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Thread: Normal Probability and finding standard deviation

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    Normal Probability and finding standard deviation




    "The weekly earnings of fastfood restaurant employees are normally distributed with a mean of $395. If only 1.1% of the employees have a weekly income of more than $429.35, what is the value of the standard deviation of the weekly earnings of the employees?"

    Can someone not just provide and answer, but HOW to come to this answer? I am confused. I can't even find this type of question answered in my college book. Calculations would be great.

    Thanks,

    Lorelei7

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    Re: Normal Probability and finding standard deviation

    Does your book have a section on Normal Distributions?

    You need to find the Z-score from a table like this:

    http://www.beanactuary.org/exams/pre...ion_Table.html

    Find the Z-score for 1 - 0.011 = .989

    Then solve Z = \frac{X - \mu}{\sigma}

    Z = \frac{429.35 - 395.00}{\sigma}

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    Lorelei7 (10-22-2013)

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    Re: Normal Probability and finding standard deviation

    If you knew the mean and standard deviation would you be able to find the proportion of employees that have a weekly income of more than 429.35?
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    Re: Normal Probability and finding standard deviation

    so;
    0989=429.35-395/stand dev
    .989 = 34.35/stand dev then multiply stand dev on both sides = .989(std dv) = 34.35
    .989/.989 (to get std. dv alone) = 34.35/.989 = std dev=34.7321 ?
    Is this correct? From this I am still unsure if the 34.7321 is correct?

    Please forgive me as IDH the sigma sign handy; as well as being confused if this is the answer?

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    Re: Normal Probability and finding standard deviation


    Sorry but no. .989 is not the Z-score. You need to look up .989 in the Normal Distribution Table and find the Z-score. Then use that number in the equation Z = \frac{X - \mu}{\sigma}

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