Answer:
The mean is 45.178
The standard deviation is 10.3325
Step-by-step explanation:
The mean is 45.178
The standard deviation is 10.3325
Answer:
[tex]\bar X = \frac{\sum_{i=1}^n X_i f_i}{N}= \frac{5267.7}{116.6}= 45.178[/tex]
And the sample variance can be calculated with this formula:
[tex] s^2 = \frac{\sum fx^2 -\frac{(\sum x*f)^2}{n}}{n-1}= \frac{250323.15 - \frac{(5267.7)^2}{116.6}}{116.6-1} = 106.7601[/tex]
And the deviation would be:
[tex] s = \sqrt{106.7601}= 10.332[/tex]
Step-by-step explanation:
For this case we can calculate the mean with the following table
Age Number (f) Midpoint x*f x^2 *f
25-34 22.1 29.5 651.95 19232.525
35-44 31.5 39.5 1244.25 49147.875
45-54 37.7 49.5 1866.15 92374.425
55-64 25.3 59.5 1505.35 89568.325
Total 116.6 5267.7 250323.15
And the mean can be calculatd with this formula:
[tex]\bar X = \frac{\sum_{i=1}^n X_i f_i}{N}= \frac{5267.7}{116.6}= 45.178[/tex]
And the sample variance can be calculated with this formula:
[tex] s^2 = \frac{\sum fx^2 -\frac{(\sum x*f)^2}{n}}{n-1}= \frac{250323.15 - \frac{(5267.7)^2}{116.6}}{116.6-1} = 106.7601[/tex]
And the deviation would be:
[tex] s = \sqrt{106.7601}= 10.332[/tex]
