Respuesta :
We start with 12 samples with an average of 34 and add two more samples of 34, so the mean remains 34.
For the variance (squared standard deviation) we add up the squared differences between the sample and the mean and divide by n-1=11. With two additional samples exactly at the mean, the sum won't change but now we divide by 13. So our new standard deviation must be
[tex] \sqrt{\dfrac{11}{13}} (4.75) \approx 4.37[/tex]
Answer: 4.37
Following are the calculation to the Mean and standard deviation:
Given :
[tex]n_{old}=12\\\\ \bar{X}_{old}=34\\\\ \sigma_{old}=4.75[/tex]
[tex]\to \bar{X}_{old}=\frac{\sum X_{old}}{n_{old}}\Rightarrow \sum X_{old}=n_{old} \times \bar{X}_{old}=12\times 34=408[/tex] [tex]\to s_{old}=4.75 \\\\\to\sqrt{\frac{\sum X_{old}^2-\frac{(\sum X_{old})^2}{n_{old}}}{n_{old}-1}}=4.75[/tex]
[tex]\to \frac{\sum X_{old}^2-\frac{(\sum X_{old})^2}{n_{old}}}{n_{old}-1}=4.75^2=22.5625[/tex]
[tex]\to \frac{ \sum X_{old}^2-\frac{408^2}{12}}{12-1} =22.5625\\\\\to \sum X_{old}^2=22.5625*11+\frac{408^2}{12}=14120.1875[/tex]
After adding additional two common values of 34 each.
[tex]\to \sum X_{new}=\sum X_{old}+34+34=408+34+34=476\\\\ \to \sum X^2_{new}=\sum X^2_{old}+34^2+34^2=14120.1875+1,156+1,156=16432.1875[/tex]
For point a:
The mean of the sample of 14 measurements:
[tex]\to \bar{X}_{new}=\frac{\sum X_{new}}{n_{new}}=\frac{476}{14}=34[/tex]
For point b:
The standard deviation of the sample of 14 measurements:
[tex]\to s_{new}=\sqrt{\frac{\sum X_{new}^2-\frac{(\sum X_{new})^2}{n_{new}}}{n_{new}-1}}=\sqrt{\frac{16432.1875-\frac{476^2}{14}}{14-1}}=4.37[/tex]
Therefore the final answer is "34 and 4.37".
Learn more:
brainly.com/question/14859731
