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1. nick was looking for apartments in two different communities. the co…

Question

  1. nick was looking for apartments in two different communities. the cost for various apartments in the two communities is shown below.

community a: $605, $610, $620, $625, $640, $650
community b: $640, $680, $580, $620, $700, $530
a. find the mean apartment price for each community. how do they compare?
b. find the range of the apartment prices for each community.
c. why is only comparing the means of the data sets misleading?

Explanation:

Step1: Calculate mean for Community A

The formula for mean is $\frac{\sum_{i = 1}^{n}x_{i}}{n}$. For Community A, $n = 6$, and $\sum_{i=1}^{6}x_{i}=605 + 610+620 + 625+640+650=3750$. So, mean $=\frac{3750}{6}=625$.

Step2: Calculate mean for Community B

For Community B, $n = 6$, and $\sum_{i = 1}^{6}x_{i}=640+680 + 580+620+700+530=3750$. So, mean $=\frac{3750}{6}=625$.

Step3: Calculate range for Community A

The formula for range is $max - min$. For Community A, $max = 650$, $min = 605$. Range $=650 - 605 = 45$.

Step4: Calculate range for Community B

For Community B, $max = 700$, $min = 530$. Range $=700 - 530=170$.

Answer:

a. Mean of Community A: $\$625$, Mean of Community B: $\$625$. They are equal.
b. Range of Community A: $\$45$, Range of Community B: $\$170$.
c. Only comparing the means is misleading because the spread (variability) of the data (as measured by range) is very different. Community A has a more consistent set of apartment prices, while Community B has a wider range of prices.