Find the standard deviation for the set of grouped sample data. IntervalFrequency0.5−3.533.5−6.556.5−9.549.5−12.55 s=?

Step 1 Solution: Class(1)Frequency(f)(2)Midvalue(x)(3)f⋅x(4)=(2)×(3)f⋅x2=(f⋅x)×(x)(5)=(4)×(3)cf(6)0.5−3.53261233.5−6.5552512586.5−9.54832256129.5−12.55115560517n=17∑f⋅x=118∑f⋅x2=998 Step 2 Sample Standard deviation S=∑f⋅x2−(∑f⋅x)2nn−1 =998−11821716 =998−819.058816 =178.941216 =11.1838 s=3.3442=3.34

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Answered question

2021-12-03

Find the standard deviation for the set of grouped sample data.

\begin{array}{cc} I n t e r v a l & F r e q u e n c y \\ 0.5 - 3.5 & 3 \\ 3.5 - 6.5 & 5 \\ 6.5 - 9.5 & 4 \\ 9.5 - 12.5 & 5 \end{array}

s = ?

Answer & Explanation

Heack1991

Beginner2021-12-04Added 13 answers

Step 1
Solution:

\begin{array}{cc} C l a s s (1) & F r e q u e n c y (f) (2) & M i d v a l u e (x) (3) & f \cdot x (4) = (2) \times (3) & f \cdot x^{2} = (f \cdot x) \times (x) (5) = (4) \times (3) & c f (6) \\ 0.5 - 3.5 & 3 & 2 & 6 & 12 & 3 \\ 3.5 - 6.5 & 5 & 5 & 25 & 125 & 8 \\ 6.5 - 9.5 & 4 & 8 & 32 & 256 & 12 \\ 9.5 - 12.5 & 5 & 11 & 55 & 605 & 17 \\ n = 17 & \sum f \cdot x = 118 & \sum f \cdot x^{2} = 998 \end{array}

Step 2
Sample Standard deviation

S = \sqrt{\frac{\frac{\sum f \cdot x^{2} - {(\sum f \cdot x)}^{2}}{n}}{n - 1}}

= \sqrt{\frac{\frac{998 - 118^{2}}{17}}{16}}

= \sqrt{\frac{998 - 819.0588}{16}}

= \sqrt{\frac{178.9412}{16}}

= \sqrt{11.1838}

s = 3.3442 = 3.34

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