Discover new knowledge and insights with IDNLearn.com's extensive Q&A database. Join our knowledgeable community and access a wealth of reliable answers to your most pressing questions.
Sagot :
To determine the standard deviation of the sample data summarized in the frequency distribution table, follow these steps:
1. Recognize the Class Intervals and Frequencies:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|} \hline \text{Interval} & 30-36 & 37-43 & 44-50 & 51-57 & 58-64 & 65-71 \\ \hline \text{Frequency} & 2 & 24 & 44 & 15 & 6 & 4 \\ \hline \end{array} \][/tex]
2. Calculate the Class Midpoints:
[tex]\[ \text{Midpoint} = \frac{\text{Lower Class Limit} + \text{Upper Class Limit}}{2} \][/tex]
- For [tex]\(30-36\)[/tex]: [tex]\(\frac{30 + 36}{2} = 33\)[/tex]
- For [tex]\(37-43\)[/tex]: [tex]\(\frac{37 + 43}{2} = 40\)[/tex]
- For [tex]\(44-50\)[/tex]: [tex]\(\frac{44 + 50}{2} = 47\)[/tex]
- For [tex]\(51-57\)[/tex]: [tex]\(\frac{51 + 57}{2} = 54\)[/tex]
- For [tex]\(58-64\)[/tex]: [tex]\(\frac{58 + 64}{2} = 61\)[/tex]
- For [tex]\(65-71\)[/tex]: [tex]\(\frac{65 + 71}{2} = 68\)[/tex]
3. Calculate the Mean of the Distribution:
[tex]\[ \text{Mean} (\bar{x}) = \frac{\sum f \cdot x}{\sum f} \][/tex]
Where [tex]\(f\)[/tex] is the frequency, and [tex]\(x\)[/tex] is the class midpoint.
[tex]\[ \begin{align*} \sum f \cdot x &= 2 \cdot 33 + 24 \cdot 40 + 44 \cdot 47 + 15 \cdot 54 + 6 \cdot 61 + 4 \cdot 68 \\ &= 66 + 960 + 2068 + 810 + 366 + 272 \\ &= 4542 \end{align*} \][/tex]
[tex]\[ \sum f = 2 + 24 + 44 + 15 + 6 + 4 = 95 \][/tex]
[tex]\[ \text{Mean} (\bar{x}) = \frac{4542}{95} \approx 47.81 \][/tex]
4. Calculate the Variance:
[tex]\[ s^2 = \frac{\sum f(x - \bar{x})^2}{\sum f - 1} \][/tex]
Where [tex]\((x - \bar{x})\)[/tex] is the deviation of the midpoint from the mean.
[tex]\[ \begin{align*} \sum f(x - \bar{x})^2 &= 2(33 - 47.81)^2 + 24(40 - 47.81)^2 + 44(47 - 47.81)^2 + 15(54 - 47.81)^2 + 6(61 - 47.81)^2 + 4(68 - 47.81)^2 \\ &= 4693.632 \end{align*} \][/tex]
[tex]\[ s^2 = \frac{4693.632}{94} \approx 55.11 \][/tex]
5. Calculate the Standard Deviation:
[tex]\[ s = \sqrt{s^2} = \sqrt{55.11} \approx 7.4 \][/tex]
Therefore, the standard deviation [tex]\( s \)[/tex] of the sample data is approximately [tex]\( 7.4 \)[/tex] when rounded to one decimal place.
1. Recognize the Class Intervals and Frequencies:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|} \hline \text{Interval} & 30-36 & 37-43 & 44-50 & 51-57 & 58-64 & 65-71 \\ \hline \text{Frequency} & 2 & 24 & 44 & 15 & 6 & 4 \\ \hline \end{array} \][/tex]
2. Calculate the Class Midpoints:
[tex]\[ \text{Midpoint} = \frac{\text{Lower Class Limit} + \text{Upper Class Limit}}{2} \][/tex]
- For [tex]\(30-36\)[/tex]: [tex]\(\frac{30 + 36}{2} = 33\)[/tex]
- For [tex]\(37-43\)[/tex]: [tex]\(\frac{37 + 43}{2} = 40\)[/tex]
- For [tex]\(44-50\)[/tex]: [tex]\(\frac{44 + 50}{2} = 47\)[/tex]
- For [tex]\(51-57\)[/tex]: [tex]\(\frac{51 + 57}{2} = 54\)[/tex]
- For [tex]\(58-64\)[/tex]: [tex]\(\frac{58 + 64}{2} = 61\)[/tex]
- For [tex]\(65-71\)[/tex]: [tex]\(\frac{65 + 71}{2} = 68\)[/tex]
3. Calculate the Mean of the Distribution:
[tex]\[ \text{Mean} (\bar{x}) = \frac{\sum f \cdot x}{\sum f} \][/tex]
Where [tex]\(f\)[/tex] is the frequency, and [tex]\(x\)[/tex] is the class midpoint.
[tex]\[ \begin{align*} \sum f \cdot x &= 2 \cdot 33 + 24 \cdot 40 + 44 \cdot 47 + 15 \cdot 54 + 6 \cdot 61 + 4 \cdot 68 \\ &= 66 + 960 + 2068 + 810 + 366 + 272 \\ &= 4542 \end{align*} \][/tex]
[tex]\[ \sum f = 2 + 24 + 44 + 15 + 6 + 4 = 95 \][/tex]
[tex]\[ \text{Mean} (\bar{x}) = \frac{4542}{95} \approx 47.81 \][/tex]
4. Calculate the Variance:
[tex]\[ s^2 = \frac{\sum f(x - \bar{x})^2}{\sum f - 1} \][/tex]
Where [tex]\((x - \bar{x})\)[/tex] is the deviation of the midpoint from the mean.
[tex]\[ \begin{align*} \sum f(x - \bar{x})^2 &= 2(33 - 47.81)^2 + 24(40 - 47.81)^2 + 44(47 - 47.81)^2 + 15(54 - 47.81)^2 + 6(61 - 47.81)^2 + 4(68 - 47.81)^2 \\ &= 4693.632 \end{align*} \][/tex]
[tex]\[ s^2 = \frac{4693.632}{94} \approx 55.11 \][/tex]
5. Calculate the Standard Deviation:
[tex]\[ s = \sqrt{s^2} = \sqrt{55.11} \approx 7.4 \][/tex]
Therefore, the standard deviation [tex]\( s \)[/tex] of the sample data is approximately [tex]\( 7.4 \)[/tex] when rounded to one decimal place.
Thank you for joining our conversation. Don't hesitate to return anytime to find answers to your questions. Let's continue sharing knowledge and experiences! Thank you for choosing IDNLearn.com for your queries. We’re committed to providing accurate answers, so visit us again soon.