A bottler of drinking water fills plastic bottles with a mean volume of 1006 milliliters (mL) and standard deviation 7 mL. The fill volumes are normally distributed. What proportion of bottles have volumes between 996 mL and 1001 mL.
1001 is 5/7 s.d. below the mean
996 is 10/7 s.d. below the mean
use a z-score table to find the portion of the population this represents
To find the proportion of bottles with volumes between 996 mL and 1001 mL, we will use the concept of the standard normal distribution.
Step 1: Convert the given values to Z-scores.
To do this, we will use the formula:
Z = (X - μ) / σ
Where:
Z = Z-score
X = Value
μ = Mean
σ = Standard deviation
For 996 mL:
Z1 = (996 - 1006) / 7 = -10 / 7 ≈ -1.43
For 1001 mL:
Z2 = (1001 - 1006) / 7 = -5 / 7 ≈ -0.71
Step 2: Use a standard normal distribution table or calculator.
We need to find the proportion between the Z-scores -1.43 and -0.71.
Using a standard normal distribution table or calculator, we can find the proportion associated with Z = -1.43 (let's call it P1) and Z = -0.71 (let's call it P2).
Step 3: Calculate the final proportion.
The proportion of bottles with volumes between 996 mL and 1001 mL is given by the difference between P2 and P1:
Proportion = P2 - P1
This final proportion represents the area under the normal distribution curve between -1.43 and -0.71.
Note: If you are using a standard normal distribution table, find the values closest to -1.43 and -0.71, and then subtract the corresponding proportion values.
By following these steps, you can find the proportion of bottles with volumes between 996 mL and 1001 mL using the given mean and standard deviation.