Outliers can cause the _________ to be misleading when summarizing a data set.

Mathematics

1 answer:

gladu [14]2 years ago

8 0

Hello!

An outlier can cause the mean to be misleading. If you have an outlier when finding the median or IQR, it will not throw off the data by much. But adding a very large or small number to the mean, and dividing it by an extra number, it will throw the number off by a large amount.

I hope this helps!

You might be interested in

Is y=x^2 a proportional relationship?

TiliK225 [7]

Answer:

is y=x^2 a proportional relationship?

${ \sf{yes. \: constant \: of \: proportionality = 1}}$

is y=2+x a proportional relationship?

${ \sf{no. \: unless \: y \: is \: proportinal \: to \: (2 + x)}}$

is y=2/x a proportional relationship?

${ \sf{yes. \: where \: proportianality \: constant \: is \: 2}}$

is y=2x a proportional relationship?

${ \sf{yeah. \: constant \: is \: 2}}$

6 0

2 years ago

Which fraction is less than 1 chose all of the correct answer, and explain why you chose and write it down and explain your answ

EleoNora [17]

Answer:

$(3/4)$ and $(2/3)$ .

Step-by-step explanation:

For a fraction of the form $(p / q)$ , the number in front of the division ( $p$ ) is the numerator. The number after the division ( $q$ ) is the denominator.

$\begin{aligned} \frac{p}{q} \quad \genfrac{}{}{0}{}{(\text{numerator})}{(\text{denominator})}\end{aligned}$ .

A positive fraction $(p/q)$ ( $p,\, q > 0$ ) is less than $1$ as long as the numerator $p$ is smaller than the denominator $q$ (that is, $p < q$ .) The reason is that as long as $p,\, q > 0$ :

$\begin{aligned} & \frac{p}{q} < 1 \\ \iff & (q)\, \left(\frac{p}{q}\right) < (q)\, (1) && (\text{given that $q > 0$}) \\ \iff & p < q\end{aligned}$ .

For example, in the fraction $(2/3)$ , $2$ is the numerator while $3$ is the denominator. Both the numerator and the denominator are positive, while the numerator $2\!$ is smaller than the denominator $3\!$ . Therefore, $(2/3)$ would be less than $1$ .