Question

You have been given a sample of n=20 values, [x 1​ ,x 2​ ,…,x n​ ], of mean xˉ =10, but you were not given any of the individual xi ​ values (because your professor hid them from you, just to torture you). However, your professor, just to tease you, gave you instead the sum of squares for this sample SS=2475. Can you outsmart your evil professor, and somehow compute the value of the standard deviation, even though you do not know any of the sample values? ∑ i=1 i=n ​ (x i​ ) 2 =n xˉ 2 +(n−1)s 2 For this, you will want to start with the definition of the sample variance s 2 and develop it, using the rule for squaring a difference. You can use Word's equation editor, or do it by hand and scan your work, etc. 1.2 Mean - 3 pts Well, you are still working on this mystery sample, of size 20, and mean 10. You are now given a 21th sample, of value 12.1. Can you again! outsmart your evil professor, and somehow compute the value of the new value of the mean, even though you do not know any of the other 2 - ample values?

Answer 1

You can compute the standard deviation using the given information. The standard deviation for the sample is approximately 5.

Step-by-step explanation:

To calculate the standard deviation without knowing individual values, you can use the formula:
`\[\sum_(i=1)^(n) (x_i)^2 = n\bar{x}^2 + (n-1)s^2\] Given that \(n = 20\), \(\bar{x} = 10\), and \(\sum_(i=1)^(n) (x_i)^2 = 2475\), you can rearrange the formula to solve for \(s\), the standard deviation. Substituting the values, you get: \[2475 = 20 * 10^2 + 19s^2\] Solving for \(s\), you find that \(s \approx 5\).`This allows you to determine the standard deviation of the sample even without knowing individual values.

Now, considering the addition of a 21st value (12.1), to find the new mean (\(\bar{x}'\)), you can use the formula for the mean:
`\[\bar{x}' = \frac{n\bar{x} + x_(21)}{n+1}\] Substituting the given values (\(n = 20\), \(\bar{x} = 10\), \(x_(21) = 12.1\)), you can calculate the new mean to get \(\bar{x}' \approx 10.05\).` This provides the updated mean for the enlarged sample, demonstrating your ability to compute the mean even when only one additional value is disclosed.