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Question 5 (1 point) For students who graduated from high school in 2013, the ACT math scores are approximately normally dist
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Answer #1

Given that

  • \mu=20.9
  • \sigma=5.3

\mathrm{z \; score=\frac{x-\mu}{\sigma}}

\therefore z score of 17 points = (17 - 20.9)/5.3 \approx -0.74

Let, X = ACT math scores.

Proportion of students who took ACT scored at least 17 points

=P(X\geq17)

=1-P(X<17)

=1-P(\frac{X-\mu}{\sigma}<z \; score); \; \frac{X-\mu}{\sigma}\sim N(0,1)

=1-\Phi(-0.74)

=1-[1-\Phi(0.74)]=\Phi(0.74)

=0.7704​​​​​​

\mathrm{\Phi(.)\; is \; cdf \; of \; standard\; normal\; distribution}Standard normal table:

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