R2.4 Aussie, Aussie, Aussie A group of Australian students were asked to estimate the width of their classroom in feet. Use the dot plot and summary statistics below to answer the following questions.

(a) Suppose we converted each student's guess from feet to meters $(3.28\mathrm{ft}=1\mathrm{m})$. How would the shape of the distribution be affected? Find the mean, median, standard deviation, and IQR for the transformed data.

(b) The actual width of the room was $42.6$feet. Suppose we calculated the error in each student's guess as follows: guess $-42.6$. Find the mean and standard deviation of the errors. How good were the students' guesses? Justify your answer.

Approximately locate the median (equal-areas point) and the mean (balance point) on a density curve.

A group of Australian students were asked to estimate the width of their classroom in feet.

Converted each student’s guess from feet to meters $(3.28\mathrm{ft}=1\mathrm{m})$

The following are the measurements of the classroom's width in feet:

$\overline{x}=43.70$

localid="1649330618988" $MEDIAN=42.00$

$s=12.50$

$Q_1=35.50$

$Q_3=44.00$

Relationship feet - meter:

localid="1649916559499" $$\begin{gathered} 3.28ft=1m \\ \Rightarrow 1ft \\ =\frac{1}{3.28}m \end{gathered}$$

The interquartile range is the area between the third and first quantiles:

localid="1649916567621" $$\begin{gathered} IQR=Q_3-Q_1 \\ =44.00-35.50 \\ =8.50 \end{gathered}$$

To convert these measurements from feet to meter, multiply them by $3.28:$

Let's find the median:

localid="1649916583777" $$\begin{gathered} MEDIAN=\frac{42.00}{3.28} \\ \approx 12.8049 \end{gathered}$$

Compute the value of$s$:

localid="1649916591708" $$\begin{gathered} s=\frac{12.50}{3.28} \\ \approx 3.8110 \end{gathered}$$

localid="1649916599850" $$\begin{gathered} IQR=\frac{8.50}{3.28} \\ \approx 2.5915 \end{gathered}$$

The distribution shape will not be affected because every value in the data set is translated in the same way.

The actual width of the room was $42.6feet$.

Each student's guess as:$-42.6feet$

The following are the feet measurements of their classroom:

$\stackrel{-}{x}=43.70$

$s=12.50$

The number of errors in the distribution is reduced by$42.6$percent. By the same amount, the mean is reduced:

localid="1649916610130" $$\begin{gathered} \overline{x}=43.70-42.60 \\ =1.1 \end{gathered}$$

If all values decline by the same amount, the standard deviation will remain unchanged. (since the spread will not change):

$s=12.50$