Answer:
The new mean is 5.
The new standard deviation is also 2.
Step-by-step explanation:
Let the sample space of hours be as follows, S = {x₁, x₂, x₃...xₙ}
The mean of this sample is 4. That is,![\bar x=\frac{x_{1}+x_{2}+x_{3}+...+x_{n}}{n}=4](https://tex.z-dn.net/?f=%5Cbar%20x%3D%5Cfrac%7Bx_%7B1%7D%2Bx_%7B2%7D%2Bx_%7B3%7D%2B...%2Bx_%7Bn%7D%7D%7Bn%7D%3D4)
The standard deviation of this sample is 2. That is,
.
Now it is stated that each of the sample values was increased by 1 hour.
The new sample is: S = {x₁ + 1, x₂ + 1, x₃ + 1...xₙ + 1}
Compute the mean of this sample as follows:
![\bar x_{N}=\frac{x_{1}+1+x_{2}+1+x_{3}+1+...+x_{n}+1}{n}\\=\frac{(x_{1}+x_{2}+x_{3}+...+x_{n})}{n}+\frac{(1+1+1+...n\ times)}{n}\\=\bar x+1\\=4+1\\=5](https://tex.z-dn.net/?f=%5Cbar%20x_%7BN%7D%3D%5Cfrac%7Bx_%7B1%7D%2B1%2Bx_%7B2%7D%2B1%2Bx_%7B3%7D%2B1%2B...%2Bx_%7Bn%7D%2B1%7D%7Bn%7D%5C%5C%3D%5Cfrac%7B%28x_%7B1%7D%2Bx_%7B2%7D%2Bx_%7B3%7D%2B...%2Bx_%7Bn%7D%29%7D%7Bn%7D%2B%5Cfrac%7B%281%2B1%2B1%2B...n%5C%20times%29%7D%7Bn%7D%5C%5C%3D%5Cbar%20x%2B1%5C%5C%3D4%2B1%5C%5C%3D5)
The new mean is 5.
Compute the standard deviation of this sample as follows:
![s_{N}=\frac{1}{n-1}\sum (x_{i}-\bar x)^{2}\\=\frac{1}{n-1}\sum ((x_{i}+1)-(\bar x+1))^{2}\\=\frac{1}{n-1}\sum (x_{i}+1-\bar x-1)^{2}\\=\frac{1}{n-1}\sum (x_{i}-\bar x)^{2}\\=s](https://tex.z-dn.net/?f=s_%7BN%7D%3D%5Cfrac%7B1%7D%7Bn-1%7D%5Csum%20%28x_%7Bi%7D-%5Cbar%20x%29%5E%7B2%7D%5C%5C%3D%5Cfrac%7B1%7D%7Bn-1%7D%5Csum%20%28%28x_%7Bi%7D%2B1%29-%28%5Cbar%20x%2B1%29%29%5E%7B2%7D%5C%5C%3D%5Cfrac%7B1%7D%7Bn-1%7D%5Csum%20%28x_%7Bi%7D%2B1-%5Cbar%20x-1%29%5E%7B2%7D%5C%5C%3D%5Cfrac%7B1%7D%7Bn-1%7D%5Csum%20%28x_%7Bi%7D-%5Cbar%20x%29%5E%7B2%7D%5C%5C%3Ds)
The new standard deviation is also 2.