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. 49.23

Explanation:

Step1: Identify the distribution parameters

Let $\mu = 49.9$ cm be the mean and $\sigma=3.74$ cm be the standard - deviation. The random variable $X$ (length of a randomly selected Atlantic cod) is normally distributed, i.e., $X\sim N(49.9,3.74^{2})$. We use the z - score formula $z=\frac{x - \mu}{\sigma}$.

Step2: Solve part b

For $x = 40.58$ cm, the z - score is $z=\frac{40.58 - 49.9}{3.74}=\frac{-9.32}{3.74}\approx - 2.49$. Using the standard normal table, $P(X\geq40.58)=1 - P(X < 40.58)=1 - P(Z < - 2.49)=1-0.0064 = 0.9936$.

Step3: Solve part c

For $x = 49.28$ cm, the z - score is $z=\frac{49.28 - 49.9}{3.74}=\frac{-0.62}{3.74}\approx - 0.17$. Using the standard normal table, $P(X\leq49.28)=P(Z\leq - 0.17)=0.4325$.

Step4: Solve part d

We already know $P(X\geq40.58) = 0.9936$ and $P(X\leq49.28)=0.4325$. So $P(40.58 < X < 49.28)=P(X < 49.28)-P(X < 40.58)=0.4325-(1 - 0.9936)=0.4261$.

Step5: Solve part e

For $x = 40.55$ cm, the z - score is $z=\frac{40.55 - 49.9}{3.74}=\frac{-9.35}{3.74}\approx - 2.5$. Using the standard normal table, $P(X\leq40.55)=P(Z\leq - 2.5)=0.0062$.

Step6: Solve part f

A probability of $0.0062\leq0.05$. So a length of at most $40.55$ cm is unusual for a randomly selected Atlantic cod.

Step7: Solve part g

If $57\%$ of all Atlantic cod have more than a certain length $x$, then $P(X > x)=0.57$, so $P(X\leq x)=1 - 0.57 = 0.43$. Looking up the z - score in the standard normal table corresponding to a probability of $0.43$, we find $z\approx - 0.18$. Then, using the z - score formula $z=\frac{x - \mu}{\sigma}$, we have $-0.18=\frac{x - 49.9}{3.74}$. Solving for $x$ gives $x=49.9+(-0.18)\times3.74=49.9 - 0.6732\approx49.23$ cm.

Answer:

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 $0.05$
g) $49.23$ cm