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. rv x = the length of a randomly selected atlantic cod b) find the probability that a randomly selected atlantic cod has a length of 40.58 cm or more. .9936 c) find the probability that a randomly selected atlantic cod has a length of 49.28 cm or less. .4325 d) find the probability that a randomly selected atlantic cod has a length between 40.58 and 49.28 cm. .4261 e) find the probability that randomly selected atlantic cod has a length that is at most 40.55 cm. .0062 f) is a length of at most 40.55 cm unusual for a randomly selected atlantic cod? why or why not? yes, since the probability of having a value of length at the most that value is less than or equal to 0.05 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: Recall the z - score formula

The z - score is given by $z=\frac{x-\mu}{\sigma}$, where $\mu$ is the mean, $\sigma$ is the standard deviation, and $x$ is the value from the data set. Here, $\mu = 49.9$ cm and $\sigma=3.74$ cm. We want to find the value $x$ such that $P(X > x)=0.57$. This is equivalent to $P(X\leq x)=1 - 0.57=0.43$.

Step2: Look up the z - score in the standard normal table

Looking up the probability $0.43$ in the standard - normal table (the cumulative distribution function of the standard normal distribution $\varPhi(z)$), the corresponding z - score $z\approx - 0.18$.

Step3: Solve for $x$ using the z - score formula

We know that $z=\frac{x-\mu}{\sigma}$. Rearranging for $x$ gives $x=\mu+z\sigma$. Substituting $\mu = 49.9$, $z=-0.18$, and $\sigma = 3.74$ into the formula:

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Answer:

49.23 cm