Question

Suppose a certain study reported that 27.7% of high school students smoke.

Answer 1

Given:

Probability of student smoke,

P = 27.7%

  = 0.277

Number of students,

n = 632

$q = 1-p$

  $=1-0.277$

  $=0.723$

(i)

Here,

Number of students (n) = 60

then,

⇒ $n_P=60\times 0.277$

         $=16.62$

⇒ $n_q=60\times 0.723$

        $=43.38$

We can see that $n_P > 10$ and $n_q>10$ so the normal approximation condition are met.

Now,

$\mu = n_P= 16.62$

$\sigma = \sqrt{n_{Pq}}$

  $= \sqrt{60\times 0.277\times 0.723}$

  $=3.9664$

Now,

⇒ $P(X$

                      $=P(Z_{18.5})$

The Z-score is:

= $\frac{18.5-16.62}{3.4664}$

= $0.5423$

hence,

The probability will be:

⇒ $P(Z_{18.5}) = 0.7062$

or,

⇒ $P(Z$

(ii)

Here,

Number of students (n) = 75

$\mu = n_P = 75\times 0.277$

            $=20.775$

$\sigma = \sqrt{n_{Pq}}$

   $=\sqrt{75\times 0.277\times 0.723}$

   $=3.8756$

Now,

⇒ $P(X>17) = P(X> 17.5)$

                      $=1-P(X \leq 17.5)$

                      $=1-P(Z_{17.5})$

The Z-score is:

= $\frac{17.5-20.775}{3.8756}$

= $-0.9740$

then, $P(Z_{17.5}) = 0.165$

hence,

The probability will be:

⇒ $P(X>17) = 1-0.165$

                      $=0.835$    

Learn more about probability here:

brainly.com/question/9825651