Answer:
The probability that the average of the scores of all 400 students exceeds 19.0 is larger than the probability that a single student has a score exceeding 19.0
Explanation:
Xi~N(18.6, 6.0), n=400, Yi~Ber(p); Z~N(0, 1);
![P(0\leq X\leq 19.0)=P((0-\mu)/(\sigma) \leq (X-\mu)/(\sigma)\leq (19-\mu)/(\sigma)), Z= (X-\mu)/(\sigma), \mu=18.6, \sigma=6.0](https://img.qammunity.org/2020/formulas/mathematics/college/7nnju4lyfveiw3syj7wt7v6zjh7cpqsio2.png)
![P(-3.1\leq Z\leq 0.0667)=\Phi(0.0667)-\Phi (-3.1)=\Phi(0.0667)-(1-\Phi (3.1))=0.52790+0.99903-1=0.52693](https://img.qammunity.org/2020/formulas/mathematics/college/y05uph89fuctxt48xtd1hs141r6cp7wbe2.png)
P(Xi≥19.0)=0.473
![\{Yi=0, Xi< 19\\Yi=1, Xi\geq 19\}](https://img.qammunity.org/2020/formulas/mathematics/college/kegbalzu0vgt2ourv6ick4igwn7qwi6t8s.png)
p=0.473
Yi~Ber(0.473)
![P((1)/(n)\displaystyle\sum_(i=1)^(n)X_i\geq 19)=P(\displaystyle\sum_(i=1)^(400)X_i\geq 7600)](https://img.qammunity.org/2020/formulas/mathematics/college/4my1uavym7p6vnvae7lw8z355iu9x3we6v.png)
Based on the Central Limit Theorem:
![\displaystyle\sum_(i=1)^(n)X_i\~{}N(n\mu, √(n)\sigma),\displaystyle\sum_(i=1)^(400)X_i\~{}N(7440, 372)](https://img.qammunity.org/2020/formulas/mathematics/college/3rszquiujwqzisk9z6nz696p2wzxh7ex7c.png)
Then:
![P(\displaystyle\sum_(i=1)^(400)X_i\geq 7600)=1-P(0<\displaystyle\sum_(i=1)^(400)X_i<7600)=\\1-P(-20<Z<0.43)=1-(\Phi(0.43)-(1-\Phi(20)))=0.334](https://img.qammunity.org/2020/formulas/mathematics/college/nus7vq8wpxrjoqdne63drcxk43newyfh9w.png)
![P(\displaystyle\sum_(i=1)^(n)Y_i=1)=P(\displaystyle\sum_(i=1)^(400)Y_i=1)](https://img.qammunity.org/2020/formulas/mathematics/college/fm0hguj7iuoayi3a4x4022inzvzem1e9cr.png)
Based on the Central Limit Theorem:
![\displaystyle\sum_(i=1)^(400)Y_i\~{}N(400* 0.473, √(400)* 0.499)=\displaystyle\sum_(i=1)^(400)Y_i\~{}N(189.2; 9.98)](https://img.qammunity.org/2020/formulas/mathematics/college/2m84i1jgsrmx99z7547r1lxu9573zyp5ag.png)
![P(\displaystyle\sum_(i=1)^(400)Y_i=1)\~{=}P(0.5<\displaystyle\sum_(i=1)^(400)Y_i<1.5)=P(-18.9<Z<-18.8)\~{=}0](https://img.qammunity.org/2020/formulas/mathematics/college/fqeaimfewchxqgke5xx8v1h99xua29u7ek.png)
Then:
the probability that the average of the scores of all 400 students exceeds 19.0 is larger than the probability that a single student has a score exceeding 19.0