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Math Help - Percentage of Grades

  1. #1
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    Percentage of Grades

    On a standardized test with a normal distribution, the mean is 88. If the standard deviation is 4, the percentage of grades that would be expected to lie between 80 and 96 is closest to:
    (1) 5 (3) 68
    (2) 34 (4) 95


    I'm not too clear on standard deviation questions.
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  2. #2
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    The interval 80-96% is 2 standard deviations either side of the mean.

    If you have learnt confidence intervals you would know that for a 95% confidence interval Z = 1.96, which is almost two standard deviations away from the mean. (For Z, mean is 0 and standard deviation is 1). So 95% would be the answer.

    You could standardise the X value (the test scores) to the Z value and use a normal distribution table otherwise. To convert to Z, remember Z = \frac{X - \mu}{\sigma}

    Finding half of the interval:

    P(88 < X < 96)

    = P\left( \frac{88 - 88}{4} < Z < \frac{96 - 88}{4} \right)

    = P(0 < Z < 2)

    Looking at a normal distribution table, the probability of this is 0.4772. Remember we just found the probability of the upper half of the interval in the question. So multiplying by 2 gives a probability very close to 0.95, i.e. 95%.

    Hope you can understand.
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  3. #3
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    ok........

    Quote Originally Posted by nzmathman View Post
    The interval 80-96% is 2 standard deviations either side of the mean.

    If you have learnt confidence intervals you would know that for a 95% confidence interval Z = 1.96, which is almost two standard deviations away from the mean. (For Z, mean is 0 and standard deviation is 1). So 95% would be the answer.

    You could standardise the X value (the test scores) to the Z value and use a normal distribution table otherwise. To convert to Z, remember Z = \frac{X - \mu}{\sigma}

    Finding half of the interval:

    P(88 < X < 96)

    = P\left( \frac{88 - 88}{4} < Z < \frac{96 - 88}{4} \right)

    = P(0 < Z < 2)

    Looking at a normal distribution table, the probability of this is 0.4772. Remember we just found the probability of the upper half of the interval in the question. So multiplying by 2 gives a probability very close to 0.95, i.e. 95%.

    Hope you can understand.
    Interesting stuff.
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