When the outliers are removed, how does the mean change? dot plot with 1 on 50, 1 on 76, 1 on 78, 2 on 79, 1 on 80, 1 on 81, 2 o
n 82, 1 on 83
2 answers:
We can plot this data on MS Excel and determine the distribution of these data reflected on the graph. Among these numbers, 50 is the outlier since it is very far from the other numbers ranging from 76 to 83. We can perform interquartile range to determine or verify the outliers in the data set. In this respect, we can see that there is not much distribution seen. The average of all data sets is equal to 96.25. When the outlier (50) is removed, we expect the mean to become higher since a low number was ommitted including high numbers only. Outliers are obtained from special causations such as human errors.
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9514 1404 393
Answer:
-3 ≤ x ≤ 19/3
Step-by-step explanation:
This inequality can be resolved to a compound inequality:
-7 ≤ (3x -5)/2 ≤ 7
Multiply all parts by 2.
-14 ≤ 3x -5 ≤ 14
Add 5 to all parts.
-9 ≤ 3x ≤ 19
Divide all parts by 3.
-3 ≤ x ≤ 19/3
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<em>Additional comment</em>
If you subtract 7 from both sides of the given inequality, it becomes ...
|(3x -5)/2| -7 ≤ 0
Then you're looking for the values of x that bound the region where the graph is below the x-axis. Those are shown in the attachment. For graphing purposes, I find this comparison to zero works well.
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For an algebraic solution, I like the compound inequality method shown above. That only works well when the inequality is of the form ...
|f(x)| < (some number) . . . . or ≤
If the inequality symbol points away from the absolute value expression, or if the (some number) expression involves the variable, then it is probably better to write the inequality in two parts with appropriate domain specifications:
|f(x)| > g(x) ⇒ f(x) > g(x) for f(x) > 0; or -f(x) > g(x) for f(x) < 0
Any solutions to these inequalities must respect their domains.
