15. there were riders in a dirt bike competition with the following birth years: 2008, 2000, 1999, 2001…

15. there were riders in a dirt bike competition with the following birth years: 2008, 2000, 1999, 2001, 2005, 2001, 2000, 2006, 2007. calculate the mean, median, mode, and range for the birth years of the riders. (you don’t have to use a complete sentence in your answer.) mean: median: mode: range:

15. there were riders in a dirt bike competition with the following birth years: 2008, 2000, 1999, 2001, 2005, 2001, 2000, 2006, 2007. calculate the mean, median, mode, and range for the birth years of the riders. (you don’t have to use a complete sentence in your answer.) mean: median: mode: range:

Answer

Explanation:

Step1: Calculate the mean

The mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$, where $x_{i}$ are the data - points and $n$ is the number of data - points. Here, $n = 9$, and $\sum_{i=1}^{9}x_{i}=2008 + 2000+1999+2001+2005+2001+2000+2006+2007=18027$. So, $\bar{x}=\frac{18027}{9}=2003$.

Step2: Calculate the median

First, arrange the data in ascending order: $1999,2000,2000,2001,2001,2005,2006,2007,2008$. Since $n = 9$ (an odd number), the median is the $\left(\frac{n + 1}{2}\right)$-th value. $\frac{9+1}{2}=5$ - th value, which is $2001$.

Step3: Calculate the mode

The mode is the value that appears most frequently in the data - set. The number $2000$ and $2001$ both appear twice, while other numbers appear once. So, the modes are $2000$ and $2001$.

Step4: Calculate the range

The range is the difference between the maximum and minimum values in the data - set. The maximum value is $2008$ and the minimum value is $1999$. So, the range is $2008−1999 = 9$.

Answer:

Mean: $2003$ Median: $2001$ Mode: $2000,2001$ Range: $9$