The annual snowfall in a town can be represented by the Normal model N(46, 6.8).
In about 2.5% of the years, the annual snowfall in this town will exceed:
A) 46 inches
B) 39.2 inches
C) 52.8 inches
D) 59.6 inches
E) 66.4 inches
I don't know how to set this problem up to choose an answer? I think I would use the empirical rule, but I don't know how to use that for this problem...
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (.025) related to a Z score. Insert the Z score and other data in the equation above and solve for the score.
To solve this problem, you can use the concept of z-scores and the normal distribution. Here's how you can set it up:
First, let's define the z-score formula:
z = (x - μ) / σ
Where:
- z is the z-score
- x is the value you want to find the probability for
- μ is the mean of the distribution (in this case, the mean annual snowfall of 46 inches)
- σ is the standard deviation of the distribution (in this case, 6.8 inches)
To find the probability of a certain value exceeding a given value, you need to find the area under the curve beyond that value. Here's how you can do that:
1. Calculate the z-score for each given value using the formula mentioned above.
2. Look up the corresponding cumulative probability in the z-table (also known as the standard normal distribution table).
3. Subtract the cumulative probability from 1 to get the probability of the value being exceeded.
Let's calculate this for each of the given options:
A) 46 inches (the mean)
The z-score is (46 - 46) / 6.8 = 0. This means there is no area beyond the mean, so the probability is 0%.
B) 39.2 inches
The z-score is (39.2 - 46) / 6.8 ≈ -0.912. Looking up this value in the z-table, we find that the cumulative probability is approximately 18.27%. Subtracting this from 1, we get 81.73%.
C) 52.8 inches
The z-score is (52.8 - 46) / 6.8 ≈ 0.941. Looking up this value in the z-table, we find that the cumulative probability is approximately 83.94%. Subtracting this from 1, we get 16.06%.
D) 59.6 inches
The z-score is (59.6 - 46) / 6.8 ≈ 2.00. Looking up this value in the z-table, we find that the cumulative probability is approximately 97.72%. Subtracting this from 1, we get 2.28%.
E) 66.4 inches
The z-score is (66.4 - 46) / 6.8 ≈ 2.941. Looking up this value in the z-table, we find that the cumulative probability is approximately 99.58%. Subtracting this from 1, we get 0.42%.
Now, based on the calculated probabilities, you can choose the answer that corresponds to the value that is exceeded with a probability of approximately 2.5%. Comparing the probabilities calculated:
B) 39.2 inches: 81.73%
C) 52.8 inches: 16.06%
D) 59.6 inches: 2.28%
E) 66.4 inches: 0.42%
The answer is D) 59.6 inches, as it is the only value that is exceeded with a probability close to 2.5%.