Question

In a recent issue of the IEEE Spectrum, $84$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. Let X = the length (in days) of an engineering conference.

a. Organize the data in a chart.

b. Find the median, the first quartile, and the third quartile.

c. Find the $65th$percentile.

d. Find the $10th$percentile.

e. Construct a box plot of the data.

f. The middle $50\%$of the conferences last from _______ days to _______ days.

g. Calculate the sample mean of days of engineering conferences.

h. Calculate the sample standard deviation of days of engineering conferences.

i. Find the mode.

j. If you were planning an engineering conference, which would you choose as the length of the conference: mean; median; or mode? Explain why you made that choice.

k. Give two reasons why you think that three to five days seem to be popular lengths of engineering conferences.

Short answer

(a)

X	f
2	4
3	36
4	18
5	19
6	4
7	1
8	1
9	1
Total	84

(b)

$$\begin{gathered} \text{Median}=4 \\ \text{FirstQuartile}=3 \\ \text{ThirdQuartile}=5 \end{gathered}$$

$\text{(c)}65th\text{percentile}=4$

$\text{(d)The10thpercentile}=3$

(e)

(f) The middle $50\%$of the conference last from $3$days to $5$days.

(g) Sample mean$=3.94$

(h) The sample standard deviation of days of engineering conferences$=1.28$

(i) Mode$=3$

(j) The length of the conference is $3$days.

Step-by-step solution

Given information (part a)

$84$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.

Explanation (part a)

X	f
2	4
3	36
4	18
5	19
6	4
7	1
8	1
9	1
Total	84

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

Given information (part b)

$84$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.

Explanation (part b)

X	f	cf
2	4	4
3	36	40
4	18	58
5	19	77
6	4	81
7	1	82
8	1	83
9	1	84
Total	84

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

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:

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

Given information (part c) 

$84$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.

Explanation (part c)

The $65th$percentile:

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

Given information (part d)

$84$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.

Explanation (part d)

The $10th$percentile

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

Explanation (part e)

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

Explanation (part e)

Given information (part f)

$84$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.

Explanation (part f)

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

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.

Given information (part g)

$84$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.

Explanation (part g)

X	f	fx
2	4	8
3	36	108
4	18	72
5	19	95
6	4	24
7	1	7
8	1	8
9	1	9
Total	84	331

$$\begin{gathered} \overline{x}=\frac{\sum fx}{\sum f} \\ =\frac{331}{84} \\ =3.94 \end{gathered}$$

Given information (part h)

$84$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.

Explanation (part h)

X	f	$(x-\overline{x}{)}^2$	$f(x-\overline{x}{)}^2$
2	4	3.7636	15.0544
3	36	0.8836	31.8096
4	18	0.0036	0.0648
5	19	1.1236	21.3484
6	4	4.2436	16.9744
7	1	9.3636	9.3636
8	1	16.4836	16.4836
9	1	25.6036	25.6036
Total	84		136.7024

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

Given information (part i)

$84$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.

Explanation (part i)

X	f
2	4
3	36
4	18
5	19
6	4
7	1
8	1
9	1
Total	84

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

Given information (part j)

$84$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.

Explanation (part j)

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.

Given information (part k)

$84$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.

Explanation (part k)

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.