Chapter 10.3, Problem 12PPS

a.

To determine

To find outliers in the given set of data

Answer to Problem 12PPS

0.204 is the outlier in the given data set.

Explanation of Solution

Given information:Data set is

0.267,0.305,0.304,0.201,0.284,0.302,0.311,0.289,0.300 and 0.292

Calculation:

Sorting data in an ascending order,

0.201,0.267,0.284,0.289,0.292,0.300,0.302,0.304,0.305, 0.311

An outlier is an observation that lies an abnormal distance from other values in a random sample from a population. Analysing the given data set.

Hence, Outlier is 0.201. It is because this value is very far away from all data points in the given dataset.

b.

To determine

To calculate range and 5 number summary.

Answer to Problem 12PPS

The range is 0.110.

The five number summary is 0.201, 0.304, 0.146, 0.289, 0.311

Explanation of Solution

Given information:Dataset is

0.267, 0.305, 0.304, 0.201, 0.284, 0.302, 0.311, 0.289, 0.300 and 0.292

Formula used:

Range= Maximum- minimum

Lower quartile= $\frac{(N+1)}{4}$

Median = $\frac{(\frac{N^{th}}{2})ob+{(\frac{N}{2}+1)}^{th}ob}{2}$

Upper quartile= $\frac{3(N+1)}{4}$

Where, N is number of tiles/data points

Calculation:

Sorting data in an ascending order,

0.201, 0.267, 0.284, 0.289, 0.292, 0.300, 0.302, 0.304, 0.305, 0.311

Minimum is the lowest value in a dataset. Hence,

Minimum is 0.201

Maximum is the highest value in the dataset. Hence,

Maximum is 0.311

For range, substituting the value in the formula of range,

  $Range=0.311-0.201$

On solving,

  $Range=0.110$

Hence, range is 0.110

For median, substituting the value in the formula of median,

  $=\frac{(\frac{{10}^{th}}{2})ob+{(\frac{10}{2}+1)}^{th}ob}{2}$

On solving,

  $=\frac{5^{th}+6^{th}}{2}$

Substituting the value of 5^th and 6^th observations,

  $=\frac{0.292+0.300}{2}$

On solving,

  $=0.146$

Hence, median is 0.146

For Lower quartile, substituting the value,

  $=\frac{{(10+1)}^{th}}{4}ob$

So,

  $=\frac{11}{4}ob$

On solving,

  $=2.75$

On solving,

  $=2.75$

On rounding off the value,

  $\approx 3$

Substituting the value,

  $Q1=0.284$

For Upper quartile, substituting the values,

  $=\frac{3{(10+1)}^{th}}{4}ob$

So,

  $={\frac{33}{4}}^{th}ob$

On solving,

  $=8.25$

On rounding off the value,

  $\approx 8^{th}ob$

Substituting the value,

  $Q3=0.304$

The five number summary is 0.201, 0.284, 0.146, 0.304, 0.311

c.

To determine

To identify outlier and find mean and median

Answer to Problem 12PPS

Outlier is 0.201. Mean is 0.28 and Median is 0.29

Explanation of Solution

Given information: the given dataset is,

0.267, 0.305, 0.304, 0.201, 0.284, 0.302, 0.311, 0.289, 0.300, 0.292

Formula used:

Median formula is,

Median= $\frac{{(n+1)}^{th}}{2}observation$

Mean formula is,

Mean= $\frac{{\displaystyle \sum _{i=1}^n{X}_i}}{n}$

Where, n=total number of terms,

Calculation:

Sorting data,

0.201, 0.267, 0.284, 0.289, 0.292, 0.300, 0.302, 0.304, 0.305 and 0.311

Total number of data points= 10

Outlier is 0.201

Removing the outlier, now the new data set is

0.267, 0.284, 0.289, 0.292, 0.300, 0.302, 0.304, 0.305, 0.311

Now, the number of data points =9

For median, substitutingthe values,

  = ${(\frac{9+1^{}}{2})}^{th}ob$

On solving,

  =5^thob

Substituting the value of 5^th observation,

  =0.292

Hence, median is 0.292

For mean, substituting the values,

  $=\frac{0.267+0.284+0.289+0.292+0.300+0.302+0.304+0.305+0.311}{9}$

On solving,

  $=0.26$

Hence, mean is 0.26

d.

To determine

To find lowest batting average.

Answer to Problem 12PPS

0.267 is the lowest batting average.

Explanation of Solution

Given information:The data set is

0.201, 0.267, 0.284, 0.289, 0.292, 0.300, 0.302, 0.304, 0.305, 0.311

0.201 is an outlier

Analysing the dataset, 0.267 is the first observation after the outlier. Hence, 0.267 is thelowest batting average. As after an outlier, it is least batting average and is close to other batting averages.

Hence, 0.267 is the lowest batting average.