Respuesta :
Let's find the mean.
[tex]\begin{gathered} \bar{x}=\frac{\Sigma(x)}{N}=\frac{9+15+15+21+23+31+33+37+45+51}{10} \\ \bar{x}=\frac{280}{10}=28 \end{gathered}[/tex]The mean is 28.
Then, we find the standard deviation
[tex]\sigma=\sqrt[]{\frac{\Sigma(x-\bar{x})^2}{N}}[/tex]Let's find the difference between each value and the mean
9-28 = -19
15-28=-13
15-28=-13
21-28=-7
23-28=-5
31-28=3
33-28=5
37-28=9
45-28=17
51-28=23
Then, we add the square power of each subtraction
[tex]\begin{gathered} \sigma=\sqrt[]{\frac{(-19)^2+(-13)^2+(-13)^2+(-7)^2+(-5)^2+3^2+5^2+9^2+17^2+23^2}{10}} \\ \sigma=\sqrt[]{\frac{1706}{10}} \\ \sigma\approx13.1 \end{gathered}[/tex]The standard deviation is 13.1.
On the other hand, the five-number summary refers to the minimum, the first quartile, the median, the third quartile, and the maximum.
According to the given data set, we have
• The minimum is 9.
,• The maximum is 51.
,• The first quartile is 15. (the middle value between the first 5 numbers)
,• The third quartile is 37. (the middle value between the second 5 numbers)
,• The median is between 23 and 31.
[tex]Me=\frac{23+31}{2}=\frac{54}{2}=27[/tex]• The median is 27.
