The first quartile of a data set is 32 and third quartile is 52. Which of these values in the data set is an outlier? 63 23 3 83

2 answers:

4 0

First we find the interquartile range which is Q3 (third quartile) - Q1 (first quartile)

so the IQR = 52 - 32 = 20

now u multiply ur IQR by 1.5
20 * 1.5 = 30

now subtract that from Q1 and add that to Q3
32 - 30 = 2....so anything below 2 is an outlier
52 + 30 = 82...and anything above 82 is an outlier

so ur outlier is : 83

Leno4ka [110]3 years ago

3 0

Hope this helps a little but your outlier is 83

You might be interested in

The picture below shows the graph of which inequality

hichkok12 [17]

The formula of a slope:

$m=\dfrac{y_2-y_1}{x_2-x_1}$

We have points:

$(-2,\ 4)\to x_1=-2;\ y_1=4\\(3;\ -6)\to x_2=3;\ y_2=-6$

substitute:

$m=\dfrac{-6-4}{3-(-2)}=\dfrac{-10}{3+2}=\dfrac{-10}{5}=-2$

Shaded left from the line therefore the inequality sign is ≤.

Your answer: C. y ≤ -2x

7 0

4 years ago

Consider the differential equation: (y^2+2*x)dx+(2*x*y-1)dy = 0. (a) show that the equation is exact by evaluating (\partial m)/

ruslelena [56]

The given ODE

$\underbrace{(y^2+2x)}_M\,\mathrm dx+\underbrace{(2xy-1)}_N\,\mathrm dy=0$

is exact if $\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$ . It so happens that we have $\frac{\partial M}{\partial y}=2y=\frac{\partial N}{\partial x}$ , so it is indeed exact. For such an ODE, we're looking for a solution of the form $\Psi(x,y)=C$ , for which the differential is