Question

show work to receive full credit. when necessary, put your final answer, with appropriate units, in the provided space. unless otherwise stated, round answers to two decimal places. you may use the desmos calculator. below is the cost per student in thousands of dollars from a sample of 10 universities. use this information to answer the following questions. 23, 26, 20, 18, 22, 24, 19, 25, 21, 20 1. find the following summary statistics: mean____ median__ mode__ range__ interquartile range__ standard deviation____ 2. another college is asked and says students pay $15,000. what is the effect on the mean and standard deviation?

Explanation:

Step1: Arrange data in ascending order

18, 19, 20, 20, 21, 22, 23, 24, 25, 26

Step2: Calculate the mean

The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$, where $n = 10$ and $x_{i}$ are the data - points.
$\sum_{i=1}^{10}x_{i}=18 + 19+20+20+21+22+23+24+25+26=218$
$\bar{x}=\frac{218}{10}=21.8$

Step3: Calculate the median

Since $n = 10$ (even), the median is the average of the $\frac{n}{2}$th and $(\frac{n}{2}+1)$th ordered values.
The 5th value is 21 and the 6th value is 22. Median $=\frac{21 + 22}{2}=21.5$

Step4: Calculate the mode

The mode is the most frequently occurring value. Here, 20 occurs twice, so mode = 20

Step5: Calculate the range

Range = Maximum value - Minimum value. Maximum value = 26, Minimum value = 18. Range = 26 - 18=8

Step6: Calculate the quartiles

The lower half of the data is 18, 19, 20, 20, 21. The first - quartile $Q_{1}$ is the median of the lower half. So $Q_{1}=20$
The upper half of the data is 22, 23, 24, 25, 26. The third - quartile $Q_{3}$ is the median of the upper half. So $Q_{3}=24$
Inter - quartile range $IQR=Q_{3}-Q_{1}=24 - 20 = 4$

Step7: Calculate the standard deviation

The formula for the sample standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n - 1}}$
$(x_{1}-\bar{x})^{2}=(18 - 21.8)^{2}=(-3.8)^{2}=14.44$
$(x_{2}-\bar{x})^{2}=(19 - 21.8)^{2}=(-2.8)^{2}=7.84$
$(x_{3}-\bar{x})^{2}=(20 - 21.8)^{2}=(-1.8)^{2}=3.24$
$(x_{4}-\bar{x})^{2}=(20 - 21.8)^{2}=(-1.8)^{2}=3.24$
$(x_{5}-\bar{x})^{2}=(21 - 21.8)^{2}=(-0.8)^{2}=0.64$
$(x_{6}-\bar{x})^{2}=(22 - 21.8)^{2}=(0.2)^{2}=0.04$
$(x_{7}-\bar{x})^{2}=(23 - 21.8)^{2}=(1.2)^{2}=1.44$
$(x_{8}-\bar{x})^{2}=(24 - 21.8)^{2}=(2.2)^{2}=4.84$
$(x_{9}-\bar{x})^{2}=(25 - 21.8)^{2}=(3.2)^{2}=10.24$
$(x_{10}-\bar{x})^{2}=(26 - 21.8)^{2}=(4.2)^{2}=17.64$
$\sum_{i = 1}^{10}(x_{i}-\bar{x})^{2}=14.44+7.84+3.24+3.24+0.64+0.04+1.44+4.84+10.24+17.64 = 63.6$
$s=\sqrt{\frac{63.6}{9}}\approx2.66$

Step8: Analyze the effect of a new data - point

The original mean is 21.8. The new sum with the new data - point $x = 15$ (since it's in thousands) is $218+15=233$ and the new $n = 11$. The new mean $\bar{x}_{new}=\frac{233}{11}\approx21.18$, so the mean decreases.
The original standard deviation is based on the spread of the data around 21.8. The new data - point 15 is further from the original mean than most of the original data - points. So the standard deviation will increase as the new data - point increases the overall spread of the data.

Answer:

Mean: 21.8
Median: 21.5
Mode: 20
Range: 8
Interquartile Range: 4
Standard Deviation: 2.66

1. The mean decreases and the standard deviation increases.