Final answer:
The probability that the snowfall was greater than 53 inches in Laytonville is greater than 99.7%, as calculated using the empirical rule and the z-score which is -3 standard deviations below the mean.
Step-by-step explanation:
To calculate the probability that in a randomly selected year the snowfall was greater than 53 inches in Laytonville, we can use the empirical rule and the given data of a mean of 92 inches and a standard deviation of 13 inches. Because we want the probability for more than 53 inches, we'll need to find the z-score for 53 inches and then use the standard normal distribution to find the probability that the value is beyond this z-score.
First, we calculate the z-score:
- Z = (X - μ) / σ
- Z = (53 - 92) / 13
- Z = -39 / 13
- Z = -3
μ (mu) is the mean, σ (sigma) is the standard deviation, and X is the value we are checking, which in this case, is 53.
According to the empirical rule:
- Approximately 68% of the data falls within 1 standard deviation of the mean.
- Approximately 95% falls within 2 standard deviations.
- Approximately 99.7% falls within 3 standard deviations.
Since a z-score of -3 indicates that we are 3 standard deviations below the mean, and the empirical rule tells us that about 99.7% of the data falls within 3 standard deviations, this means that more than 99.7% of the data is greater than 53 inches of snowfall.
Therefore, the probability that the snowfall in a year is greater than 53 inches is very high, specifically, greater than 0.997 or 99.7%.