Chapter 0.9, Problem 11E

a.

To determine

To find an outlier in the given set of data mean, median, mode, range and standard deviation of the given set of data with and without the outlier

Answer to Problem 11E

14.9

With outlier: Mean is 16.325, median is 16.5, mode is 16.5, range is 1.9 and standard deviation is 0.436

Without outlier: Mean is 16.4, median is 16.5, mode is 16.5, range is 1 and standard deviation is 0.296

Explanation of Solution

Given:

The given set of weights of cereal box is

16.7, 16.8, 15.9, 16.1, 16.5, 16.6, 16.5, 15.9, 16.7, 16.5, 16.6, 14.9, 16.5, 16.1, 15.8, 16.7, 16.2, 16.5, 16.4, 16.6

Calculation:

The given data can be arranged in ascending order to determine the outlier

  

On examination of data, it can be said that majority of the weights of cereal boxes are in the range of 15.8 to 16.8 except for one cereal which weighs 14.9 ounces.

The reason why 14.9 is considered as an outlier is because difference between 14.9 and the immediate next weight is larger as compared any other two immediate weights.

For mean of the given data, the weight can be averaged as

  $\frac{14.9+15.8+15.9+15.9+16.1+16.1+16.2+16.4+16.5+16.5+16.5+16.5+16.5+16.6+16.6+16.6+16.7+16.7+16.7+16.8}{20}$ which gives the result $16.325$

For the median of the given data, average of the middle two terms from the data arranged in ascending order has to be calculated.

  

Here, the middle two terms are $16.5$ and $16.5$ so their average will also be $16.5$ .

Mode can be calculated by finding the weight that occurs most often in the data.

By observing the data, it can be said that $16.5$ is the mode or modal value for the given data.

Range is the difference between the most extreme values. Here the two extreme values are $14.9$ and $16.8$ , so the range will be $16.8-14.9=1.9$

Standard deviation can be calculated by the formula $\sigma =\sqrt{\frac{{\displaystyle \sum _{i=1}^n{(x_i-\mu )}^2}}{N}}$

Here, $\sigma$ is the standard deviation, $x_i$ is each value of weight, $\mu$ is the mean of the weights, N is the total number of weights.

Value of $x_i-\mu$ is calculated in the given table.

  

Value of ${(x_i-\mu )}^2$ is calculated in the table below

  

The summation of the squared value is $3.805$ .

Now to put in formula it comes out to be

  $\sigma =\sqrt{\frac{3.805}{20}}$

  $\sigma =\sqrt{0.19025}$

  $\sigma =0.436$

Now after removing the outlier the mea can be given by

  $\frac{15.8+15.9+15.9+16.1+16.1+16.2+16.4+16.5+16.5+16.5+16.5+16.5+16.6+16.6+16.6+16.7+16.7+16.7+16.8}{19}$

which gives $16.4$

Here too, the median comes out to be 16.5 as it the middle value.

The mode for this set of data will also be the same, that is 16.5 as it is repeated most number of times.

The range however decreases to $16.8-15.8=1$

For standard deviation once again the Value of $x_i-\mu$ is calculated in the given table.

  

Value of ${(x_i-\mu )}^2$ is calculated in the table below

  

The summation of the squared value is $1.68$

Now to put in formula it comes out to be

  $\sigma =0.296$

  $\sigma =\sqrt{0.088}$

  $\sigma =0.296$

By removing the outlier, here the mean increased by 0.075 as the outlier was at lower extreme.

There was no effect on median because the middle value was same.

There was no effect on mode as the most repetitive value was same.

The range however decreased after removing the outlier from 1.9 to 1

Standard deviation also decreased from 0.436 to 0.296 thus affirming the continuity of the new set of data.

Conclusion:

14.9 is considered as an outlier because it has a large variation from the immediate higher weight compared to the majority of the data.

After removing the outlier generally mean increases or decrease. Median may change or not and if changed, the change isn’t very large. Mode however tends to remain same as the most repetitive value is not an outlier thus remaining unchanged. Standard deviation generally decreases as the data becomes more continuous.

(b)

To determine

To find an outlier of a new set If 17.35, which will be replaced for 14.9, will be an outlier in the new set of data.

