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 to answer the following questions.

a. Suppose we converted each student’s guess from feet to meters (3.28 ft=1 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 in feet.

c. Find the percentile for the student who estimated the classroom width as 63 feet.

$$\begin{gathered} \overline{x}=43.70 \\ MEDIN=42.00 \\ s=12.50 \\ {Q}_1=35.50 \\ {Q}_3=44.00 \end{gathered}$$

$Percentile=\frac{Numberofdatavaluesthatarelessthantheindividual\text{'}sdatavalue}{Totalnumberofdatavalues}\times 100\%$

Relationship between feet and meter:

$3.2ft=1m\Rightarrow 1ft=\frac{1}{3.28}m$

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

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

To convert the measurement from feet to meters, multiply by $3.28$:

$$\begin{gathered} \overline{x}=\frac{43.70}{3.28}=13.3232 \\ MEDIAN=\frac{42.00}{3.28}=12.8049 \\ s=\frac{12.50}{3.28}=3.8110 \\ IQR=\frac{8.50}{3.28}=2.5915 \end{gathered}$$

The distribution's shape would not be modified because every data value is translated in the same way, and so the data values would be distributed in the same way.

Every guess reduced the error distribution by $42.6$ percent. Then the mean goes down by the same amount:

$\overline{x}=43.70-42.60=1.1$

Because the spread would not vary when the value reduced by the same amount, the standard deviation would remain unchanged.

$s=12.50$

The summary statistic has a value of $66$below n, indicating that there are $66$data values in the data set.

$4$dots are to the right of $63$on the dot plot, and $1$ dot is related with $63$implying that $66-4-1=61$dots of the $66$dots are to the left of $63$

Required percentile

$Percentile=\frac{Numberofdatavaluesthatarelessthantheindividual\text{'}sdatavalue}{Totalnumberofdatavalues}\times 100\%$

$$\begin{gathered} =\frac{61}{66}\times 100\% \\ =0.92\times 100\% \\ =92\% \end{gathered}$$

Therefore the percentile associated with $63$feet is the $92nd$percentile.