Final answer:
The probability that the annual snowfall in Lake Tahoe exceeds 269 inches is approximately 0.015 or 1.5%, calculated by determining the z-score for 269 inches and referring to the standard normal distribution.
Step-by-step explanation:
The question pertains to finding the probability that the annual snowfall in Lake Tahoe exceeds 269 inches when the snowfall amounts are normally distributed with a mean of 221 inches and a standard deviation of 22 inches. We need to calculate the z-score for 269 inches and then use the standard normal distribution to find the probability that the annual snowfall exceeds this value.
To calculate the z-score, we use the formula:
Z = (X - μ) / σ
where X is the value of interest (269 inches), μ is the mean (221 inches), and σ is the standard deviation (22 inches). Substituting the values:
Z = (269 - 221) / 22 = 48 / 22 ≈ 2.18
Now we look up the z-score of 2.18 in the standard normal distribution table or use a calculator with the normal distribution function to find the area to the right of this z-score, which represents the probability of snowfall exceeding 269 inches. Since the standard normal distribution is symmetrical, the area to the right of z = 2.18 is the same as the area to the left of z = -2.18, which is approximately 0.985. Thus, the probability we're seeking is 1 - 0.985 = 0.015 or 1.5%.