Answer:
a) New mean 78.07692308 thousands of dollars
b) The median does not vary
c) New standard deviation 150.1793799 thousands of dollars
Step-by-step explanation:
We are working in units of $1,000
a)What is the value of the mean after the change?
Let
[tex]s_1,s-2,...,s_38[/tex]
be the salaries of the employees that earn less than 100 units.
The mean of the 39 salaries is 55 units so
[tex]\displaystyle\frac{s_1+s_2+...+s_{38}+100}{39}=55[/tex]
and
[tex]s_1+s_2+...+s_{38}+100=55*39=2145[/tex]
By accident, the 100 on the left is changed to 1,000
[tex]s_1+s_2+...+s_{38}+100+900=55*39=2145+900\Rightarrow \\\\\Rightarrow s_1+s_2+...+s_{38}+1000=3045[/tex]
Dividing by 39 both sides, we get the new mean
[tex]\displaystyle\frac{s_1+s_2+...+s_{38}+1000}{39}=\displaystyle\frac{3045}{39}=78.07692308[/tex]
b)What is the value of the median after the change?
Since the number of data does not change and only the right end of the range of salaries is changed, the median remains the same; 55
c)What is the value of the standard deviation after the change?
The variance is 400, so
[tex]\displaystyle\frac{(s_1-70)^2+(s_2-70)^2+...+(s_{38}-70)^2+(100-70)^2}{39}=400\Rightarrow\\\\\Rightarrow (s_1-70)^2+(s_2-70)^2+...+(s_{38}-70)^2+(100-70)^2=400*39=15600[/tex]
Adding 864,000 to both sides we get
[tex](s_1-70)^2+(s_2-70)^2+...+(s_{38}-70)^2+(100-70)^2+864000=15600+864000\Rightarrow\\\\\Rightarrow (s_1-70)^2+(s_2-70)^2+...+(s_{38}-70)^2+(1000-70)^2=879600[/tex]
Dividing by 39 and taking the square root we get the new standard deviation
[tex]\displaystyle\frac{(s_1-70)^2+(s_2-70)^2+...+(s_{38}-70)^2+(1000-70)^2}{39}=\displaystyle\frac{879600}{39}=22553.84615\Rightarrow\\\\\Rightarrow \sqrt{\displaystyle\frac{(s_1-70)^2+(s_2-70)^2+...+(s_{38}-70)^2+(1000-70)^2}{39}}=150.1793799[/tex]
