Question

Here are Xavier's bowling scores:
135, 140, 130, 190, 112, 200, 185, 172, 163, 151,
149
What is the variance rounded to the nearest tenth?

Answer 1

Answer:

The variance rounded to the nearest tenth is 691.8

Step-by-step explanation:

Xavier's bowling scores:

135, 140, 130, 190, 112, 200, 185, 172, 163, 151, 149

No. of observations n = 11

$Mean = \frac{\text{Sum of all observations}}{\text{No. of observations}}\\Mean = \frac{135+140+130+190+112+200+185+172+ 163+ 151+149}{11}\\Mean =157$

Formula of variance : $\sigma^2=\frac{\sum(x_i-\bar{x})^2}{n}$

$\sigma^2=\frac{(135-157)^2+(140-157)^2+(130-157)^2+(190-157)^2+(112-157)^2+(200-157)^2+(185-157)^2+(172-157)^2+(163-157)^2+(151-157)^2+(149-157)^2}{11}$

$\sigma^2=\frac{7610}{11}$

$\sigma^2=691.81$

Hence the variance rounded to the nearest tenth is 691.8

Answer 2

Answer:

26.3

Step-by-step explanation:

correct answer