Question

A bottled water company has found in the past that 2% of their bottled water does not meet the company’s high standards. As such periodic samples are taken and tested for their quality. If from the last batch a sample of 12 bottles are taken and tested, determine the probability:

i. Carefully defining the random variable of interest [1]

ii. What is the probability distribution? State the values of the parameters [2]

iii. Justifying the suitability of the probability distribution identified in part (ii) [4]

iv. Calculate the probability that at most 2 bottles do not meet the company’s Standards [4]

v. The expected number of bottles that do not meet the company’s standards. [2]

vi. The amount of sodium sulfate, in mg, found in the bottled water follows a normal distribution with a mean 13mg and a standard deviation of 2.5mg. It is known that 76% of the bottles have a mean that is more than x mg. Find the value of x

Answer 1

i. The variable of interest is the number of bottles that don't meet the companies high standard.

II. This the a binomial distribution with n=12 and p=0.02

III. This is binomial distribution since we have a fixed number of trials (12) ,a fixed success rate and they are only 2 possible outcomes.

Iv.

P(X=x)=$\binom{n}{x}\cdot p^x\cdot(1-p)^{n-x }$

P(x$\le$ 2)=P(x=0)+P(x=1)+P(x=2)

P(x=0)=$\binom{12}{0}\cdot0.02^0\cdot0.98^{12}$ =0.785

P(x=1)=$\binom{12}{1}\cdot0.02^1\cdot0.98^{11}$ =0.193

P(x=2)=$\binom{12}{2}\cdot0.02^2\cdot0.98^{10}$ =0.022

P(x$\le$ 2)=0.785+0.193+0.022

P(x$\le$ 2)=0.998

The probability that at most 2 don't meet the companies high standard is 0.998.

v

Expected=n$\cdot$ p

Expected =12$\cdot$0.02 =0.24

vi

mean ($\bar x)$ =13

standard deviation(s)=2.5

x=?

From the standard normal table 0.24 corresponding to z value 0.71

z=$\frac{x-\bar x}{s}$

0.71=$\frac{x-13}{2.5}$

x=0.71*2.5+13

x=14.775