Respuesta :
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
Step-by-step explanation:
Xi~N(18.6, 6.0), n=400, Yi~Ber(p); Z~N(0, 1);
[tex]P(0\leq X\leq 19.0)=P(\frac{0-\mu}{\sigma} \leq \frac{X-\mu}{\sigma}\leq \frac{19-\mu}{\sigma}), Z= \frac{X-\mu}{\sigma}, \mu=18.6, \sigma=6.0[/tex]
[tex]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[/tex]
P(Xi≥19.0)=0.473
[tex]\{Yi=0, Xi< 19\\Yi=1, Xi\geq 19\}[/tex]
p=0.473
Yi~Ber(0.473)
[tex]P(\frac{1}{n}\displaystyle\sum_{i=1}^{n}X_i\geq 19)=P(\displaystyle\sum_{i=1}^{400}X_i\geq 7600)[/tex]
Based on the Central Limit Theorem:
[tex]\displaystyle\sum_{i=1}^{n}X_i\~{}N(n\mu, \sqrt{n}\sigma),\displaystyle\sum_{i=1}^{400}X_i\~{}N(7440, 372)[/tex]
Then:
[tex]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[/tex]
[tex]P(\displaystyle\sum_{i=1}^{n}Y_i=1)=P(\displaystyle\sum_{i=1}^{400}Y_i=1)[/tex]
Based on the Central Limit Theorem:
[tex]\displaystyle\sum_{i=1}^{400}Y_i\~{}N(400\times 0.473, \sqrt{400}\times 0.499)=\displaystyle\sum_{i=1}^{400}Y_i\~{}N(189.2; 9.98)[/tex]
[tex]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[/tex]
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