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

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

(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.

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 =
`(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(
`(X-\mu)/(\sigma)` >
`(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(
`(X-\mu)/(\sigma)` <
`(257-252)/(20)` ) = P(Z < 0.25) = 0.59871

P(X
`\leq` 227) = P(
`(X-\mu)/(\sigma)`
`\leq`
`(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(
`(X-\mu)/(\sigma)`
`\leq`
`(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(
`(X-\mu)/(\sigma)` <
`(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.