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anonymous

  • 5 years ago

Could someone please help me? A set of five natural numbers has a mean, median and mode. The mean is 100. Two of the numbers are 19 and 96. If the median is two less than the mode, what is the value of the mode?

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  1. anonymous
    • 5 years ago
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    You have five natural numbers that sum to 500: a + b + c + d + e = 500 You're told two are 19 and 96, leaving c, d and e, say: 19 + 96 + c + d + e = 500 The median, c, say, is two less than the mode: c = d - 2 = e - 2 e and d will be the mode (i.e. d = e) since 19, 96 and c will be three distinct numbers and the mode is the score with the highest frequency (the other scores have frequency 1, and since there are only two numbers left, these two scores (i.e. frequency two) must be the mode). So we have, 19 + 96 + c + d + e = 19 + 96 + (d-2) + d + d = 19 + 96 + 3d - 2 = 500 Hence, 3d = 387, which means d = 129. Since d was one of the scores that comprised the mode, the mode is 129.

  2. anonymous
    • 5 years ago
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    To check, the scores are: 19, 96, 127, 129, 129 They sum to 500, 127 is indeed the median, 129 the mode.

  3. radar
    • 5 years ago
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    A stupid question I'm sure, but how did you know the sum was 500?

  4. anonymous
    • 5 years ago
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    Oh, because the mean is 100 and there are five numbers: \[\frac{a+b+c+d+e}{5}=100 \iff a+b+c+d+e=500\]

  5. radar
    • 5 years ago
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    AAHHHH so. thanks, you are already a fan. Like the new icon for your profile.

  6. anonymous
    • 5 years ago
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    Thank you :) At least brick96 appreciated the help.

  7. radar
    • 5 years ago
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    I'm sure he did, and I learned something to. Have a good day.

  8. anonymous
    • 5 years ago
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    You too.

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