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? select an answer g) select an answer yes, since the probability of having a value of length at least that high is less than or equal to 0.05 yes, since the probability of having a value of length at the most that value is less than or equal to 0.05 no, since the probability of having a value of length at least that high is less than or equal to 0.05 no, since the probability of having a value of length at the most that value is less than or equal to 0.05 yes, since the probability of having a value of length at least that high is greater than 0.05 yes, since the probability of having a value of length at the most that value is greater than 0.05 no, since the probability of having a value of length at least that high is greater than 0.05 no, since the probability of having a value of length at the most that value is greater than 0.05

Explanation:

Step1: Calculate z - score formula

The z - score is calculated using the formula $z=\frac{x-\mu}{\sigma}$, where $x$ is the value from the data set, $\mu$ is the mean, and $\sigma$ is the standard deviation. Given $\mu = 49.9$ cm and $\sigma=3.74$ cm.

Step2: For part b

We want to find $P(X\geq40.58)$. First, calculate the z - score: $z=\frac{40.58 - 49.9}{3.74}=\frac{-9.32}{3.74}\approx - 2.49$. Then $P(X\geq40.58)=P(Z\geq - 2.49)=1 - P(Z < - 2.49)$. From the standard normal table, $P(Z < - 2.49)=0.0064$, so $P(X\geq40.58)=1 - 0.0064 = 0.9936$.

Step3: For part c

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

Step4: For part d

We use the results from part b and c. $P(40.58

Step5: For part e

We want to find $P(X\leq40.55)$. Calculate the z - score: $z=\frac{40.55 - 49.9}{3.74}=\frac{-9.35}{3.74}\approx - 2.5$. From the standard normal table, $P(X\leq40.55)=P(Z\leq - 2.5)=0.0062$.

Step6: For part f

An event is considered unusual if its probability is less than or equal to 0.05. Since $P(X\leq40.55)=0.0062\leq0.05$, a length of at most 40.55 cm is unusual. The correct answer is: yes, since the probability of having a value of length at the most that value is less than or equal to 0.05.

Answer:

a) $rv\ X =$ the length of a randomly selected Atlantic cod
b) $0.9936$
c) $0.4325$
d) $0.4261$
e) $0.0062$
f) yes, since the probability of having a value of length at the most that value is less than or equal to 0.05