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Math Help - Question on normal distribution #2

  1. #1
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    Question on normal distribution #2

    A manufacturer of electric light bulbs finds that his bulbs have an average life span of 1200 hours and a S.D of 200 hours.Assume that the distribution of lifetimes is normal.

    What should be the guaranteed life of the bulbs if the manufacturer is prepared to replace 5% of the bulbs sold?

    How should I go about doing this question?
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  2. #2
    MHF Contributor Unknown008's Avatar
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    So, let X be the number of hours of life span of a bulb.

    X ~ N(1200, 200^2)

    P(X < x) = P(Z < z) = 0.05

    Find the z value that corresponds to this probability, then convert it into hours using:

    z = \dfrac{x - \mu}{\sigma}
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    I tried doing that but the answer was wrong.The solution said 0.5-0.05=0.45 and find the correspoding Z value,they also said that this Z value would be negative.I don't really understand it.
    The answer is 870hours btw.
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    Quote Originally Posted by mngeow View Post
    A manufacturer of electric light bulbs finds that his bulbs have an average life span of 1200 hours and a S.D of 200 hours.Assume that the distribution of lifetimes is normal.

    What should be the guaranteed life of the bulbs if the manufacturer is prepared to replace 5% of the bulbs sold?

    How should I go about doing this question?
    X is normally distributed random variable X\sim(1200,200^2) measuring the life span of light bulbs.

    You have to determine the value A (number of hours such) that the life span of 95% of light bulbs will last longer than A hours. That means that only 5% of light bulbs will have life span shorter than A and thus such A should be the guaranteed life of light bulbs.

    P(X>A)=0.95
    1-P(X<A)=0.95
    P(X<A)=0.05
    F_X(A)=0.05
    \Phi(\frac{A-1200}{200})=0.05

    Now use the table values for standard normal distribution and you'll see that \Phi(-1.65)\approx 0.05 so you form the equation
    \frac{A-1200}{200}=-1.65
    A=870.
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    Quote Originally Posted by MathoMan View Post
    X is normally distributed random variable X\sim(1200,200^2) measuring the life span of light bulbs.

    You have to determine the value A (number of hours such) that the life span of 95% of light bulbs will last longer than A hours. That means that only 5% of light bulbs will have life span shorter than A and thus such A should be the guaranteed life of light bulbs.

    P(X>A)=0.95
    1-P(X<A)=0.95
    P(X<A)=0.05
    F_X(A)=0.05
    \Phi(\frac{A-1200}{200})=0.05

    Now use the table values for standard normal distribution and you'll see that \Phi(-1.65)\approx 0.05 so you form the equation
    \frac{A-1200}{200}=-1.65
    A=870.
    I only understood until the P(X<A)=0.05 part,I got lost after that.What does Fx(A)=0.05 mean? and whats the funny symbol in the line after that?
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    MHF Contributor Unknown008's Avatar
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    Quote Originally Posted by mngeow View Post
    I tried doing that but the answer was wrong.The solution said 0.5-0.05=0.45 and find the correspoding Z value,they also said that this Z value would be negative.I don't really understand it.
    The answer is 870hours btw.
    That means that you are asked to use a z table different from mine.

    So, from where I left:

    P(Z < z) = 0.05

    Is your graph like this: http://p1cture.me/images/51394697395543890592.png
    Or like this: http://p1cture.me/images/88238005700832427228.png

    ?
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    Its the one in the 2nd link,where the shaded area is on the right hand side.
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  8. #8
    MHF Contributor Unknown008's Avatar
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    Ok.

    You need to find the z value for this area: http://p1cture.me/images/17640253637132137970.png

    But you don't have this in your table. The equivalent of this area is this area: http://p1cture.me/images/50823189974773945850.png

    However, this isn't in your table either. But you have this area: http://p1cture.me/images/30470114981255486093.png

    The probability of the last area is 0.5 - 0.05 = 0.45, okay?

    What is the value of z for this area? in your table, you should read z = 1.65.

    But when you look at the original distribution (first graph), you know that z must be negative as it's value is on the left of the mean, so, the z value is -1.65.

    From there, you can then work back the value of X, the maximum lifetime of the bulb for it to fall in the 5% worst quality.

    Is it clear now?
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    Yes.I understand now.Thanks
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  10. #10
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    Quote Originally Posted by mngeow View Post
    I only understood until the P(X<A)=0.05 part,I got lost after that.What does Fx(A)=0.05 mean? and whats the funny symbol in the line after that?
    F_X(A) is the value of the distribution function of the random variable X at point A.
    \Phi a uppercase Greek letter phi, commonly used to denote the distribution function of the standard normal random variable.

    If X is normal random variable with parameters called expectation and variance,  X\sim {\cal N}(\mu, \sigma ^2) then distribution function of that random variable is usually denoted with F_X and it is defined as F_X(a)=P(X<a)
    Standard normal random variable is normal random variable with parameters 0 and 1, and to distinct it from other normal variables its distribution function, although defined in the same manner, is usually denoted with \Phi:
    X\sim {\cal N}(0, 1) ,  F_X(a)=\Phi(a)=P(X<a)
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