Question

Personal daily usage of water in Zambia has been found to be normally distributed with a mean of 24 litres a variance of 39 litres.
i. What percentage of the population uses more than 33 litres?

ii. What percentage of the population uses between 16 and 27 litres?

iii. What is the probability of finding a person who uses less than 12 litres?

iv. As there is always a shortage of water in Zambia, the Finance Minister has decided to give a tax rebate to the 22% of the population who use the least amount of water. What should the Finance Minister set as the maximum water usage for a person to qualify for a tax rebate?

Answer 1

Let X=personal daily usage of water,

use $z=\frac{x-\bar{x}}{\sigma}=\frac{x-24}{\sqrt{39}}$ to convert standard normal form,

i)

$P(X>33)=P(Z>\frac{33-24}{\sqrt{39}})\\ P(X>33)=P(Z>1.44)\\$

using table or calculator,

$P(X>33)=0.0749\\$

Percentage of the population uses more than 33 liters=7.49%

ii)

$P(16<X<27)=P(\frac{16-24}{\sqrt{39}}<Z<\frac{27-24}{\sqrt{39}})\\ P(16<X<27)=P(-1.28<Z<0.48)\\ P(16<X<27)=P(-1.28<Z<0)\\ \hspace{12 em}+P(0<Z<0.48)\\ P(16<X<27)=P(0<Z<1.28)\\ \hspace{12 em}+P(0<Z<0.48)\\$

using table or calculator,

$P(16<X<27)=0.3997+0.1844=0.5841$

Percentage of the population uses between 16 and 27 liters=58.41%

iii)

$P(12>X)=P(\frac{12-24}{\sqrt{39}}>Z)\\ P(12>X)=P(-1.92>Z)\\$

using table or calculator,

$P(12>X)=0.0274\\$

Probability of finding a person who uses less than 12 liters=0.0274

iv)

here the task is to find z, such that

$P(z<Z)=0.22\\$

using table,

$z=-0.772\\ z=\frac{x-24}{\sqrt{39}}\\ \frac{x-24}{\sqrt{39}}=-0.772\\ x=19.179liters$

The maximum water usage for a person to qualify for a tax rebate=19.179 liters