Respuesta :
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,[tex]\bar x=\frac{x_{1}+x_{2}+x_{3}+...+x_{n}}{n}=4[/tex]
The standard deviation of this sample is 2. That is, [tex]s=\frac{1}{n-1}\sum (x_{i}-\bar x)^{2}=2[/tex].
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:
[tex]\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[/tex]
The new mean is 5.
Compute the standard deviation of this sample as follows:
[tex]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[/tex]
The new standard deviation is also 2.
