The weight of the fish in a certain lake are normally distributed with a mean of
ID: 3292663 • Letter: T
Question
The weight of the fish in a certain lake are normally distributed with a mean of 11lbs. and a standard deviation of 6lbs.a) If a fish is randomly selected, what is the probability that it's weight will be between 8.6 and 14.6 lbs.?
b) If a sample of 4 fish is randomly selected, what is the probability that the sample weight will be between 8.6lbs and 14.6lbs.
c) Find the weight that separates the heaviest 25% (I.e P75) of fish from the rest. The weight of the fish in a certain lake are normally distributed with a mean of 11lbs. and a standard deviation of 6lbs.
a) If a fish is randomly selected, what is the probability that it's weight will be between 8.6 and 14.6 lbs.?
b) If a sample of 4 fish is randomly selected, what is the probability that the sample weight will be between 8.6lbs and 14.6lbs.
c) Find the weight that separates the heaviest 25% (I.e P75) of fish from the rest.
a) If a fish is randomly selected, what is the probability that it's weight will be between 8.6 and 14.6 lbs.?
b) If a sample of 4 fish is randomly selected, what is the probability that the sample weight will be between 8.6lbs and 14.6lbs.
c) Find the weight that separates the heaviest 25% (I.e P75) of fish from the rest.
Explanation / Answer
as we know for normal distribution z score =(X-mean)/std deviation
a) P(8.6<X<14.6)=P((8.6-11)/6<z<(14.6-11)/6)=P(-0.4<Z<0.6)=0.7257-0.3446=0.3812
b) for n=4 ; std error =std deviation/(n)1/2 =6/(4)1/2 =3
therefore P(8.6<X<14.6)=P((8.6-11)/3<z<(14.6-11)/3)=P(-0.8<Z<1.2)=0.8849-0.2119=0.6731
c) for 75th percentile ; z=0.6745
therefore corresponding weight =mean +z*std deviaiton =11+0.6745*6=15.047
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