Điểm kiểm tra môn Toán của lớp  $10B$  cuối học kỳ I được cho trong bảng sau:

a) Tính điểm trung bình.
b) Tính phương sai và độ lệch chuẩn, hệ số biến thiên.
a) Điểm trung bình:  $\overline{x}=\frac{1}{40}(1.3+4.3+5.15+6.6+10.7+8.8) = 6,5 $
b) Ta tính phương sai theo công thức:
$s^2=\frac{1}{n}\sum_{i=1}^{k}n_i(x- \overline{x})^2 $
$s^2=\frac{1}{40}[1.(3-6,5)^2+3.(4-6,5)^2+15.(5-6,5)^2+6.(6-6,5)^2$
                                                                 $+8.(8-6,5)^2+7.(10-6,5)^2] $
$\Rightarrow   s^2=\frac{170}{40}    \Rightarrow    s^2=4,25 $.
Độ lệch chuẩn:  $s=\sqrt{4,25} \approx 2,06 $
Hệ số biến thiên:  $v=\frac{s}{x}  \approx \frac{2,06}{6,5}.100\%   \Rightarrow v \approx 38\%$

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