Question

he lengths of a particular​ animal's pregnancies are approximately normally​ distributed, with mean muequals252 days and standard deviation sigmaequals20 days. ​(a) What proportion of pregnancies lasts more than 262 ​days? ​(b) What proportion of pregnancies lasts between 227 and 257 ​days? ​(c) What is the probability that a randomly selected pregnancy lasts no more than 237 ​days? ​(d) A​ "very preterm" baby is one whose gestation period is less than 202 days. Are very preterm babies​ unusual?

Answer 1

Answer:

Step-by-step explanation:

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

X is N(252, 20)

​(a) What proportion of pregnancies lasts more than 262 ​days? ​

$P(X>262) = 0.3085$

(b) What proportion of pregnancies lasts between 227 and 257 ​days?

$P(227<x<257)\\= F(257)-F(227)\\=0.5987-0.1056\\=0.4931$

​(c) What is the probability that a randomly selected pregnancy lasts no more than 237 ​days?

P(X<237) = 0.2266

​(d) A​ "very preterm" baby is one whose gestation period is less than 202 days. Are very preterm babies​ unusual?

$P(X<202) = 0.00621$

Yes very unusual since probability is very near to 0.

Answer 2

Answer:

(a) P(X > 262) = 0.30854

(b) P(227 < X < 257) = 0.49306

(c)  P(X $\leq$ 237) = 0.22663

(d) Yes, a very preterm babies​ are unusual.

Step-by-step explanation:

We are given that the lengths of a particular​ animal's pregnancies are approximately normally​ distributed, with mean = 252 days and standard deviation = 20 days.

Let X = lengths of a particular​ animal's pregnancies

So, X ~ N($\mu = 252, \sigma^{2} =20^{2}$)

The z score probability distribution is given by;

                 Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)  

where, $\mu$ = population mean

            $\sigma$ = population standard deviation

(a) Probability that pregnancies lasts more than 262 ​days is given by = P(X > 262)

P(X > 262) = P( $\frac{X-\mu}{\sigma}$ > $\frac{262-252}{20}$ ) = P(Z > 0.5) = 1 - P(Z $\leq$ 0.5)

                                                 = 1 - 0.69146 = 0.30854

(b) Probability that pregnancies lasts between 227 and 257 ​days is given by = P(227 < X < 257) = P(X < 257) - P(X $\leq$ 227)

    P(X < 257) = P( $\frac{X-\mu}{\sigma}$ < $\frac{257-252}{20}$ ) = P(Z < 0.25) = 0.59871

    P(X $\leq$ 227) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{227-252}{20}$ ) = P(Z $\leq$ -1.25) = 1 - P(Z < 1.25)

                                                       = 1 - 0.89435 = 0.10565

Therefore, P(227 < X < 257) = 0.59871 - 0.10565 = 0.49306

(c) Probability that a randomly selected pregnancy lasts no more than 237 ​days = P(X $\leq$ 237)

     P(X $\leq$ 237) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{237-252}{20}$ ) = P(Z $\leq$ -0.75) = 1 - P(Z < 0.75)

                                                        = 1 - 0.77337 = 0.22663

(d) Firstly, we will find the probability that gestation period is less than 202 days = P(X < 202)

    P(X < 202) = P( $\frac{X-\mu}{\sigma}$ < $\frac{202-252}{20}$ ) = P(Z $\leq$ -2.5) = 1 - P(Z < 2.5)

                                                      = 1 - 0.99379 = 0.00621

Since, we got a P(X < 202) of 0.00621 which is very small or nearly close to 0, so it means that Yes, a very preterm babies​ are unusual.