Chapter 2.1, Problem 21E

(a)

To determine

Shape of the distribution of corrected long jump distance.

Answer to Problem 21E

The distribution is roughly symmetric with a single peak.

Explanation of Solution

Given information:

  

The corrected long − jump distances when decreased by 20 cm, they become uncorrected long − jump distances.

When each data value of the distribution gets decreased by 20 cm, the shape of the distribution will not be affected.

Thus,

Both corrected and uncorrected long − jump distances will have same shape of distribution.

Note that

Since most of the data values roughly lie in the middle of the dotplot, the distribution is roughly symmetric.

Also note that

There is a single peak in the dotplot, which is at 576.

Thus,

The shape of the distribution of corrected long − jump distances will be roughly symmetric with a single peak.

(b)

To determine

Mean and median of the distribution of corrected long − jump distance.

Answer to Problem 21E

New mean,

  ${\overline{x}}_{\text{N}}=557.3\text{cm}$

New median,

  ${\left|x\right|}_{\text{N}}=557\text{cm}$

Explanation of Solution

Given information:

  

The corrected long − jump distances when decreased by 20 cm, they become uncorrected long − jump distances.

Since mean and median represent the center of data set, both of them will also decrease by 20 cm.

New Mean,

  ${\overline{x}}_{\text{N}}=\overline{x}-20\text{cm}=577.3\text{cm}-20\text{cm}=557.3\text{cm}$

New Median,

  ${\left|x\right|}_{\text{N}}=\left|x\right|-20\text{cm}=577\text{cm}-20\text{cm}=557\text{cm}$

Thus,

The corrected long − jump distances will new mean of 557.3 cm and new median of 557 cm.

(c)

To determine

Standard deviation and interquartile range of the distribution of corrected long − jump distance.

Answer to Problem 21E

New standard deviation,

  ${\text{SD}}_{\text{N}}=4.713\text{cm}$

Interquartile range,

  $IQR=7.0\text{cm}$

Explanation of Solution

Given information:

  

Difference between 3^rd and 1^st quartile gives interquartile range.

Interquartile range,

  $IQR=Q_3-Q_1=581.5\text{cm}-574.5\text{cm}=7.0\text{cm}$

When every data value gets decreased by 20, the data values variability will not be affected. Thus, variation measures such as IQR and standard deviation remain unchanged.

New standard deviation,

  ${\text{SD}}_{\text{N}}=\text{SD}=4.713\text{cm}$

Thus,

The corrected long − jump distances will have an interquartile range of 7.0 cm and standard deviation of 4.713 cm.