Answer:The standard deviation of the difference between the means of the two populations is 2.
Step-by-step explanation:
The given table :
Today Five years ago
[tex]\overline{x}=\ \ \ \ 82\ \ \ \ \ 88\\\sigma^2=\ \ 112.5\ \ 54\\n=\ \ \ \ 45\ \ \ \ \ 36[/tex]
Let A denotes the scores from students enrolled today and B denotes the scores from students enrolled five years ago.
Then, the standard deviation of the difference between the means of the two populations would be :-
[tex]S.E._{\overline{x}_A-\overline{x}_B}}=\sqrt{\dfrac{\sigma^2_A}{n_1}+\dfrac{\sigma^2_B}{n_2}}\\\\=\sqrt{\dfrac{112.5}{45}+\dfrac{54}{36}}\\\\=\sqrt{4}=2[/tex]
Hence, the standard deviation of the difference between the means of the two populations is 2.
