Answer: A. 0.0164
Step-by-step explanation:
Given : The lengths of human pregnancies are normally distributed with a mean [tex]\mu=268\text{ days}[/tex]
Standard deviation : [tex]\sigma=15\text{ days}[/tex]
Let X be the random variable that represents the length of pregnancy of a randomly selected human .
z-score : [tex]z=\dfrac{X-\mu}{\sigma}[/tex]
For X = 300
[tex]z=\dfrac{300-268}{15}\approx2.13[/tex]
Now, the probability that a pregnancy last at least 300 days will be :-
[tex]P(X\geq300)=P(z\geq 2.1333)=1-P(z<2.1333)\\\\=1- 0.9835513=0.0164487\approx0.0164[/tex]
Hence, the probability that a pregnancy last at least 300 days =0.0164
