Question

#2 check up

1. enter the following nobel laureates’ ages into a column of the geogebra spreadsheet tool:

nobel laureates

name	age
wolfgang ketterle	44
joseph leonard goldstein	45
aung san suu kyi	46
kenneth joseph arrow	51
barry james marshall	54
stanley ben prusiner	55
torsten nils wiesel	57
richard axel	58
robert coleman richards	59
james alexander mirrlees	60
robert merton solow	63
stanley cohen	64
peter mansfield	70
vernon lomax smith	75
richard fred heck	79

32, 44, 45, 46, 51, 54, 55, 57, 58, 59, 60, 63, 64, 70, 75, 79

1. generate the five - number summary that describes the data set’s distribution:

minimum value: 32
1st quartile (q1): 48
median (q2): 58
3rd quartile (q3): 67
maximum value: 79
(please label these five values on the box plot, below.)

1. calculate the interquartile range (i.q.r.)=(q3 - q1): i.q.r. = 67 - 48 = 19
2. calculate the upper (outlier) fence: q3+1.5(i.q.r.) = q3+1.5×(i.q.r.)
3. calculate the lower (outlier) fence: q1 - 1.5(i.q.r.) = q1 - 1.5×(i.q.r.) 46.5
4. label the five - number summary and draw the outlier fences on the box - plot number line, below:
5. list any outliers:

if they exist, then label them on the box - plot, above.

1. the data is:

skewed - left
symmetric (circle your choice)
skewed - right

Explanation:

Step1: Organize data in ascending order

32, 44, 45, 46, 51, 54, 55, 57, 58, 59, 60, 63, 64, 70, 75, 79

Step2: Find minimum value

The smallest number in the data - set is 32.

Step3: Calculate first quartile ($Q_1$)

There are $n = 16$ data points. The position of $Q_1$ is $\frac{n + 1}{4}=\frac{16+1}{4}=4.25$. So, $Q_1=46+(0.25)\times(51 - 46)=46 + 1.25=48$.

Step4: Calculate median ($Q_2$)

The position of the median for $n = 16$ is $\frac{n}{2}=8$ and $\frac{n}{2}+1 = 9$. So, $Q_2=\frac{57 + 58}{2}=57.5\approx58$.

Step5: Calculate third quartile ($Q_3$)

The position of $Q_3$ is $\frac{3(n + 1)}{4}=\frac{3\times(16 + 1)}{4}=12.75$. So, $Q_3=64+(0.75)\times(70 - 64)=64+4.5 = 67$.

Step6: Find maximum value

The largest number in the data - set is 79.

Step7: Calculate inter - quartile range (IQR)

$IQR=Q_3 - Q_1=67 - 48 = 19$.

Step8: Calculate upper outlier fence

Upper fence $=Q_3+1.5\times IQR=67+1.5\times19=67 + 28.5=95.5$.

Step9: Calculate lower outlier fence

Lower fence $=Q_1-1.5\times IQR=48-1.5\times19=48 - 28.5 = 19.5$.

Step10: Check for outliers

Since all data points are within the range of 19.5 and 95.5, there are no outliers.

Step11: Determine skewness

Since $Q_1$ is closer to the median than $Q_3$ is to the median, the data is skewed - right.

Answer:

1. Minimum value: 32
2. $1^{st}$ Quartile ($Q_1$): 48
3. Median ($Q_2$): 58
4. $3^{rd}$ Quartile ($Q_3$): 67
5. Maximum value: 79
6. Inter - quartile range (IQR): 19
7. Upper outlier fence: 95.5
8. Lower outlier fence: 19.5
9. Outliers: None
10. The data is: skewed - right