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denpristay [2]
2 years ago
10

I need help quick please

Mathematics
1 answer:
Sati [7]2 years ago
3 0

Answer:

search it up its there

Step-by-step explanation:

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The scores of individual students on the American College Testing (ACT) Program College Entrance Exam have a normal distribution
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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);

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

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

P(Xi≥19.0)=0.473

\{Yi=0, Xi<  19\\Yi=1, Xi\geq  19\}

p=0.473

Yi~Ber(0.473)

P(\frac{1}{n}\displaystyle\sum_{i=1}^{n}X_i\geq 19)=P(\displaystyle\sum_{i=1}^{400}X_i\geq 7600)

Based on the Central Limit Theorem:

\displaystyle\sum_{i=1}^{n}X_i\~{}N(n\mu, \sqrt{n}\sigma),\displaystyle\sum_{i=1}^{400}X_i\~{}N(7440, 372)

Then:

P(\displaystyle\sum_{i=1}^{400}X_i\geq 7600)=1-P(0

P(\displaystyle\sum_{i=1}^{n}Y_i=1)=P(\displaystyle\sum_{i=1}^{400}Y_i=1)

Based on the Central Limit Theorem:

\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)

P(\displaystyle\sum_{i=1}^{400}Y_i=1)\~{=}P(0.5

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

7 0
3 years ago
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