June 2011 p62 q3
2530
A sample of 36 data values, \(x\), gave \(\Sigma(x - 45) = -148\) and \(\Sigma(x - 45)^2 = 3089\).
- Find the mean and standard deviation of the 36 values.
- One extra data value of 29 was added to the sample. Find the standard deviation of all 37 values.
Solution
(i) To find the mean, use the formula:
\(\text{Mean} = 45 + \frac{-148}{36} = 40.9\)
To find the variance, use:
\(\text{Var} = \frac{3089}{36} - \left(\frac{-148}{36}\right)^2 = 68.9\)
The standard deviation is:
\(\text{sd} = \sqrt{68.9} = 8.30\)
(ii) With the new data value of 29 added:
New \(\Sigma(x - 45) = -148 + (29 - 45) = -164\)
New \(\Sigma(x - 45)^2 = 3089 + (29 - 45)^2 = 3345\)
The new standard deviation is:
\(\text{New sd} = \sqrt{\frac{3345}{37} - \left(\frac{-164}{37}\right)^2} = 8.41\)
Log in to record attempts.