I need a step by step solution to this:
(The answers are already given; I just need the computations)
Data indicates an average population weight of a fish of 40 lbs and a standard deviation of 4 lbs. Assume weight is normally distributed in the population. How probable was it to catch a tuna:
a. greater than 50 lbs Answer: 0.62%
b. less than 35 lbs Answer: 10.56%
c. between 32 and 36 lbs Answer: 13.59%
d. 36 and 48 lbs Answer: 81.85%

Thank you so much!!

Data indicates an average population weight of a fish of 40 lbs and a standard deviation of 4 lbs. Assume weight is normally distributed in the population. How probable was it to catch a tuna:
a. greater than 50 lbs
b. less than 35 lbs
c. between 32 and 36 lbs
d. 36 and 48 lbs

Data indicates an average population weight of a fish of 40 lbs and a standard deviation of 4 lbs. Assume weight is normally distributed in the population. How probable was it to catch a fish:
a. greater than 50 lbs
b. less than 35 lbs
c. between 32 and 36 lbs
d. 36 and 48 lbs

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You'll have to define your variable first.

Let X be the weight of a fish.

Then, X \sim N(\mu , \sigma^2)

You should know what are mu and sigma.

Then, the probability that the weight is greater than a certain value x,

P(X > x) = P(Z > \frac{x - \mu}{\sigma})

Then you look in your normal table and look for the probability associated with the z value.

Can you use this hint and solve your problem, then post the work that you did?