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

i)

µ =    24                  
σ = √39 = 6.244997998                  
                      
P ( X ≥   33   ) = P( (X-µ)/σ ≥ (33-24) / 6.2450)              
= P(Z ≥   1.44   ) = P( Z <   -1.441   ) =    0.0748   (answer)

ii)

we need to calculate probability for ,                                      
P (   16   < X <   27   )                      
=P( (16-24)/6.245 < (X-µ)/σ < (27-24)/6.245)                                      
                                      
P (    -1.281   < Z <    0.480   )                       
= P ( Z <    0.480   ) - P ( Z <   -1.28   ) =    0.6845   -    0.1001   =    0.5844   (answer)

iii)

P( X ≤    12   ) = P( (X-µ)/σ ≤ (12-24) /6.245)      
=P(Z ≤   -1.92   ) =   0.0273   (answer)

iv)

P(X≤x) =   0.22                  
                      
Z value at    0.22   =   -0.7722   (excel formula =NORMSINV(   0.22   ) )
z=(x-µ)/σ                      
so, X=zσ+µ=   -0.772   *   6.245 +   24  
X   =   19.18   (answer)