Question

Environmentalists are concerned about emissions of sulfur dioxide into the air. The average number of days per year in which sulfur dioxide levels exceed 150 milligrams per cubic meter in Milan, Italy, is 29. The number of days per year in which emission limits are exceeded is normally distributed with a standard deviation of 4.0 days 95.45 a. What percentage of the years would average between 21 and 37 days of excess emissions of sulfur dioxide? b. What percentage of the years would exceed 37 days? c. What percentage of the years would exceed 41 days? d. In what percentage of the years would there be fewer than 25 days with excess sulfur dioxide emissions? 2.29 0.14 15.88

Answer 1

Your answers are in lieu of what is being calculated using Z tables. If we use the emperical rule of 68-95-99.7 where

μ = 29, σ = 4.

The 68-95-99.7 Rule states that

From μ - 1σ to μ + 1σ, 68% of the values are contained. (i.e. 34% to the left of the mean and 34% to the right)

From μ - 2σ to μ + 2σ, 95% of the values are contained. (i.e. 47.5% to the left of the mean and 47.5% to the right)

From μ - 3σ to μ + 3σ, 99.7% of the values are contained. (i.e. 49.85% to the left of the mean and 49.85% to the right).

To the right we have

μ + 1σ = 29 + 4 = 33,

μ + 2σ = 29 + (2*4) = 37 and

μ + 3σ = 29 + (3*4) = 41.

Similarly on the left side we have

μ - 1σ = 29 - 4 = 25,

μ - 2σ = 29 - (2*4) = 21 and

μ - 3σ = 29 - (3*4) = 17

(a) The % of years between 21 and 37 days i.e between μ - 2σ and μ + 2σ = 95.00%

(b) % of years that exceed 37 days that is exceeds μ + 2σ. Upto μ + 2σ we have 50 % of the left + 47.5% of the right = 97.5%. Therefore the % that exceeds = 100 - 97.5 = 2.50%.

(c) % of years that exceed 41 days that is exceeds μ + 3σ. Upto μ + 3σ we have 50 % of the left + 49.85% of the right = 99.85%. Therefore the % that exceeds = 100 - 99.85 = 0.15%.

(d) % of years that is fewer than 25 days that is lesser than μ - σ. = 50% - 34% = 16.00%