Answer to Problem 11E

Yes, 17.35 will be considered as an outlier in the new set of data.

With outlier: Mean is 16.447, median is 16.5, mode is 16.5, range is 1.55 and standard deviation is 0.357

Without outlier: Mean is 16.4, median is 16.5, mode is 16.5, range is 1 and standard deviation is 0.296

Explanation of Solution

Given:

A new set of data is formed by removing outlier from the original data and replacing it with 17.35.

  

Calculation:

On examining the data in ascending order, 17.35 seems to be an outlier as every two adjacent weights don’t differ by more than two whereas 17.35 differs by more five making it an outlier

For mean of the given data, the weight can be averaged as

  $\frac{15.8+15.9+15.9+16.1+16.1+16.2+16.4+16.5+16.5+16.5+16.5+16.5+16.6+16.6+16.6+16.7+16.7+16.7+16.8+17.35}{20}$ which gives the result $16.447$

For the median of the given data, average of the middle two terms from the data arranged in ascending order has to be calculated.

  

Here, the middle two terms are $16.5$ and $16.5$ so their average will also be $16.5$ .

Mode can be calculated by finding the weight that occurs most often in the data.

By observing the data, it can be said that $16.5$ is the mode or modal value for the given data.

Range is the difference between the most extreme values. Here the two extreme values are $17.35$ and $15.8$ , so the range will be $17.35-15.8=1.55$

Standard deviation can be calculated by the formula $\sigma =\sqrt{\frac{{\displaystyle \sum _{i=1}^n{(x_i-\mu )}^2}}{N}}$

Here, $\sigma$ is the standard deviation, $x_i$ is each value of weight, $\mu$ is the mean of the weights, N is the total number of weights.

Value of $x_i-\mu$ is calculated in the given table.

  

Value of ${(x_i-\mu )}^2$ is calculated in the table below

  

The summation of the squared value is $2.554$ .

Now to put in formula it comes out to be

  $\sigma =\sqrt{\frac{2.554}{20}}$

  $\sigma =\sqrt{0.1277}$

  $\sigma =0.357$

Now after removing the outlier the mean can be given by

  $\frac{15.8+15.9+15.9+16.1+16.1+16.2+16.4+16.5+16.5+16.5+16.5+16.5+16.6+16.6+16.6+16.7+16.7+16.7+16.8}{19}$ which gives $16.4$

Here too, the median comes out to be 16.5 as it the middle value.

The mode for this set of data will also be the same, that is 16.5 as it is repeated most number of times.

The range however decreases to $16.8-15.8=1$

For standard deviation once again the Value of $x_i-\mu$ is calculated in the given table.

  

Value of ${(x_i-\mu )}^2$ is calculated in the table below

  

The summation of the squared value is $1.68$

Now to put in formula it comes out to be

  $\sigma =\sqrt{0.088}$

  $\sigma =0.296$

By removing the outlier, here the mean decreased by 0.047 as the outlier was at higher extreme.

There was no effect on median because the middle value was same.

There was no effect on mode as the most repetitive value was same.

The range however decreased after removing the outlier from 1.55 to 1

Standard deviation also decreased from 0.357 to 0.296 thus affirming the continuity of the new set of data.

Conclusion:

17.35 is considered as an outlier because of the large difference from the immediate lower weight as compared to majority of the weights.

After removing the outlier generally mean increases or decrease. Median may change or not and if changed, the change isn’t very large. Mode however tends to remain same as the most repetitive value is not an outlier thus remaining unchanged. Standard deviation generally decreases as the data becomes more continuous.

c.

To determine

To find the possible cause of outlier in the given situation.

Answer to Problem 11E

Maybe caused due to lesser quantity of cereal in boxes or due to error in measurement of weights of the boxes.

Explanation of Solution

Given set of data is of weights of cereal boxes.

Generally filling of cereal boxes is mechanized which fill exact weight of cereal in each boxes.

An outlier is an unusually high or low data as compared to majority of the data.

During filling of boxes due to some error in machine lesser weight of cereal may have been filled in one of the boxes.

Another reason could be error in measurement of weight of the boxes.

Conclusion:

Outlier may be caused due to some error in processing of cereal or during measuring the weight of the cereal.