Question

6.2.31 Question Help O 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 307 days or longer. b. If the length of pregnancy is in the lowest 3%, then the baby is premature. Find the length that separates premature babies from those who are not premature. Click to view page 1 of the table. Click to view page 2 of the table. - a. The probability that a pregnancy will last 307 days or longer is (Round to four decimal places as needed.)

Answer 1

Solution :

Given that ,

mean = $\mu$ = 268

standard deviation = $\sigma$ = 15

a)

P(x $\geq$ 307) = 1 - P(x  $\leq$ 307)

= 1 - P((x - $\mu$ ) / $\sigma$ $\leq$ (307 - 268) / 15)

= 1 -  P(z $\leq$ 2.6)  

= 1 - 0.9953 Using standard normal table.   

Probability = 0.0047

b)

The z - distribution of the % is,

P( Z < z ) = 3%

P( Z < z ) = 0.03

P( Z < -1.88 ) = 0.03

z = -1.88

Using z - score formula,

X = z * $\sigma$ + $\mu$

= -1.88 * 15 + 268

= 239.8