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anonymous

  • one year ago

The life expectancy of a typical lightbulb is normally distributed with a mean of 2,000 hours and a standard deviation of 27 hours. What is the probability that a lightbulb will last between 1,975 and 2,050 hours? I figure that since one standard deviation is 27, we are looking at 3 standard deviations away from the mean(1 less than the mean and 2 more than the mean). Is this right? Where do I go from here?

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  1. anonymous
    • one year ago
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    The answer options are as follows: A 0.17619 B 0.32381 C 0.79165 D 0.96784

  2. anonymous
    • one year ago
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    I have no idea how to do this, will fan and medal

  3. MrNood
    • one year ago
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    no - it's not 3sd from the mean it's from -1 to +2sd from mean |dw:1433548569142:dw| So you know tharea between mean and -1 you know th earea between mean and +2 so add them up (use 68, 95, 99.7 rule)

  4. MrNood
    • one year ago
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    (however - the answers are more accurate than the approximations you have used - so choose the answer nearest to your approximation

  5. IrishBoy123
    • one year ago
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    first you need to calc the z scores/ ie normalise. you want: \( z_{lower} = \frac{1,975 - \mu}{\sigma}; \ z_{upper} = \frac{2,050- \mu}{\sigma}\) \( \mu = 2,000; \ \sigma = 27\) then you need to look \( z_{lower}\) and \( z_{upper}\) up in your tables. or on your calculator/ spread sheet. or here is a fun visual interactive: https://www.mathsisfun.com/data/standard-normal-distribution-table.html

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