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 $(μ)=0.789$

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

Solution:

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

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