We need to use the formula provided. The formula is:
[tex]s=\sqrt{\frac{1}{n-1}\sum_{i\mathop{=}1}^n(X_i-\bar{X})^2}[/tex]
Where:
[tex]\bar{X}=mean\text{ }data\text{ }set[/tex][tex]X_i=i\text{-}th\text{ }value\text{ }of\text{ }the\text{ }data\text{ }set[/tex]
Then, first, we need to find the mean:
[tex]\bar{X}=\frac{\sum_{i\mathop{=}1}^nX_i}{n}[/tex]
The data set is: 2, 3, 4, 4,7. Since there are 5 values, n = 5.
Now calculate:
[tex]\bar{X}=\frac{2+3+4+4+7}{5}=\frac{20}{5}=4[/tex]
Now, we need to find the square of the difference between each value of the data set and the mean:
[tex]\begin{gathered} (X_i-\bar{X})^2: \\ (2-4)^2=(-2)^2=4 \\ (3-4)^2=(-1)^2=1 \\ (4-4)^2=0^2=0 \\ (4-4)^{2}=0^{2}=0 \\ (7-4)^2=3^2=9 \end{gathered}[/tex]
And now we need to find the sum of those numbers:
[tex]\sum_{i\mathop{=}1}^n(X_i-\bar{X})^2=4+1+0+0+9=14[/tex]
Now, we can calculate the standard deviation:
[tex]s=\sqrt{\frac{1}{n-1}\sum_{i\mathop{=}1}^n(X_i-\bar{X})^2}=\sqrt{\frac{1}{5-1}\cdot14}=\sqrt{\frac{14}{4}}=\sqrt{\frac{7}{2}}\approx1.870828[/tex]
Thus, the correct answer is option a.) 1.87
