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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.

Short Answer

Expert verified

Part (a) The distribution is roughly symmetric with a single peak.

Part (b) New mean,xN=557.3cm

New median, |x|N=557cm

Part (c) New standard deviation,SDN=4.713cm

Interquartile range,IQR=7.0cm

Step by step solution

01

Part (a) Step 1: Given information

02

Part (a) Step 2: Concept

The formula used:IQR=Q3-Q1

03

Part (a) Step 3: Explanation

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.

04

Part (b) Step 1: Calculation

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,

xN=x−20cm=577.3cm−20cm=557.3cm

New Median,

|x|N=|x|-20cm=577cm−20cm=557cm

Thus,

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

05

Part (c) Step 1: Calculation

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

IQR=Q3-Q1=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,

SDN=SD=4.713cm

Thus,

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

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