Question

The amount of water a construction worker consumes in a day is normally distributed with a mean of 2.7 litres, and a standard deviation of 0.4 litres. Construction workers carry water bottles containing 3 litres.

(a) What is 20th percentile of this distribution (i.e. 80% of workers will consume more than this volume)?

(b) What is the probability of a randomly selected construction worker consuming his/her bottle of water in a day?

(c) Management is considering letting teams of two construction workers share their two bottles of water. If this is done, what proportion of teams (or two) will consume all of their water in a day?

(d) What is the most important assumption you have made in answering part (c)? Comment briefly on whether you think this is a reasonable assumption or not.

Answer 1

here we use standard normal variate z=(x-mean)/sd=(x-2.7)/0.4

(a)answer is 2.3634 Liters

here we want to find x such that P(X<x)=0.2

first we find z such that P(Z<z)=0.2 and z=-0.8416

now x=mean+z*sd=2.7-0.8416*0.4=2.3634

(b) answer is 0.2266

for x=3, z=(3-2.7)/0.4=0.75

P(X>3)=P(Z>0.75)=1-P(Z<0.75)=1-0.7734=0.2266

(c) answer is 0.903

mean of of two person will be =2,7 and sd will be 0.4/sqrt(2)

so z=(3-2.7)/(0.4/sqrt(3))=1.299

required probability=P(Z<1.299)=0.903

(d) here we use central limit theorem in which

If X has a distribution with mean μ, and standard deviation σ, then sample mean will be normally distributed with mean μ and standard deviation