The table shows the $10$ countries with the highest GDPs in the world in $2018$ as well as their GDPs in trillions of dollars. If the United States were removed from the table, how would the mean and standard deviation of the GDPs change? $\newline$ Choose $1$ answer: $\newline$ (A) The mean would stay the same, and the standard deviation would increase. $\newline$ (B) The mean would stay the same, and the standard deviation would decrease. $\newline$ (C) The mean would decrease, and the standard deviation would increase. $\newline$ (D) Both the mean and the standard deviation would decrease.

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One-step inequalities: word problems

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Q. The table shows the

10

countries with the highest GDPs in the world in

2018

as well as their GDPs in trillions of dollars. If the United States were removed from the table, how would the mean and standard deviation of the GDPs change?

\newline

Choose

1

answer:

\newline

(A) The mean would stay the same, and the standard deviation would increase.

\newline

(B) The mean would stay the same, and the standard deviation would decrease.

\newline

\newline

(D) Both the mean and the standard deviation would decrease.

Calculate Mean of GDPs: Calculate the current mean of the GDPs. Assume the GDPs are as follows (in trillions): USA $21$ , China $14$ , Japan $5$ , Germany $4$ , India $3$ , UK $3$ , France $3$ , Italy $2$ , Brazil $2$ , Canada $2$ . Total GDP = $14$ $0$ trillions. Number of countries = $14$ $1$ . Mean = Total GDP / Number of countries = $14$ $2$ trillions.
Remove USA's GDP: Remove the USA's GDP and recalculate the mean. New total GDP = $59 - 21 = 38$ trillions. New number of countries = $10 - 1 = 9$ . New mean = New total GDP / New number of countries = $\frac{38}{9} \approx 4.22$ trillions.
Analyze Mean Change: Analyze the change in mean. Original mean was $5.9$ trillions, new mean is approximately $4.22$ trillions. The mean decreases.
Consider Standard Deviation Effect: Consider the effect on standard deviation. Removing a value that is significantly higher than the mean generally decreases the spread of the remaining data, thus reducing the standard deviation.