91影视

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

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

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

$\stackrel{炉}{x}=43.70$

localid="1649330618988" $MEDIAN=42.00$

$s=12.50$

$Q_1=35.50$

$Q_3=44.00$

Relationship feet - meter:

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

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" $$\begin{gathered} MEDIAN=\frac{42.00}{3.28} \\ 鈮�12.8049 \end{gathered}$$

Compute the value of$s$:

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

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

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" $$\begin{gathered} \stackrel{炉}{x}=43.70-42.60 \\ =1.1 \end{gathered}$$

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

$s=12.50$