Answer:
Step-by-step explanation:
Given that many biological measurements on the same species follow a Normal distribution quite closely. The weights of seeds of a variety of winged bean are approximately Normal with mean 525 milligrams (mg) and standard deviation 110 mg.
X = biological measurements on same species
is N(525, 110)
a) P(X>500) = 1-0.4101 = 0.5899
b) Lighest weight are those which are below z = -1.28
Corresponding x score = [tex]525-1.28*110\\= 384.20[/tex]
c) P(435<X<620) = F(620) -F(435)
= 0.8061-0.2066
= 0.5995
In percent = 59.95%
d) Heavies 25% seeds correspond to z = 0.675
X score = [tex]525+0.675*110\\=599.25[/tex]
Above 599.25 mg are the heaviest of the seeds.