Question

In a certain population of herring, Pomolobus Aestivalis, the lengths of the individual fish are normally distributed. The mean length of a fish is 54 mm and the standard deviation is 4mm.

a.) What is the probability that the first fish capture is less than 62mm long?
b.) What percentage of the fish are longer than 59 mm?
c.) What is the 25th Percentile value?
d.) What is the 70th Percentile value?

Answer 1

Answer:

0.9773,10.57%,,51.31, 56.69

Step-by-step explanation:

Given that in a certain population of herring, Pomolobus Aestivalis, the lengths of the individual fish are normally distributed. The mean length of a fish is 54 mm and the standard deviation is 4mm.

Let X represent the length of fish caught

X is N(54, 4)

a.)  the probability that the first fish capture is less than 62mm long

$=P(X<62) \\=0.9773$

b.)  percentage of the fish are longer than 59 mm

=$100*P(X>59)\\=100(0.10565)\\=10.57%$%

c.)  the 25th Percentile value

=51.31

d) 75th percentile value

=56.69

d.) What is the 70th Percentile value?