Two professors teach two sections A and B of the same class. Thecourse scores in section A follow a Gaussian random variable of mean 76 and variance64. The course scores in section B follow a Gaussian random variable of mean 67 andvariance 36. What is the probability that the scores of two students, one in sectionA and one in section B, differ by no more than 1 point in absolute value

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.

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