Question

The lengths of pregnancies are normally distributed with a mean of 268 days and a standard deviation of 15 days.
a. Find the probability of a pregnancy lasting 309 days or longer.
b. If the length of pregnancy is in the lowest 4​%, then the baby is premature. Find the length that separates premature babies from those who are not premature.

Answer 1

Answer:

probability of a pregnancy lasting 309 days is 0.0032

the length of pregnancy is in the lowest 4​% is 241.74 days

Step-by-step explanation:

Given data

mean = 268

standard deviation = 15

to find out

probability of a pregnancy lasting 309 days or longer and the length of pregnancy is in the lowest 4​%

solution

we know mean (M)  is 268 and standard deviation (SD) is 15

so probability will be  in 1st part where pregnancy lasting 309 days

P(X  >  309  )  =  P(  X−  mean  > 309  −  mean  )

we know  

Z   =   (309 −  mean)  / standard deviation ,  

it will be Z =  309−268/15  =  2.73  

we can say these both are equal

P(X  >  309  )  =  P(  Z  >  2.73)

now we use the standard normal z-table i.e.

P(  Z  >  2.73  )  =  0.0032

so P(  X  >  309  ) is 0.0032

probability of a pregnancy lasting 309 days is 0.0032

and in 2nd part z value with 4% .i.e 0.04

z = -1.7507 from the standard table

so days = z × standard deviation + mean

days = -1.7507  × 15 + 268

days = 241.74

the length of pregnancy is in the lowest 4​% is 241.74 days