Question

The owner of a fish market determined that the mean weight for a catfish is 3.2 pounds assuming that the weight of a catfish follows a normal distribution and its standard deviation is unknown. He also knew that that probability of a randomly selected catfish that would weigh more than 3.8 pounds is 20% and the probability that a randomly selected catfish that would weigh less than 2.8 pounds is 30%. What is the probability that a randomly selected catfish will weigh between 2.6 and 3.6 pounds?

Answer 1

Assume that the average distribution of the data is an average of 3.2 pounds with unknown standard deviation.

The normal distribution is symmetric.

It means, $\mathrm{P}(\mathrm{X}>(\mathrm{mean}+\mathrm{A}))=\mathrm{P}(\mathrm{X}<(\mathrm{mean}-\mathrm{A}))$

So. If $\mathrm{P}(\mathrm{X}>3.8)=0.2$

$\mathrm{P}(\mathrm{X}>(3.2+0.6))=0.2$

So, $\mathrm{P}(\mathrm{X}<(3.2-0.6))=0.2$

$P(X<2.6)=0.2$

It is give that $\mathrm{P}(\mathrm{X}<2.8)=0.3$

$\mathrm{P}(\mathrm{X}<(3.2-0.4)=0.3$

So, $P(X>(3.2+0.4))=0.3$

$P(X<3.6)=1-0.3=0.7$

Probability that a randomly selected catfish will weigh between 2.6 and 3.6 pounds $=\mathrm{P}(\mathrm{X}<3.6)-\mathrm{P}(\mathrm{X}<2.6)$

$=0.7-0.2$

$=0.5$