Given:
The scores that Jenna has received on her seven quizzes is 16, 13, 20, 16, 12, 17, 18.
We need to determine the variance.
Mean:
The mean of the data is given by
![Mean=(16+13+20+16+12+17+18)/(7)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/2cu4kqix9fmm8alf9t04lt9z6syplwmqsh.png)
![Mean=(112)/(7)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/dllkliyv4xwu1zobludhtsk7q423leauwp.png)
![Mean=16](https://img.qammunity.org/2021/formulas/mathematics/middle-school/rkrsw5o82x93gr7up7mjolkmxt77fnqy3h.png)
Thus, the mean of the given data is 16.
Variance:
The variance of the data can be determined using the formula,
![variance =(\sum(X-\mu)^(2))/(n)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/48kt1dpxj7porvoc4a1sa79xsy48mt8wvv.png)
where
is the mean and n is the number of terms in the distribution.
Thus, we have;
![variance=((16-16)^2+(13-16)^2+(20-16)^2+(16-16)^2+(12-16)^2+(17-16)^2+(18-16)^2)/(7)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/sfby5q651mrvgd623de6w9lvezukf3gtxg.png)
Simplifying the terms, we get;
![variance=((0)^2+(-3)^2+(4)^2+(0)^2+(4)^2+(1)^2+(2)^2)/(7)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/td9an18ny8wboe74khfi0yn1n6fd1e7qxk.png)
![variance=(0+9+16+0+16+1+4)/(7)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/7693e7vw6eu1b4qp1g5lr3xs0judke3kkt.png)
![variance=(46)/(7)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/55cqfkc8shqia90z6gce98x8ks46xri5j6.png)
![variance=6.57](https://img.qammunity.org/2021/formulas/mathematics/middle-school/5rdkrz5yl2qve7abi7xygrdfk5hsvzp7bq.png)
Rounding off to the nearest tenth, we get;
![variance=6.6](https://img.qammunity.org/2021/formulas/mathematics/middle-school/43sp47yqypdejxpj4f68xhsuxo4luaryy6.png)
Thus, the variance is 6.6