Given:
The IQ of 10 randomly selected senior high school students are given as:
102,125,120,128,116,108,124,115,109,99.
Required:
To compute the sample mean, variance and standard deviation.
Explanation:
First, we will calculate the sample mean .
Here, the IQ of 10 students is given.
Therefore, total number of students (n) = 10.
Thus,
[tex]\begin{gathered} \bar{x}=\frac{102+125+120+128+116+108+124+115+109+99}{10} \\ \bar{x}=\frac{1146}{10} \\ \bar{x}=114.6 \end{gathered}[/tex]Next, the variance is given as follows:
[tex]S^2=\frac{\sum_{i\mathop{=}1}^n(x_i-\bar{x})^2}{n-1}[/tex]Now, we have,
[tex]\begin{gathered} x_1-\bar{x}=102-114.6=-12.6 \\ x_2-\bar{x}=125-114.6=10.4 \\ x_3-\bar{x}=120-114.6=5.4 \\ x_4-\bar{x}=128-114.6=13.4 \\ x_5-\bar{x}=116-114.6=1.4 \\ x_6-\bar{x}=108-114.6=-6.6 \\ x_7-\bar{x}=124-114.6=9.4 \\ x_8-\bar{x}=115-114.6=0.4 \\ x_9-\bar{x}=109-114.6=-5.6 \\ x_{10}-\bar{x}=99-114.6=-15.6 \end{gathered}[/tex]Thus,
[tex]\begin{gathered} S^2=\frac{158.76+108.16+29.16+179.56+1.96+43.56+88.36+0.16+31.36+243.36}{9} \\ S^2=\frac{884.4}{9} \\ S^2=98.27 \end{gathered}[/tex]Next, the standard deviation is,
[tex]\begin{gathered} S=\sqrt{98.27} \\ S=9.9131 \end{gathered}[/tex]Final Answer:
The mean is 114.6.
The variance is 98.27.
The standard deviation is 9.9131.