Question

the size of fish is very important to commercial fishing. a study conducted in 2012 found the length of atlantic cod caught in nets in karlskrona to have a mean of 49.9 cm and a standard deviation of 3.74 cm. round the probabilities to four decimal places. it is possible with rounding for a probability to be 0.0000. a) state the random variable. b) find the probability that a randomly selected atlantic cod has a length of 40.58 cm or more. c) find the probability that a randomly selected atlantic cod has a length of 49.28 cm or less. d) find the probability that a randomly selected atlantic cod has a length between 40.58 and 49.28 cm. e) find the probability that randomly selected atlantic cod has a length that is at most 40.55 cm. f) is a length of at most 40.55 cm unusual for a randomly selected atlantic cod? why or why not? g) what length do 57% of all atlantic cod have more than? round your answer to 2 decimal places in the first box./p> put the correct units in the second box.

Explanation:

Step1: Define the random variable

Let \(X\) be the length of an Atlantic - cod caught in nets in Karlskrona.

Step2: Calculate the z - score formula

The z - score is given by \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 49.9\) cm (mean) and \(\sigma = 3.74\) cm (standard deviation).

Step3: Solve part b

First, calculate the z - score for \(x = 40.58\):
\(z=\frac{40.58 - 49.9}{3.74}=\frac{-9.32}{3.74}\approx - 2.49\)
\(P(X\geq40.58)=P(Z\geq - 2.49)=1 - P(Z\lt - 2.49)\)
From the standard normal table, \(P(Z\lt - 2.49)=0.0064\), so \(P(X\geq40.58)=1 - 0.0064 = 0.9936\)

Step4: Solve part c

Calculate the z - score for \(x = 49.28\):
\(z=\frac{49.28 - 49.9}{3.74}=\frac{-0.62}{3.74}\approx - 0.17\)
\(P(X\leq49.28)=P(Z\leq - 0.17)\)
From the standard normal table, \(P(Z\leq - 0.17)=0.4325\)

Step5: Solve part d

We already know \(z_1\) for \(x_1 = 40.58\) is approximately \(-2.49\) and \(z_2\) for \(x_2 = 49.28\) is approximately \(-0.17\)
\(P(40.58\lt X\lt49.28)=P(-2.49\lt Z\lt - 0.17)=P(Z\lt - 0.17)-P(Z\lt - 2.49)\)
\(P(-2.49\lt Z\lt - 0.17)=0.4325 - 0.0064 = 0.4261\)

Step6: Solve part e

Calculate the z - score for \(x = 40.55\):
\(z=\frac{40.55 - 49.9}{3.74}=\frac{-9.35}{3.74}\approx - 2.50\)
\(P(X\leq40.55)=P(Z\leq - 2.50)\)
From the standard normal table, \(P(Z\leq - 2.50)=0.0062\)

Step7: Solve part f

A probability is considered unusual if it is less than \(0.05\). Since \(P(X\leq40.55)=0.0062\lt0.05\), a length of at most \(40.55\) cm is unusual for a randomly selected Atlantic cod.

Step8: Solve part g

If \(57\%\) of all Atlantic cod have more than a certain length \(x\), then \(P(X\gt x)=0.57\), so \(P(X\leq x)=1 - 0.57 = 0.43\)
Looking up the z - score in the standard normal table for a probability of \(0.43\), the z - score \(z\approx - 0.18\)
Using the z - score formula \(z=\frac{x-\mu}{\sigma}\), we can solve for \(x\):
\(-0.18=\frac{x - 49.9}{3.74}\)
\(x-49.9=-0.18\times3.74\)
\(x-49.9=-0.6732\)
\(x = 49.9-0.6732\approx49.23\) cm

Answer:

a) The random variable \(X\) is the length of an Atlantic - cod caught in nets in Karlskrona.
b) \(0.9936\)
c) \(0.4325\)
d) \(0.4261\)
e) \(0.0062\)
f) Yes, because \(P(X\leq40.55) = 0.0062\lt0.05\)
g) \(49.23\) cm