Question

A bottler of drinking water fills plastic bottles with a mean volume of 1,000 milliliters (mL) and standard deviation The fill volumes are normally distributed. What proportion of bottles have volumes greater than

Answer 1

Answer:

84.13% of bottles will have volume greater than 994 mL

Step-by-step explanation:

Mean volume = u = 1000

Standard deviation = $\sigma$ = 6

We need to find the proportion of bottles with volume greater than 994. So our test value is 994. i.e.

x = 994

Since the data is normally distributed we will use the concept of z-score to find the required proportion. First we convert 994 to its equivalent z-score, then using the z-table we will find the corresponding value of proportion. The formula to calculate the z score is:

$z=\frac{x-u}{\sigma}$

Substituting the values, we get:

$z=\frac{994-1000}{6}=-1$

This means 994 is equivalent to a z score of -1. Now we will find the proportion of z values which are greater than -1 from the z table.

i.e. P(z > -1)

From the z-table this value comes out to be:

P(z >- 1) = 1 - P(z < -1) = 1 - 0.1587 = 0.8413

Since, 994 is equivalent to a z score of -1, we can write that proportion of values which will be greater than 994 would be:

P( X > 994 ) = P( z > -1 ) = 0.8413 = 84.13%