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In 2017 a pollution index was calculated for a sample of cities in the eastern states using data on air and water pollution. Assume the distribution of pollution indices is unimodal and symmetric. The mean of the distribution was 35.9 points with a standard deviation of 11.6 points.

Answer :

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The percentage of cities with pollution index between 13.0 and 58.5 is 95%

Given that :

Mean = 35.9

Mean = 35.9Standard deviation = 11.6

The percentage of of the cities with pollution index between 13.0 and 58.5 points can be calculated thus :

Calculate the Zscore of both pollution index given :

Zscore = (x - mean) / standard deviation

x = raw score

For x = 13.0

Zscore = (13 - 35.9) / 11.6

Zscore = - 1.974

P(Z < - 1.974) = 0.024191

For x = 58.5

Zscore = (58.5 - 35.9) / 11.6

Zscore = - 1.948

P(Z < - 1.948) = 0.97429

The percentage of cities with pollution index between 13.0 and 58.5 :

P(Z < 1.948) - P(Z < - 1.974)

0.97429 - 0.024191 = 0.9500

0.9500 × 100 = 95%

Hence, The percentage of cities with pollution index between 13.0 and 58.5 is 95%

Learn more : Zscore : https://brainly.com/question/8165716

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