91Ó°ÊÓ

$84$engineering conferences were announced. Four conferences lasted two days. Thirty-six lasted three days. Eighteen lasted four days. Nineteen lasted five days. Four lasted six days. One lasted seven days. One lasted eight days. One lasted nine days.

X is the length (in days) of an engineering conference and f is the number of engineering conferences.

$84$engineering conferences were announced. Four conferences lasted two days. Thirty-six lasted three days. Eighteen lasted four days. Nineteen lasted five days. Four lasted six days. One lasted seven days. One lasted eight days. One lasted nine days.

$$\begin{gathered} M=\frac{\left(n+1\right)}{2}th\text{observation} \\ =\frac{84+1}{2}th\text{observation} \\ =\frac{85}{2}th\text{observation} \\ =42.5th\text{observation} \\ =\frac{43th+42th}{2}\text{observation} \\ M=\frac{4+4}{2} \\ M=2 \end{gathered}$$

First quartile:

localid="1648403247345" $$\begin{gathered} Q_1=\frac{\left(n+1\right)}{4}th\text{observation} \\ =\frac{84+1}{4}th\text{observation} \\ =\frac{85}{4}th\text{observation} \\ =21.25th\text{observation} \\ =21th+\frac{22th-21th}{4}th\text{observation} \\ =3+\frac{3-3}{4} \\ =3 \end{gathered}$$

Third quartile:

$$\begin{gathered} Q_3=\frac{3\left(n+1\right)}{4}th\text{observation} \\ =\frac{3\left(84+1\right)}{4}th\text{observation} \\ =\frac{3×85}{4}th\text{observation} \\ =63.75th\text{observation} \\ =63th+\frac{3\left(64th-63th\right)}{4}\text{observation} \\ =5+\frac{3\left(5-5\right)}{4} \\ =5 \end{gathered}$$

$84$engineering conferences were announced. Four conferences lasted two days. Thirty-six lasted three days. Eighteen lasted four days. Nineteen lasted five days. Four lasted six days. One lasted seven days. One lasted eight days. One lasted nine days.

The $65th$percentile:

role="math" localid="1648403626322" $$\begin{gathered} P_{65}=\frac{65\left(n+1\right)}{100}th\text{observation} \\ =\frac{65\left(84+1\right)}{100}th\text{observation} \\ =\frac{65\left(85\right)}{100}th\text{observation} \\ =55.25th\text{observation} \\ =55th+0.25\left(56th-55th\right)\text{observation} \\ {P}_{65}=4+0.25\left(4-4\right) \\ =4 \end{gathered}$$

$84$engineering conferences were announced. Four conferences lasted two days. Thirty-six lasted three days. Eighteen lasted four days. Nineteen lasted five days. Four lasted six days. One lasted seven days. One lasted eight days. One lasted nine days.

The $10th$percentile

$$\begin{gathered} P_{10}=\frac{10\left(n+1\right)}{100}th\text{observation} \\ =\frac{\left(84+1\right)}{10}th\text{observation} \\ =\frac{85}{100}th\text{observation} \\ =8.5th\text{observation} \\ =8th+0.5\left(9th-8th\right)\text{observation} \\ {P}_{10}=3+0.5\left(3-3\right) \\ =3 \end{gathered}$$

$$\begin{gathered} \text{Lowestvalue}=2 \\ \text{FirstQuartile}=3 \\ \text{Median}=4 \\ \text{Thirdquartile}=5 \\ \text{Highestvalue}=9 \end{gathered}$$

$84$engineering conferences were announced. Four conferences lasted two days. Thirty-six lasted three days. Eighteen lasted four days. Nineteen lasted five days. Four lasted six days. One lasted seven days. One lasted eight days. One lasted nine days.

$$\begin{gathered} \text{FirstQuartile=}3 \\ \text{ThirdQuartile}=5 \end{gathered}$$

The range of middle $50\%$data can be expressed by the area between the third quartile and first quartile of a given data set. hence the middle $50\%$of the conference last from $3$days to $5$days.

$84$engineering conferences were announced. Four conferences lasted two days. Thirty-six lasted three days. Eighteen lasted four days. Nineteen lasted five days. Four lasted six days. One lasted seven days. One lasted eight days. One lasted nine days.

$$\begin{gathered} \stackrel{¯}{x}=\frac{∑fx}{∑f} \\ =\frac{331}{84} \\ =3.94 \end{gathered}$$

$84$engineering conferences were announced. Four conferences lasted two days. Thirty-six lasted three days. Eighteen lasted four days. Nineteen lasted five days. Four lasted six days. One lasted seven days. One lasted eight days. One lasted nine days.

$$\begin{gathered} S=\sqrt{\frac{∑f(x-\stackrel{¯}{x}{)}^2}{∑f-1}} \\ =\sqrt{\frac{136.7024}{84-1}} \\ =\sqrt{\frac{136.7024}{83}} \\ =\sqrt{1.647} \\ =1.2834 \end{gathered}$$

$84$engineering conferences were announced. Four conferences lasted two days. Thirty-six lasted three days. Eighteen lasted four days. Nineteen lasted five days. Four lasted six days. One lasted seven days. One lasted eight days. One lasted nine days.

According to the given frequency table, the mode of the given data is $3$.

$84$engineering conferences were announced. Four conferences lasted two days. Thirty-six lasted three days. Eighteen lasted four days. Nineteen lasted five days. Four lasted six days. One lasted seven days. One lasted eight days. One lasted nine days.

Since the majority should participate in the conference, it is reasonable to choose the value of mode as the length of the conference. The length of the conference is $3$days, the value of the mode.

$84$engineering conferences were announced. Four conferences lasted two days. Thirty-six lasted three days. Eighteen lasted four days. Nineteen lasted five days. Four lasted six days. One lasted seven days. One lasted eight days. One lasted nine days.

Reason $1$: According to the box-and-whisker plot in part (e), it is clear that the middle $505$of data are plotted in the range of $3-5$days. Hence half of the conferences lasted from three to five days. So that range of das should be the popular length of days.

Reason $2$: when considering the frequencies, there are $73$conference that lasted in $3-5$days. It is $86.9\%$from all conferences held. Hence that range of days should be the populated length of days.