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anonymous

  • one year ago

The time required to finish a test normally distributed with a mean of 60 minutes and a standard deviation of 10 minutes. What is the probability that a student will finish the test between 40 and 70 minutes?

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  1. misssunshinexxoxo
    • one year ago
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    P(40<X<70) = P[(40-60)/10 < z < (70-60)/10] = P(-2 < z < 1) Next re-write so that it can be looked up in a Standard Normal table P(-2 < z < 1) = P(z < 1) - P(z < -2) = 0.8413 - 0.0228 = 0.8185

  2. anonymous
    • one year ago
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    is .8185 the answer?

  3. misssunshinexxoxo
    • one year ago
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    Yes

  4. anonymous
    • one year ago
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    does that need to have a percentage answer? Should it be in percentage form?

  5. misssunshinexxoxo
    • one year ago
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    No just like that

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