Question

The lengths of a particular​ animal's pregnancies are approximately normally​ distributed, with mean muμequals=265265 days and...

The lengths of a particular​ animal's pregnancies are approximately normally​ distributed, with mean

muμequals=265265

days and standard deviation

sigmaσequals=1616

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

285285

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

253253

and

273273

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

237237

​days?​(d) A​ "very preterm" baby is one whose gestation period is less than

229229

days. Are very preterm babies​ unusual?

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Click the icon to view a table of areas under the normal curve.

​(a) The proportion of pregnancies that last more than

285285

days is

nothing.

​(Round to four decimal places as​ needed.)

​(b) The proportion of pregnancies that last between

253253

and

273273

days is

nothing.

​(Round to four decimal places as​ needed.)

​(c) The probability that a randomly selected pregnancy lasts no more than

237237

days is

nothing.

​(Round to four decimal places as​ needed.)

​(d) A​ "very preterm" baby is one whose gestation period is less than

229229

days. Are very preterm babies​ unusual?The probability of this event is

nothing​,

so it

would

would not

be unusual because the probability is

greater

less

than 0.05.

​(Round to four decimal places as​ needed.)

Homework Answers

Answer #1

a)
P(X >= 285) = P(z <= (285 - 265)/16)
= P(z >= 1.25)
= 1 - 0.8944
= 0.1056

b)
P(253 <= X <= 273) = P((273 - 265)/16) <= z <= (273 - 265)/16)
= P(-0.75 <= z <= 0.5) = P(z <= 0.5) - P(z <= -0.75)
= 0.6915 - 0.2266
= 0.4649

c)
P(X <= 237) = P(z <= (237 - 265)/16)
= P(z <= -1.75)
= 0.0401

d)
P(X <= 229) = P(z <= (229 - 265)/16)
= P(z <= -2.25)
= 0.0122

This is very unusual
The probability of this event is 0.0122 so it would be unusual because the probability is less than 0.05

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