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) select an answer select an answer rv x = the length of a randomly selected atlantic cod rv x = length is normally distributed rv x = a randomly selected atlantic cod c) rv x = the mean length of a sample of atlantic cod tic cod has a length of 40.58 cm or more. rv x = the mean length of all atlantic cod tic 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? select an answer 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

The random variable $X$ is the length of a randomly - selected Atlantic cod.

Step2: Standardize the values

The formula for standardizing a value $x$ in a normal distribution with mean $\mu$ and standard deviation $\sigma$ is $z=\frac{x - \mu}{\sigma}$. Here, $\mu = 49.9$ cm and $\sigma=3.74$ cm.

Step3: Calculate $P(40.58

First, find the $z$ - scores:
For $x = 40.58$, $z_1=\frac{40.58 - 49.9}{3.74}=\frac{-9.32}{3.74}\approx - 2.49$.
For $x = 49.28$, $z_2=\frac{49.28 - 49.9}{3.74}=\frac{-0.62}{3.74}\approx - 0.17$.
Then, $P(40.58Using the standard normal table, $P(-2.49$\varPhi(-0.17) = 0.4325$ and $\varPhi(-2.49)=0.0064$.
So, $P(-2.49

Step4: Calculate $P(X\leq40.55)$

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

Step5: Determine if $x = 40.55$ is unusual

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

Step6: Find the length $x$ such that $P(X>x)=0.57$

If $P(X>x)=0.57$, then $P(X\leq x)=1 - 0.57 = 0.43$.
Looking up the $z$ - score in the standard normal table corresponding to a probability of 0.43, we get $z\approx - 0.18$.
Using the 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) rv $X$ = the length of a randomly selected Atlantic cod
d) $0.4261$
e) $0.0062$
f) Yes, because the probability $P(X\leq40.55) = 0.0062<0.05$.
g) $49.23$ cm