Long jump A member of a track team was practicing the long jump and

recorded the distances (in centimeters) shown in the dot plot. Some numerical summaries of the data are also provided.

After chatting with a teammate, the jumper realized that he measured his jumps from the back of the board instead of the front. Thus, he had to subtract $20$ centimeters from each of his jumps to get the correct measurement for each jump.

a. What shape would the distribution of corrected long jump distance have?

b. Find the mean and median of the distribution of corrected long-jump distance.

c. Find the standard deviation and interquartile range (IQR) of the distribution of corrected long-jump distance.

The formula used:$IQR=Q_3-Q_1$

When the corrected long-jump lengths are reduced by $20$ cm, the corrected long-jump distances become uncorrected long-jump distances.

The form of the distribution will not be modified if each data value of the distribution is reduced by $20$ cm.

Thus,

The distribution will have the same form for both corrected and uncorrected long jump distances.

Keep in mind that

The distribution is essentially symmetric since the majority of the data values are roughly in the center of the dot plot.

Also, keep in mind that

The dot-plot has only one peak, which is at $576$

Thus,

The corrected long-jump distance distribution will have a fairly symmetrical structure with a single peak.

When the corrected long-jump lengths are reduced by $20$ cm, the corrected long-jump distances become uncorrected long-jump distances.

Because the mean and median represent the data set's center, they will both decrease by $20$ cm.

New Mean,

$\overline{x_N}=\overline{x}-20cm=577.3cm-20cm=557.3cm$

New Median,

${\left|x\right|}_N=\left|x\right|-20cm=577cm-20cm=557cm$

Thus,

Long-jump distances will be corrected to a new mean of $20$cm and a new median of $557$cm.

The interquartile range is the difference between the third and first quartiles.

$IQR=Q_3-Q_1=581.5cm-574.5cm=7.0cm$

When each data value is reduced by a factor of $20$, the data value variability is unaffected. As a result, variation measurements like IQR and standard deviation are unaffected.

New standard deviation,

${SD}_N=SD=4.713cm$

Thus,

The interquartile range for adjusted long jump distances will be $7.0$cm, with a standard deviation of $4.713$ cm.