Question

MyLê This Question: 1 pt Stude 2 of 12 (1 complete) This Quiz: 12 pts possible о му с Question Help Cours The lengths of a particular animal's pregnancies are approximately normally distributed, with mean y = 270 days and standard deviation o 8 days. (a) What proportion of pregnancies lasts more than 276 days? Assig (b) What proportion of pregnancies lasts between 268 and 274 days? (c) What is the probability that a randomly selected pregnancy lasts no more than 258 days? (d) A very preterm" baby is one whose gestation period is less than 252 days. Aro very preterm babies unusual? (a) The proportion of pregnancies that last more than 276 days is eText (Round to four decimal places as needed.) Study (b) The proportion of pregnancies that last between 268 and 274 days is (Round to four decimal places as needed.) Video (e) The probability that a randomly selected pregnancy lasts no more than 258 days is (Round to four decimal places as needed) Class (d) The probability of a very preterm baby is This event unusual because the probability is than 0.05. Stude (Round to four decimal places as needed.) Work Ubra Data State Acces Purch Enter your answer in each of the answer boxes. Come ? https://www.matal.com/Student/My Dashboard.aspx?productiecongs handler urn. Pearsonsfeerg.mxX2link-pearson-cong.mm x&learningproduct VL# 31

Answer 1

Let X be the lengths of a particular animal's pregnancies.

X ~ N($\small \mu$ = 270, = 8)

0

Note that,

$\small \frac{X-\mu}{\sigma}\sim N(0,1) \\ Let \ Z=\frac{X-\mu}{\sigma} \\ i.e. \ P(Z\le z)=\Phi(z) \ and \ \Phi(-z)=1-\Phi(z)$

(a) We have to find P(X>276)

PX > 276) = 1- P(X < 276)

                       $\small =1-P(\frac{X-\mu}{\sigma}\le \frac{276-270}{8})$

                        $\small =1-P(Z\le 0.75)=1-\Phi(0.75)$

                                                            $\small =1-0.7764=0.2236$

The proportion of pregnancies that last more than 276 days is 0.2236 or 22.36%       (Answer)

(b) We have to find,

$\small P(268\le X\le 274)=P(X\le 274)-P(X\le 268)$

                                   $\small =P(\frac{X-\mu}{\sigma}\le \frac{274-270}{8})-P(\frac{X-\mu}{\sigma}\le \frac{268-270}{8})$

                                   $\small =P(Z\le 0.5)-P(Z\le -0.25)$

                                    $\small =\Phi(0.5)-\Phi(-0.25)=\Phi(0.5)-1+\Phi(0.25)$

                                                                           $\small =0.6915-1+0.5987=0.2904$

The proportion of pregnancies that last between 268 and 274 days is 0.2904 or 29.04%       (Answer)

(c) We have to find,

$\small P(X\le 258)=P(\frac{X-\mu}{\sigma}\le \frac{258-270}{8})$

                        $\small =P(Z\le -1.5)=\Phi(-1.5)=1-\Phi(1.5)$

                                                                          $\small =1-0.9332=0.0668$

The probability that a randomly selected pregnancy lasts no more than 258 days is 0.0668 or 6.68%       (Answer)

(d) Probability of a very preterm baby is -

$\small P(X<252)=P(\frac{X-\mu}{\sigma}<\frac{252-270}{8})$

                        $\small =P(Z< -2.25)=\Phi(-2.25)=1-\Phi(2.25)$

                                                                               $\small =1-0.9878=0.0122$

The probability of a "very preterm" baby is 0.0122. This event would be unusual because the probability is less than 0.05.                  (Answer)