Respuesta :
Answer:
Mean = 45.178
Standard Deviation = 10.33
Step-by-step explanation:
class m frequency, f m [tex]\times[/tex] f f [tex]\times[/tex] [tex]m^{2}[/tex]
25 - 34 29.5 22.1 651.95 19232.525
35 - 44 39.5 31.5 1244.25 49147.875
45 - 54 49.5 37.7 1866.15 92374.425
55 - 64 59.5 25.3 1505.35 89568.352
116.6 5267.7 250323.15
mean = [tex]\frac{5267.7}{116.6} = 45.178[/tex] standard deviation = [tex]\sqrt{\frac{250323.15 - \frac{(5267.7)^2}{116.6} }{(116.1 - 1)} } = 10.33[/tex]
The mean of a distribution is the average of the distribution ; it is the sum of observation divided by the count of the observation.
- The mean age is 45.18
- The standard deviation is 10.29
Given that:
[tex]\begin{array}{ccccc}{Age}&25-34&35-44&45-45&55-64\\{Number}&22.1&31.5&37.7&25.3\end{array}[/tex]
First, we calculate midpoint of each class.
This is the sum of the class interval divided by 2.
So, we have:
[tex]x_1 = \frac{25+34}{2} = 29.5[/tex]
[tex]x_2 = \frac{35+44}{2} = 39.5[/tex]
And so on...
The table becomes:
[tex]\begin{array}{ccccc}{Age}&25-34&35-44&45-45&55-64&{Number(f)}&22.1&31.5&37.7&25.3&{x}&29.5&39.5&49.5&59.5\end{array}[/tex]
The mean of the distribution is:
[tex]\bar x = \frac{\sum fx}{\sum f}[/tex]
[tex]\bar x = \frac{29.5 \times 22.1 + 39.5 \times 31.5 + 49.5 \times 37.7 +59.5 \times 25.3}{22.1 + 31.5 + 37.7 +25.3}[/tex]
[tex]\bar x = \frac{5267.7}{116.6}[/tex]
[tex]\bar x = 45.18[/tex]
The standard deviation is calculated as follows:
[tex]\sigma = \sqrt{\frac{\sum f(x - \bar x)^2}{\sum f}}[/tex]
[tex]\sigma= \sqrt{\frac{(29.5 - 45.1)^2\times 22.1 + (39.5 - 45.1)^2 \times 31.5 + (49.5 - 45.1)^2 \times 37.7 + (59.5 - 45.1)^2 \times 25.3}{22.1 + 31.5 + 37.7 +25.3}}[/tex]
[tex]\sigma= \sqrt{\frac{12342.176}{116.6}}[/tex]
[tex]\sigma= \sqrt{105.85}[/tex]
[tex]\sigma= 10.29[/tex]
Hence:
- The mean age is 45.18
- The standard deviation is 10.29
Read more about mean and standard deviation at:
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