Solve using elimination. 3x − 5y = 18 -10x + 5y = 10
2 answers:
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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.
The mean is the average of the numbers.
It is easy to calculate: add up all the numbers, then divide by how many numbers there are.
Let the number of sit-ups you have to do Sunday be x. Then the sum of all sit-ups is
100+65+120+150+90+0+x=525+x.
The number of days is 7.
Thus, the mean is
You know that the mean number of sit-ups per day is 100, equate these two means and solve the equation:
Answer: 175 sit-ups