Question

standard deviation calculation questions
show your work for each question.

1. calculate the mean and the standard deviation for the following data set of 10 quiz scores: 7, 8, 6, 9, 10, 5, 8, 7, 9, 8
2. find the standard deviation of the following set of numbers (round to two decimal places): 12, 15, 14, 10, 9, 13, 15, 14, 11, 13
3. the following data shows the number of books read by 10 students 3, 4, 2, 5, 6, 3, 4, 5, 4, 3 calculate the standard deviation.

Explanation:

Question 11

Step1: Calculate the mean

The mean $\bar{x}$ of a data - set $x_1,x_2,\cdots,x_n$ is given by $\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}$. Here, $n = 10$, and the data - set is $7,8,6,9,10,5,8,7,9,8$.
$\sum_{i=1}^{10}x_i=7 + 8+6 + 9+10+5+8+7+9+8=77$
$\bar{x}=\frac{77}{10}=7.7$

Step2: Calculate the variance

The variance $s^{2}=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n}$.
$(7 - 7.7)^2=(-0.7)^2 = 0.49$
$(8 - 7.7)^2=(0.3)^2 = 0.09$
$(6 - 7.7)^2=(-1.7)^2 = 2.89$
$(9 - 7.7)^2=(1.3)^2 = 1.69$
$(10 - 7.7)^2=(2.3)^2 = 5.29$
$(5 - 7.7)^2=(-2.7)^2 = 7.29$
$(8 - 7.7)^2=(0.3)^2 = 0.09$
$(7 - 7.7)^2=(-0.7)^2 = 0.49$
$(9 - 7.7)^2=(1.3)^2 = 1.69$
$(8 - 7.7)^2=(0.3)^2 = 0.09$
$\sum_{i = 1}^{10}(x_i - 7.7)^2=0.49+0.09+2.89+1.69+5.29+7.29+0.09+0.49+1.69+0.09 = 19.1$
$s^{2}=\frac{19.1}{10}=1.91$

Step3: Calculate the standard deviation

The standard deviation $s=\sqrt{s^{2}}$.
$s=\sqrt{1.91}\approx1.38$

Step1: Calculate the mean

The data - set is $12,15,14,10,9,13,15,14,11,13$.
$n = 10$
$\sum_{i=1}^{10}x_i=12 + 15+14+10+9+13+15+14+11+13=126$
$\bar{x}=\frac{126}{10}=12.6$

Step2: Calculate the variance

$(12 - 12.6)^2=(-0.6)^2 = 0.36$
$(15 - 12.6)^2=(2.4)^2 = 5.76$
$(14 - 12.6)^2=(1.4)^2 = 1.96$
$(10 - 12.6)^2=(-2.6)^2 = 6.76$
$(9 - 12.6)^2=(-3.6)^2 = 12.96$
$(13 - 12.6)^2=(0.4)^2 = 0.16$
$(15 - 12.6)^2=(2.4)^2 = 5.76$
$(14 - 12.6)^2=(1.4)^2 = 1.96$
$(11 - 12.6)^2=(-1.6)^2 = 2.56$
$(13 - 12.6)^2=(0.4)^2 = 0.16$
$\sum_{i = 1}^{10}(x_i - 12.6)^2=0.36+5.76+1.96+6.76+12.96+0.16+5.76+1.96+2.56+0.16 = 38.4$
$s^{2}=\frac{38.4}{10}=3.84$

Step3: Calculate the standard deviation

$s=\sqrt{3.84}\approx1.96$

Step1: Calculate the mean

The data - set is $3,4,2,5,6,3,4,5,4,3$.
$n = 10$
$\sum_{i=1}^{10}x_i=3 + 4+2+5+6+3+4+5+4+3=39$
$\bar{x}=\frac{39}{10}=3.9$

Step2: Calculate the variance

$(3 - 3.9)^2=(-0.9)^2 = 0.81$
$(4 - 3.9)^2=(0.1)^2 = 0.01$
$(2 - 3.9)^2=(-1.9)^2 = 3.61$
$(5 - 3.9)^2=(1.1)^2 = 1.21$
$(6 - 3.9)^2=(2.1)^2 = 4.41$
$(3 - 3.9)^2=(-0.9)^2 = 0.81$
$(4 - 3.9)^2=(0.1)^2 = 0.01$
$(5 - 3.9)^2=(1.1)^2 = 1.21$
$(4 - 3.9)^2=(0.1)^2 = 0.01$
$(3 - 3.9)^2=(-0.9)^2 = 0.81$
$\sum_{i = 1}^{10}(x_i - 3.9)^2=0.81+0.01+3.61+1.21+4.41+0.81+0.01+1.21+0.01+0.81 = 12.9$
$s^{2}=\frac{12.9}{10}=1.29$

Step3: Calculate the standard deviation

$s=\sqrt{1.29}\approx1.14$

Answer:

Mean: $7.7$, Standard Deviation: $1.38$

Question 12