Answer:
Step-by-step explanation:
Let A denote the score for students of section A.
Let B denote the score for students of section B.
It is provided that:
[tex]A\sim N(76,64)\\B\sim N(67,36)[/tex]
The scores of the students of the two section are independent.
Compute the mean and variance of A - B as follows:
[tex]E(A-B)=E(A)-E(B)=76-67=9\\\\V(A-B)=V(A)+V(B)-2Cov(A,B)=64+36-0=100[/tex]
Compute the probability that the scores of two students, one in section A and one in section B, differ by no more than 1 point in absolute value as follows:
[tex]P(|A-B|\leq 1)=P(-1\leq X-Y\leq 1)[/tex]
[tex]=P(\frac{-1}{10}\leq \frac{(X-Y)-E(X-Y)}{\sqrt{V(X-Y)}}\leq \frac{1}{10})\\\\=P(-0.10<Z<0.10)\\\\=P(Z<0.10)-P(Z<-0.10)\\\\=0.53983-0.46017\\\\=0.07966\\\\\approx 0.08[/tex]
Thus, the probability that the scores of two students, one in section A and one in section B, differ by no more than 1 point in absolute value is 0.08.