Question

Will give rating ASAP, please give explanation on how to find the answers with a TI-80 graphing calculator

3. (13 points) Sales, in dollars, for one week were collected for 18 stores in a food store chain. The data is given in the table below. The stores and towns they are located in vary in size. Round your answers to 2 decimal places. You will be graded on notation, functionality of your calculator (if applicable you must be explicit with what you input in your calculator), appropriate units, and complete sentences. 99,971 222,585 77,605 117.356 134,609 97,659 58,762 104,518 99.988 95,583 68,526 204,135 137,511 69,817 72,911 75,886 77,500 74,942 (a) (5 points) Give the 5 number summary. (b) (4 points) Identify any outliers that would appear in a box plot. You must explain how you (or your calculator if you are using its graphing capabilities) identified these outliers. (c) (2 points) Find the standard deviation. (d) (2 points) Why might you suspect the histogram for this data is skewed without graphing?

Answer 1

(a)

The five number summary is,

$\small MinValue = 58,762$

$\small Q1 = \frac{x_4+x_5}{2} = 73,927$

$\small Median=\frac{x_9+x_10}{2} = 77,605$

$\small Q3=\frac{x_{14 }+ x_{15}}{2} = 125,983$

$\small MaxValue=222,585$

(b) The box plot is,

| 2300001 | 2200001 | 210000 | 200000 | 190000- | 180000- | 170000- | 160000- | 150000- | 140000- | 130000- | 120000- | 110000- | 100000- 90000- 80000- 70000- 60000

We know that any data point that lies outside the range of [Q1 - 1.5IQR , Q3 + 1.5IQR] is an outlier.

The IQR is 52056

Therefore, the outliers are { 204135 , 222585 }
(c)
The standard deviation is,

$\small s=\sqrt\frac{\sum (x-\mu)^2}{n-1}$

where, $\small \mu = 104,992$

Therefore, the standard deviations is,

$\small s = 45,327.06$

(d)

The distribution is right skewed. As seen from the boxplot, the values stretch far to the right from median then they stretch towards left. Therefore, it can be said that the distribution is skewed to right.