Question

The weight of Bluefin Tuna is approximately normally distributed with mean 32 pounds and standard deviation 5 pounds. Answer the following questions:

1a. How much does a fish have to weigh to fall into the 75 percentile? (Inverse)

a) 32.37

b) 35.37

c) 32.75

d) 38.7

1b. The record catch for a Tuna is 511 pounds. What are the chances of that happening? Find the z score.

a) 504.6

b) 95.8

c) 94

d) 100

1c. In the sample of 30 fish what is the probability the mean weight of fish is more than 30 pounds?

a) 65

b) .015

c) .47

d) .98

Answer 1

The weight of Bluefin Tuna is approximately normally distributed with a mean of 32 pounds and standard deviation of 5 pounds.

Let, X be the random variable denoting the weight of Bluefin Tuna.

Then X follows normal with mean 32 and standard deviation of 5.

Question (a)

We have to find the 75th percentile of X.

So, we have to find m, such that

Where, Z is the standard normal variate.

Where, phi is the distribution function of the standard normal variate.

From the standard normal table, we note that

So, we can conclude that

So, the correct answer is option (b) 35.37.

Question (b)

The record catch for a Tuna is 511 pounds.

So, the z-score of this weight is given by

So, the z-score is option (b) 95.8.

Question (c)

In a sample of 30 fishes, we have to find the probability that the mean weight of this sample is more than 30 pounds.

Let, Y be the random variable denoting the average weight of these fishes.

Then Y follows normal with mean 32 and standard deviation of

Now, we have to find

Where, Z follows standard normal with mean 0 and standard deviation of 1.

Where, phi is the distribution function of the standard normal variate.

From the standard normal table, this becomes

approximately.

So, the correct answer is option (D) 0.98.