Answer: Choice B) I and II only
The mean and median are considered measures of center as they represent the average of a data set. The average basically being a point that collectively speaks for all of the data. For example, if you have a group of basketball players whose heights range from 5'11" to 6'9", then there is no single height to report; however, we can compute the average to get a basic idea of the single height. The interquartile range (IQR) is a measure of variability or spread of the data. The higher the IQR, the more spread out the data is. Recall that 50% of the data is represented by the IQR and that IQR = Q3 - Q1 where Q1 and Q3 are the first and third quartiles respectively. So because the IQR is a measure of spread, it is not considered a center point.
3 = 10%
15 = 50%
30 = 100%
30 * 1.5 = 45
Liang has D.) 45 shirts
Hi there!
So first off, here are little definitions you need to know in order to be able to answer your question :
<u>Constant</u> means that the rates of change are the same.
<u>Variable</u> means that the rates of change are different.
But from what I can see, you don't even need those definitions because you got both answers right!
Continue your great work! If there's anything just let me know! :)
Answer:
try the app socratic Step-by-step explanation:
Answer:
n = 98, that is, she scored at the 98th percentile.
Step-by-step explanation:
When the distribution is normal, we use the z-score formula.
In a set with mean 
and standard deviation 
, the zscore of a measure X is given by:
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this question, we have that:
She scored 38, so 
Test scores are normally distributed with a mean of 25 and a standard deviation of 6.4.
This means that 
Find the percentile:
We have to find the pvalue of Z. So
has a pvalue of 0.98(rounding to two decimal places).
So n = 98, that is, she scored at the 98th percentile.
