This solution explains how to find the geometric mean for a set of numbers. The geometric mean is a type of average that indicates the central tendency of a set of numbers by using the product of their values (as opposed to the arithmetic mean which uses their sum).
The geometric mean is particularly useful for data that grows exponentially, like investment returns or population growth. It is calculated by multiplying all the numbers together and then taking the n-th root of the product, where n is the count of the numbers.
For a set of n numbers, denoted as $a_1, a_2, ..., a_n$, the geometric mean (GM) is calculated using the following formula:
$ GM = \sqrt[n]{a_1 \times a_2 \times ... \times a_n} $Alternatively, this can be expressed using exponents:
$ GM = (a_1 \times a_2 \times ... \times a_n)^{\frac{1}{n}} $We need to find the geometric mean of the numbers 2, 4, and 8. Here:
First, multiply the given numbers together:
$ 2 \times 4 \times 8 $Performing the multiplication:
$ 8 \times 8 = 64 $Next, find the n-th root of the product, where n is 3 (since there are three numbers):
$ GM = \sqrt[3]{64} $To find the cube root of 64, we look for a number that, when multiplied by itself three times, equals 64. We know that:
$ 4 \times 4 \times 4 = 16 \times 4 = 64 $Therefore, the cube root of 64 is 4.
The geometric mean of 2, 4, and 8 is 4.
The calculation confirms that the geometric mean of the numbers 2, 4, and 8 is 4.
Let $X_1, X_2, X_3$ be a random sample of size 3 from an absolutely continuous distribution that is symmetric about 0. For $i=1,2,3$, let $R_i$ denote the rank of $|X_i|$ among $|X_1|, |X_2|$ and $|X_3|$.
If $T^+ = \sum_{i=1, X_i>0}^3 R_i$
is the Willcoxon signed-rank statistic, then which of the following statements are true?,