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Math Help - Basic Statistics Word Problems

  1. #1
    Guarana
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    Basic Statistics Word Problems

    I have these problems due tomorrow - any help is greatly appreciated


    1. Suppose that a brewery has a filling machine that fills 12 oz bottles of beer. It is known that the amount of beer poured by this filling machine follows a normal distribution with a mean 12.32 oz and a standard deviation of 0.04 oz. Find the probability that a bottle contains between 12.22 and 12.28 oz.


    2. One year, pro sports players salaries averaged $1.5 million with a standard deviation of $0.6 million. Suppose a sample of 400 major league players was taken. Find the approximate probability that the average salary of the 400 players exceeded $1.1 million.


    3. Assume that the salaries of elementary school teachers in the US are normally distributed with a mean of $32,000 and a standard deviation of $6,000. What is the cutoff salary for teachers in the bottom 10%?


    4. The tread life of a particular brand of tire is a random variable best described by a normal distribution with mean of 60,000 miles and a standard deviation of 1,400 miles. What warranty should the company use if they want 96% of the tires to outlast the warranty?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Guarana View Post
    I have these problems due tomorrow - any help is greatly appreciated


    1. Suppose that a brewery has a filling machine that fills 12 oz bottles of beer. It is known that the amount of beer poured by this filling machine follows a normal distribution with a mean 12.32 oz and a standard deviation of 0.04 oz. Find the probability that a bottle contains between 12.22 and 12.28 oz.
    Compute the z scores of 12.22 and 12,28 ounces, they are:

    z1 = (12.22-12.32)/0.04 = -2.5

    and:

    z2 = (12.28-12.32)/0.04 = -1

    Now look these up in a table of the cumulative standard normal distribution:

    P(z<z1) = P(z<-2.5) = 0.0062

    P(z<z2) = P(z<-1) = 0.1587

    So:

    P(z1<z<z2)= P(z<z2) - P(z<z1) = 0.1587 - 0.0062 = 0.1525

    RonL
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by Guarana View Post
    2. One year, pro sports players salaries averaged $1.5 million with a standard deviation of $0.6 million. Suppose a sample of 400 major league players was taken. Find the approximate probability that the average salary of the 400 players exceeded $1.1 million.
    For a sample of size 400 the avaerage salary will have a mean m equal to the
    population mean, and a standarn deviation equal to the population SD divided
    by the square root of the sample size.

    Therefore the avaerage of our sample is a random variable with mean $1.5 million,
    and SD $0.6/sqrt(400) ~= $ 0.03 million

    Thus the z-score corresponding to $1.1 million is:

    z1 = (1.1 - 1.5)/0.03 = -13.3

    The approximate value of P(z<z1) = P(z<-13.3) ~= 0, so P(z>z1) ~=1.

    RonL
    Last edited by CaptainBlack; August 8th 2007 at 02:23 AM.
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