Question

The lengths of brook trout caught in a certain Colorado stream are normally distributed with a mean of 14 inches and a standard deviation of 3 inches. What proportion of brook trout caught will be between 12 and 18 inches in length?

Answer 1

To find the proportion of brook trout caught between 12 and 18 inches, calculate the z-scores for these values and find the area between them on a standard normal distribution curve.

Step-by-step explanation:

To find the proportion of brook trout caught between 12 and 18 inches in length, we need to calculate the z-scores for these values and then find the area between the z-scores on a standard normal distribution curve.

First, we calculate the z-score for 12 inches: z = (12 - 14) / 3 = -2/3.

Second, we calculate the z-score for 18 inches: z = (18 - 14) / 3 = 4/3.

Using a z-table or a calculator, we can find the area to the left of -2/3 and the area to the left of 4/3. Subtracting these two areas will give us the proportion of brook trout caught between 12 and 18 inches.

Answer 2

0.6563 or 65.63% of brook trout caught will be between 12 and 18 inches

Step-by-step explanation:

Mean trout length (μ) = 14 inches

Standard deviation (σ) = 3 inches

The z-score for any given trout length 'X' is defined as:

`z=(X-\mu)/(\sigma)` e interval

For a length of X =12 inches:

`z=(12-14)/(3)\\z=-0.6667`

According to a z-score table, a score of -0.6667 is equivalent to the 25.25th percentile of the distribution.

For a length of X =18 inches:

`z=(18-14)/(3)\\z=1.333`

According to a z-score table, a score of 1.333 is equivalent to the 90.88th percentile of the distribution.

The proportion of trout caught between 12 and 18 inches, assuming a normal distribution, is the interval between the equivalent percentile of each length:

`P(12\leq X\leq 18) = 90.88\% - 25.25\%\\P(12\leq X\leq 18) = 65.63\%`