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

The mean (normal) of an informational collection is found by adding all numbers in the informational index and afterwards separating them by the number of values in the set. The median is the middle value when an informational collection is requested from least to most noteworthy. The mode is the number that happens most frequently in an informational collection

Data in chart

X = days of engineering conference

f = number of engineering conference

For finding the median,

we have

$M=\frac{(n+1)}{2}th$observation

$M=\frac{(84+1)}{2}th=42.5th$obs

Now,

$$\begin{gathered} 42.5thobs=\frac{43th+42th}{2}obs \\ \Rightarrow \frac{4+4}{2}=4 \end{gathered}$$

The median is$4$

Now, finding the first quartile

$Q_1=\frac{(n+1)}{4}th$obs

$=\frac{(84+1)}{4}th=21.25th$obs

$Q_1=21.25th\text{observation}=21th+\frac{22th-21th}{4}$obs

$Q_1=3+\frac{3-3}{4}=3$

the first quartile is $3$

finding the third quartile,

$Q_3=\frac{3(n+1)}{4}\text{th observation}=\frac{3(84+1)}{4}th$obs

$Q_3=63.75th\text{observation}=63th+\frac{3(64th-63th)}{4}$obs

$Q_3=5+\frac{3(5-5)}{4}=5$

the third quartile is$5$

Percentile for an ungrouped data is

$P_t=\frac{t(n+1)}{100}th$

For $65th$percentile,

$P_{65}=\frac{65(n+1)}{100}th=\frac{65(84+1)}{100}th=55.25th$obs

$P_{65}=55th+0.25(56th-55th)$obs

$P_{65}=4+0.25(4-4)=4$

The $65th$percentile is $4$

For $10th$percentile,

$P_{10}=\frac{10(n+1)}{100}th\text{obs}=\frac{(84+1)}{10}th$obs

$P_{10}=8.5th\text{obs}=8th+0.5(9th-8th)$obs

$P_{10}=3+0.5(3-3)=3$

Box plot of the data

The area between the third quartile and first quartile is the range of middle $50\%$data.

hence, it will last from$3-5$days.

the sample mean of days of engineering conferences

Mean= $\overline{x}=\frac{\sum fx}{\sum f}$

$$\begin{gathered} =\frac{331}{84} \\ =3.94 \end{gathered}$$

The sample standard deviation of days of engineering conferences

sample standard deviation S,

$$\begin{gathered} =\sqrt{\frac{136.7024}{84-1}} \\ =1.28 \end{gathered}$$

Mode

The value of the observation which has the highest frequency is the mode.

Hence, mode=$3$

The value of mode would be chosen as the length of the conference since the maximum should be present during the conference.

Hence the length of the conference is$3$days

Three to five days seem to be popular lengths of engineering conferences as-

1. The range of days should be the popular length of days which is from $3-5$days

since half of the data lies in this range.

2. within $3-5$days $73$conferences occurred. Hence the range of days should be the popular length of days