ANSWER
0.3783
EXPLANATION
The lengths of the pregnancies, X, is normally distributed with a mean of 246 days and a standard deviation of 13 days,
[tex]X=N(246,13)[/tex]We have to find the probability that a randomly selected pregnancy lasts less than 242 days,
[tex]P(X<242)=?[/tex]For this, we have to standardize the variable X with the formula,
[tex]Z=\frac{X-\mu}{\sigma}[/tex]So the probability we have to solve is,
[tex]P(X<242)=P\mleft(\frac{X-\mu}{\sigma}<\frac{242-\mu}{\sigma}\mright)=P\mleft(Z<\frac{242-246}{13}\mright)=P(Z<-0.31)[/tex]This is equivalent to,
[tex]P(Z<-0.31)=P(Z>0.31)[/tex]Which is also equivalent to,
[tex]P(Z>0.31)=1-P(Z<0.31)[/tex]We have to find these equivalences because, usually, normal distribution tables show the probabilities for positive z-scores and to the left of those values - i.e. less than those values. Find z = 0.31 in a z-table,
So, the probability is,
[tex]P(X<242)=1-P(Z<0.31)=1-0.6217=0.3783[/tex]Hence, the probability that a randomly selected pregnancy lasts less than 242 days is 0.3783.
