Chapter 10.4, Problem 10PPS

a.

To determine

To create a box and whisker plot.

Answer to Problem 10PPS

Median is 12.34, lower quartile is 12.18, upper quartile is 12.93, minimum is 11.96, and maximum is 13.60

Explanation of Solution

Given information:

S’s 100 meter dash (seconds)

12.20, 12.18, 12.87, 12.30, 12.50, 12.46, 12.35, 12.06, 12.04, 13.27, 12.14, 12.33, 13.60, 12.41, 12.38, 12.38, 12.93, 11.97, 13.57, 12.24, 12.28, 12.20, 12.16, 12.24, 11.96, 12.72, 13.06, 13.06, 13.12, 12.02, 13.09, 13.34

Formula Used: Median is,

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

Lower quartile is,

  $Q1=(n+1)\frac{1}{4}$

Upper quartile is,

  $Q3=(n+1)\frac{3}{4}$

Calculation:

Re-arranging the values in an ascending order,

11.96, 11.97, 12.02, 12.04, 12.06, 12.14, 12.16, 12.18, 12.20, 12.20, 12.24, 12.24, 12.28, 12.30, 12.33, 12.35, 12.38, 12.41, 12.72, 12.48, 12.50, 12.87, 12.93, 13.06, 13.09, 13.12, 13.27, 13.34, 13.57, 13.60

For median, substituting the values,

  $Median=\frac{{(\frac{30}{2})}^{th}+{(\frac{30}{2}+1)}^{th}}{2}term$

On solving,

  $Median=\frac{{15}^{th}+{16}^{th}}{2}term$

Substituting the values,

  $Median=\frac{12.33+12.35}{2}$

On solving,

  $median=12.34$

Hence, median is 12.34

For lower quartile substituting the values,

  $Q1=(30+1)\frac{1}{4}$

On solving,

  $Q1=7.75$

On rounding off,

  $\approx 8$

Substituting the value of 8^th observation,

  $Q1=12.18$

Hence, Lower quartile is 12.18.

For upper quartile substituting the values,

  $Q3=(30+1)\frac{3}{4}$

On solving,

  $Q3=23.25$

Rounding off the value,

  $\approx 23$

Substituting the value,

  $Q3=12.93$

Hence, Upper quartile is 12.93

Minimum is the smallest value of a data set. Hence substituting the value,

Minimum is 11.96

Maximum is the biggest value of a data set. Hence, substituting the value,

Maximum is 13.60

For box and whisker,

  

b.

To determine

To create a box and whisker plot, removing the data before her injury,

Answer to Problem 10PPS

Median is 12.40, Lower quartile is 11.96, Upper quartile is 13.09, Minimum is 11.96, and Maximum is 13.57

Explanation of Solution

Given information:

12.20, 12.18, 12.87, 12.30, 12.50, 12.46, 12.35, 12.06, 12.04, 13.27, 12.14, 12.33, 13.60, 12.41, 12.38, 12.38, 12.93, 11.97, 13.57, 12.24, 12.28, 12.20, 12.16, 12.24, 11.96, 12.72, 13.06, 13.06, 13.12, 12.02, 13.09, 13.34

S’s 100 meter dash (seconds), not including after injury,

11.96, 11.97, 12.14, 12.24, 12.33, 12.46, 12.50, 13.09, 13.34, 13.57

Formula Used: Median is,

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

Lower quartile is,

  $Q1=(n+1)\frac{1}{4}$

Upper quartile is,

  $Q3=(n+1)\frac{3}{4}$

Calculation:

For median, substituting the values,

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

On solving,

  $Median=\frac{5^{th}+6^{th}}{2}term$

Substituting the values,

  $Median=\frac{12.33+12.46}{2}$

On solving,

  $median=12.40$

For lower quartile substituting the values,

  $Q1=(10+1)\frac{1}{4}$

On solving,

  $Q1=2.75$

On rounding off,

  $\approx 3$

Substituting the value,

  $Q1=12.14$

Hence, Lower quartile is 12.14

For upper quartile substituting the value,

  $Q3=(10+1)\frac{3}{4}$

On solving,

  $Q3=8.25$

Rounding off the value,

  $\approx 8$

Substituting the value,

  $Q3=13.09$

Hence, Upper quartile is 13.09

Minimum is the smallest value of a data set. Hence substituting the value,

Minimum is 11.96

Maximum is the biggest value of a data set. Hence, substituting the value,

Maximum is 13.57

For box and whisker,

  

c.

To determine

To evaluate the effect removing the times recorded show on S’s record.

Answer to Problem 10PPS

The average speed of S is increased.

Explanation of Solution

Given information:

The average speed of S with and without her injury.

Concept Used:

Comparison of the median

Calculation:

The average speed of S is increased when removed the data of her speed after her injury.

Difference is,

  $=12.40-12.34$

On solving,

  $=0.06$

The difference comes of 0.06 seconds.