Annual snowfall total in Menomonie, x, has a distribution x~N(40,6).

What percent of years will have snowfall totals between 30 and 60 inches?

Give your answer as a percent rounded to two decimal places (so of the form XY.WZ%), but do not enter the % symbol.

Question

Annual snowfall total in Menomonie, x, has a distribution x~N(40,6).

What percent of years will have snowfall totals between 30 and 60 inches?

Give your answer as a percent rounded to two decimal places (so of the form XY.WZ%), but do not enter the % symbol.

BlackTom AI · Accepted Answer

We are given a normal distribution x ~ N(40, 6), where 40 is the mean and 6 is the standard deviation. The task is to find the probability that snowfall totals fall between 30 and 60 inches.

First, convert the raw values to z-scores using z = (value - mean) / standard deviation.
- For the lower bound 30: z1 = (30 - 40) / 6 = -10 / 6 ≈ -1.6667.
- For the upper bound 60: z2 = (60 - 40) / 6 = 20 / 6 ≈ 3.3333.

Next, use the standard normal distribution to find P(-1.6667 ≤ Z ≤ 3.3333).
- Φ(z) denotes the CDF of the standard normal distribution.
- P(-1.6667 ≤ Z ≤ 3.3333) = Φ(3.3333) − Φ(-1.6667).

Look up or approximate the values:
- Φ(3.3333) is very close to 1. A typical approximation is Φ(3.33) ≈ 0.9996.
- Φ(-1.6667) = 1 − Φ(1.6667). Since Φ(1.6667) ≈ 0.9525, we get Φ(-1.6667) ≈ 1 − 0.9525 = 0.0475.

Subtracting gives: 0.9996 − 0.0475 ≈ 0.9521.

Convert to a percentage by multiplying by 100: 0.9521 × 100 ≈ 95.21%.

Finally, round to two decimal places as required: 95.21%.

STAT 130-007 STAT 130 Midterm Exam - Spring 2025

Annual snowfall total in Menomonie, x, has a distribution x~N(40,6). What percent of years will have snowfall totals between 30 and 60 inches? Give your answer as a percent rounded to two decimal places (so of the form XY.WZ%), but do not enter the % symbol.

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