Question

3. The amount of water filled in a particular brand of water bottle follows a normal distribution with a mean of 20 fl oz and variance of 0.01 fl oz2 .

(a) What is the probability that the water bottle contains between 19.9 and 20.1 fluid oz of water? (4 points)

(b) 10% of water bottles have less that what volume? (4 points)

(c) A large case contains 40 individual water bottles. What is the probability that its total volume is more than 801 fl. oz? (4 points)

(d) For the same case in the part (c), what is the probability that its average volume is less than 19.96 fl. oz? (5 points)

Answer 1

Solution:

Given, X follows Normal distribution with,

   = 20

² = 0.01

= 0.1

a)

P(19.9 < x< 20.1)

= P(X < 20.1) - P(X < 19.9)

=  P[(X - )/ <  (20.1 - 20)/0.1] -   P[(X - )/ <  (19.9 - 20)/0.1]

= P[Z < 1.00] - P[Z < -1.00]

= 0.8413 - 0.1587 ..Use z table

= 0.6826

P(19.9 < x< 20.1) = 0.6826

b)

For bottom 10% data , let x be the required cut-off.

P(X < x) = 0.10

For the standard normal variable z , P(Z < z) = 0.10

Use z table , see where is 0.10 probability and then see the corresponding z value.

P(Z < -1.282) = 0.10

So z = -1.282

Now using z score formula ,

x = + (z * ) = 20+ (-1.282* 0.1) = 19.8718

Required answer : 19.8718

c)

A sample of size n = 40 is taken from this population.

Let be the mean of sample.

The sampling distribution of the is approximately normal with

Mean = = 20

SD =      = 0.1/​40 = 0.0158113883

P(total volume of sample is  more than 801)

= P[sum > 801]

= P[sum/n > 801/40]

= P[ > 20.025]

= P[( - )/ >  (20.025 - )/]

= P[ ( - )/ > (20.025 - 20)/ 0.0158113883]

= P[Z > 1.58]

= 1 - P[Z < 1.58]

= 1 - 0.9429 ( use z table)

= 0.0571

P(total volume of sample is  more than 801) = 0.0571

d)

Find P( < 19.96 )

= P[( - )/ < (19.96 - 20)/0.0158113883]

= P[Z < -2.53]

= 0.0057

P( < 19.96 ) = 0.0057