Suppose 3 people each called 3 friends on the phone.

Explain the circumstances for which the interquartile range is the preferred measure of dispersion. What is an advantage that th

KatRina [158]

Answer:

Explain the circumstances for which the interquartile range is the preferred measure of dispersion

Interquartile range is preferred when the distribution of data is highly skewed (right or left skewed) and when we have the presence of outliers. Because under these conditions the sample variance and deviation can be biased estimators for the dispersion.

What is an advantage that the standard deviation has over the interquartile range?

The most important advantage is that the sample variance and deviation takes in count all the observations in order to calculate the statistic.

Step-by-step explanation:

Previous concepts

The interquartile range is defined as the difference between the upper quartile and the first quartile and is a measure of dispersion for a dataset.

$IQR= Q_3 -Q_1$

The standard deviation is a measure of dispersion obatined from the sample variance and is given by:

$s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

Solution to the problem

Explain the circumstances for which the interquartile range is the preferred measure of dispersion

Interquartile range is preferred when the distribution of data is highly skewed (right or left skewed) and when we have the presence of outliers. Because under these conditions the sample variance and deviation can be biased estimators for the dispersion.

What is an advantage that the standard deviation has over the interquartile range?

The most important advantage is that the sample variance and deviation takes in count all the observations in order to calculate the statistic.

8 0

2 years ago

The graph of a quadratic function contains (4,-64) and it's vertex is at the origin.write an equation for this function.

lesya692 [45]

Answer:

Step-by-step explanation:

Not sure what form you need this in, but it really doesn't matter, as you'll see in the final equation. I used the vertex form and solved for a:

$y=a(x-h)^2+k$

We are given the vertex (h, k) as the origin (0, 0), and we have a point that the graph goes through as (4, -64). That's our x and y. Plugging in what we have:

$-64=a(4-0)^2+0$ gives us