Final answer:
The trout lengths in the local lake are normally distributed with a mean of 13.3 inches and a standard deviation of 1.9 inches. The values of a and b can be determined using the properties of the normal distribution, with a representing the lower bound and b representing the upper bound within which approximately 95% of the trout lengths will fall. The distribution can be written as N(13.3, 1.9).
Step-by-step explanation:
The subject of this question is Statistics as it involves the distribution of fish lengths and the calculation of mean and standard deviation.
For this question, the mean (μ) of the trout lengths in the local lake is given as 13.3 inches and the standard deviation (σ) is given as 1.9 inches. To find the values of a and b, we need to consider the normal distribution.
In a normal distribution, approximately 68% of the data falls within 1 standard deviation of the mean, 95% falls within 2 standard deviations, and 99.7% falls within 3 standard deviations. Based on this, we can determine the values of a and b as follows:
- a = μ - 2σ = 13.3 - (2 * 1.9) = 13.3 - 3.8 = 9.5 inches
- b = μ + 2σ = 13.3 + (2 * 1.9) = 13.3 + 3.8 = 17.1 inches
The values of a and b represent the range within which approximately 95% of the trout lengths in the lake will fall.
The distribution can be written as N(13.3, 1.9), where N represents the normal distribution.
The theoretical mean can be calculated as the same as the given mean (μ), which is 13.3 inches.
The theoretical standard deviation can be calculated as the same as the given standard deviation (σ), which is 1.9 inches.