summary calculations
/ 24 pts
two statistics classes recently took a quiz.
the points that a sample of students earned in each class is shown here:
class #1: 84; 78; 91; 91; 72; 75; 80; 89; 85; 73; 78
class #2: 94; 85; 74; 81; 96; 49; 70; 97; 85; 87; 72; 97
summarize each classs performance by calculating each of the following.
be sure to round each answer appropriately*
18 points - 1 point per calculation including proper rounding.
class #1 calculation class #2
mean
median
minimum
maximum
range
lower quartile
upper quartile
iqr
standard deviation
based on your calculations above...
3 points - 2 pts for correct choice; 1 pt for good & accurate defense
which class performed best on the quiz?
defend your choice
3 points - 2 pts for correct choice; 1 pt for good & accurate defense.
which class performed most consistently on the quiz?
defend your choice

Question

summary calculations
/ 24 pts
two statistics classes recently took a quiz.
the points that a sample of students earned in each class is shown here:
class #1: 84; 78; 91; 91; 72; 75; 80; 89; 85; 73; 78
class #2: 94; 85; 74; 81; 96; 49; 70; 97; 85; 87; 72; 97
summarize each classs performance by calculating each of the following.
be sure to round each answer appropriately*
18 points - 1 point per calculation including proper rounding.
class #1 calculation class #2
mean
median
minimum
maximum
range
lower quartile
upper quartile
iqr
standard deviation
based on your calculations above...
3 points - 2 pts for correct choice; 1 pt for good & accurate defense
> which class performed best on the quiz?
>> defend your choice
3 points - 2 pts for correct choice; 1 pt for good & accurate defense.
> which class performed most consistently on the quiz?
>> defend your choice

Accepted Answer

# Explanation:
## Class #1:
### Mean:
$$\frac{84 + 78+91+91+72+75+80+89+85+73+78}{11}=\frac{916}{11}\approx83.27$$
### Median:
Sort the data: $72,73,75,78,78,80,84,85,89,91,91$. The median is the $6^{th}$ value, so $80$.
### Minimum:
$72$
### Maximum:
$91$
### Range:
$91 - 72=19$
### Lower Quartile ($Q_1$):
The lower half of the data is $72,73,75,78,78$. The median of the lower half is $75$.
### Upper Quartile ($Q_3$):
The upper half of the data is $84,85,89,91,91$. The median of the upper half is $89$.
### IQR:
$Q_3 - Q_1=89 - 75 = 14$
### Standard Deviation:
First, find the variance.
$$
\begin{align*}
\sigma^{2}&=\frac{\sum_{i = 1}^{n}(x_{i}-\mu)^{2}}{n}\
&=\frac{(84 - 83.27)^{2}+(78-83.27)^{2}+(91 - 83.27)^{2}+(91 - 83.27)^{2}+(72 - 83.27)^{2}+(75 - 83.27)^{2}+(80 - 83.27)^{2}+(89 - 83.27)^{2}+(85 - 83.27)^{2}+(73 - 83.27)^{2}+(78 - 83.27)^{2}}{11}\
&=\frac{0.5929+27.7729+60.  3529+60.3529+127.0129+68.3929+10.6929+32.8329+2.9929+105.4729+27.7729}{11}\
&=\frac{524.25}{11}\approx47.66\
\sigma&=\sqrt{47.66}\approx6.90
\end{align*}
$$

## Class #2:
### Mean:
$$\frac{94+85+74+81+96+49+70+97+85+87+72+97}{12}=\frac{987}{12}=82.25$$
### Median:
Sort the data: $49,70,72,74,81,85,85,87,94,96,97,97$. The median is the average of the $6^{th}$ and $7^{th}$ values. $\frac{85 + 85}{2}=85$
### Minimum:
$49$
### Maximum:
$97$
### Range:
$97-49 = 48$
### Lower Quartile ($Q_1$):
The lower half of the data is $49,70,72,74,81,85$. The median of the lower half is $\frac{72+74}{2}=73$
### Upper Quartile ($Q_3$):
The upper half of the data is $85,87,94,96,97,97$. The median of the upper half is $\frac{94 + 96}{2}=95$
### IQR:
$Q_3 - Q_1=95 - 73=22$
### Standard Deviation:
First, find the variance.
$$
\begin{align*}
\sigma^{2}&=\frac{\sum_{i = 1}^{n}(x_{i}-\mu)^{2}}{n}\
&=\frac{(94 - 82.25)^{2}+(85-82.25)^{2}+(74 - 82.25)^{2}+(81 - 82.25)^{2}+(96 - 82.25)^{2}+(49 - 82.25)^{2}+(70 - 82.25)^{2}+(97 - 82.25)^{2}+(85 - 82.25)^{2}+(87 - 82.25)^{2}+(72 - 82.25)^{2}+(97 - 82.25)^{2}}{12}\
&=\frac{138.0625+7.5625+68.0625+1.5625+189.0625+1078.5625+150.0625+217.5625+7.5625+22.5625+105.0625+217.5625}{12}\
&=\frac{2303.25}{12}\approx191.94\
\sigma&=\sqrt{191.94}\approx13.85
\end{align*}
$$

# Answer:
|  | Class #1 | Class #2 |
|---|---|---|
| MEAN | $83.27$ | $82.25$ |
| MEDIAN | $80$ | $85$ |
| MINIMUM | $72$ | $49$ |
| MAXIMUM | $91$ | $97$ |
| RANGE | $19$ | $48$ |
| LOWER QUARTILE | $75$ | $73$ |
| UPPER QUARTILE | $89$ | $95$ |
| IQR | $14$ | $22$ |
| STANDARD DEVIATION | $6.90$ | $13.85$ |

For which class performed best:
- Class #1 has a higher mean ($83.27>82.25$). Although Class #2 has a higher median, the mean takes into account all data points. So Class #1 performed best.

For which class performed most consistently:
- Class #1 has a smaller standard deviation ($6.90<13.85$) and a smaller range ($19 < 48$). A smaller standard deviation and range indicate less spread in the data, so Class #1 performed most consistently.

	Class #1	Class #2
MEDIAN	\(80\)	\(85\)
MINIMUM	\(72\)	\(49\)
MAXIMUM	\(91\)	\(97\)
RANGE	\(19\)	\(48\)
LOWER QUARTILE	\(75\)	\(73\)
UPPER QUARTILE	\(89\)	\(95\)
IQR	\(14\)	\(22\)
STANDARD DEVIATION	\(6.90\)	\(13.85\)

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Class #1:

Mean:

Median:

Minimum:

Maximum:

Range:

Lower Quartile (\(Q_1\)):

Upper Quartile (\(Q_3\)):

IQR:

Standard Deviation:

Class #2: