqrtA = B = standard deviation
    1 - P(z<2.5/b) = 0.123
    P(z<2.5/B) = 0.877
    Referring to statistical tables for the normal distribution I found 2.5/B to equal 1.16
    This yields a value of B (standard deviation) to be 2.155
    But my answer book says 1.72.

    Can anyone see where I'm going wrong?
  2. jcsd
  3. May 8, 2007 #2


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    Homework Helper

    I don't understand what you're doing here. Why are you "approximating" by 2.5g? And there's no need to bring in the variance at all.

    It's a very straightforward problem. You're given the probability of the washer weighing less than 3g to be 0.123 Let's say the value of 3g corresponds to a z-score of Z. Z is obviously negative since 3g is less than the mean of 5g.

    Meaning [tex]\int_{-\infty}^Zf(z)dz = 0.123[/tex]

    where f(z) is the normal distribution. Often tables are given for positive z-scores and you're supposed to use the symmetry of the normal curve to figure out the negative part. So we'll use the table from here : http://sweb.cz/business.statistics/normal01.jpg

    Look at the tail area shaded in that picture. Now picture the tail area in our problem. Imagine reflecting that tail area about the vertical axis (giving the same tail area, except now it's in the positive z-score region). To get the complementary area corresponding to the illustration in the reference, you have to subtract this area from one (the total area under the curve). With me so far?

    In mathematical terms, you want :

    [tex]\int_{-\infty}^{-Z}f(z)dz = 1 - 0.123 = 0.877[/tex]

    Right, now look at the table in the reference, search for 0.877, and you'll find it under a z-score of 1.16. So we know that (-Z) = 1.16 or Z = -1.16

    Any value x on the normal curve is given by x = mean + z*sigma, where z is the z-score and sigma is the std deviation.

    So set up the equation :

    3 = 5 + (-1.16)*sigma and solve for sigma.
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