Measurement of the density of alcohol samples provides a Gaussian distribution with an average of 0.789 mg/ml and a standard deviation of 0.0012 mg/ml. What percentage of the samples will have a density between 0.7880 mg/ml and 0.7904 mg/ml?

Answer:

Given:

The population mean $(\mu )=0.789$

The population standard deviation $(\sigma )=0.0012$

Solution:

The percentage of the samples will have a density between 0.7880 and 0.7904:

$$P(0.788<x<0.7904)=P(\frac{0.788–0.789}{0.0012}<\frac{x–\mu }{\sigma }<\frac{0.7904–0.789}{0.0012})$$ $$=P(-0.83<z<1.17)$$ $$=P(z<1.17)–P(z<-0.83)$$ $$=0.879–0.2033$$ $$=0.6757$$ \text{ Change in percentage = 0.6757\times 100 = 67.57%$$