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nignag [31]
3 years ago
15

The physical plant at the main campus of a large state university receives daily requests to replace fluorescent light bulbs. Th

e distribution of the number of daily requests is bell-shaped and has a mean of 38 and a standard deviation of 6. Using the empirical rule, what is the approximate percentage of lightbulb replacement requests numbering between 38 and 56?
Mathematics
1 answer:
ankoles [38]3 years ago
6 0

We have been given that the distribution of the number of daily requests is bell-shaped and has a mean of 38 and a standard deviation of 6. We are asked to find the approximate percentage of lightbulb replacement requests numbering between 38 and 56.

First of all, we will find z-score corresponding to 38 and 56.

z=\frac{x-\mu}{\sigma}

z=\frac{38-38}{6}=\frac{0}{6}=0

Now we will find z-score corresponding to 56.

z=\frac{56-38}{6}=\frac{18}{6}=3

We know that according to Empirical rule approximately 68% data lies with-in standard deviation of mean, approximately 95% data lies within 2 standard deviation of mean and approximately 99.7% data lies within 3 standard deviation of mean that is -3\sigma\text{ to }3\sigma.

We can see that data point 38 is at mean as it's z-score is 0 and z-score of 56 is 3. This means that 56 is 3 standard deviation above mean.

We know that mean is at center of normal distribution curve. So to find percentage of data points 3 SD above mean, we will divide 99.7% by 2.

\frac{99.7\%}{2}=49.85\%

Therefore, approximately 49.85\% of lightbulb replacement requests numbering between 38 and 56.

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Given:

The frequency distribution table.

To find:

The mean average score on a test.

Solution:

The frequency distribution table is

Marks (x_i)                   Frequency(f_i)                  f_ix_i

     x                                     a                               xa

     y                                     b                               yb

     z                                     c                               zc

  Sum                              a+b+c                          xa+yb+zc

Now, the mean average score on the test is

Mean=\dfrac{\sum f_ix_i}{\sum f_i}

Mean=\dfrac{xa+yb+zc}{a+b+c}

Therefore, the mean average score on the test is \dfrac{xa+yb+zc}{a+b+c}.

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