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Math Help - normally distributed lifespan

  1. #1
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    normally distributed lifespan

    I have the general idea to approach this problem ; however, I don't understand it completely , so I would like to get some help.

    A manufacturer knows that their items have a normally distributed lifespan, with a mean of 10.1 years, and standard deviation of 1.8 years. The lowest 7% of items will last less than how many years?
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  2. #2
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    Quote Originally Posted by skorpiox View Post
    I have the general idea to approach this problem ; however, I don't understand it completely , so I would like to get some help.

    A manufacturer knows that their items have a normally distributed lifespan, with a mean of 10.1 years, and standard deviation of 1.8 years. The lowest 7% of items will last less than how many years?
    You require the value of x* such that Pr(X < x*) = 0.07.

    Get the value of z* such that Pr(Z < z*) = 0.07 (from tables or technology).

    The it follows from Z = \frac{X - \mu}{\sigma} that z* = \frac{x* - 10.1}{1.8}.

    Substitute the value of z* and solve for x*.
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  3. #3
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by skorpiox View Post
    I have the general idea to approach this problem ; however, I don't understand it completely , so I would like to get some help.

    A manufacturer knows that their items have a normally distributed lifespan, with a mean of 10.1 years, and standard deviation of 1.8 years. The lowest 7% of items will last less than how many years?
    Compute the Z-score for 0.07 using a table or the InvNormal function on the TI graphing calculator.

    Translate this Z-score into X using:

    z = \frac{X - \mu}{\sigma}

    \mu = 10.1
    \sigma = 1.8
    X = unknown
    z = known

    My calculation indicates X = 7.4435. Can you confirm?
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  4. #4
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    Thank you guys .By the way , Apcalculus your answer is correct
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