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.

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

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.
Hello!
If Bruce observes that the number of pitches a batter hits varies and is given by the function f(x)=x-11, and the batters get {4, 12, 14, 27, 42}, then Bruce threw {15, 23, 25, 38, 53} pitches. We get this solution set by adding 11 to each element in the set {4, 12, 14, 27, 42}.
Have a nice day
Answer:
Quadrant 1
Step-by-step explanation:
The co-ordinates given match up exactly to co-ordinate 1 due to all the numbers given being positive.
There are 4 quadrants. Quadrant 1, 2, 3, and 4.
Answer:
Center = (2,5)
Radius = 10
Choice A
To find this answer, first write the equation
(x-2)^2 + (y-5)^2 = 100
into
(x-2)^2 + (y-5)^2 = 10^2
Note how the second equation is in the form
(x-h)^2 + (y-k)^2 = r^2
We see that (h,k) = (2,5) is the center
and r = 10 is the radius
<span>Given that the orbital period, P, of a planet and the planet’s distance from the sun, a, in astronomical units is related by the formula P = a^(3/2).
If Saturn’s orbital period is 29.5 years, then
29.5 = a^(3/2)
a = (29.5)^(2/3) = 9.547
Therefore, Saturn's distance from the sun is 9.547 astronomical units.</span>