Cans of soda are filled by a machine. Tests show that the amount of soda in each can varies. If the volume of soda in the cans is normally distributed with a mean of 12 and a standard deviation of 0.15, find the probability that a can chosen at random will have between 11.90 and 12.19 oz. of soda.
To find the probability that a can chosen at random will have between 11.90 and 12.19 oz. of soda, we need to find the z-scores for these two values and then use the standard normal distribution table.
First, calculate the z-score for 11.90 oz.:
\[ z_1 = \dfrac{11.90 - 12}{0.15} = -0.67 \]
Next, calculate the z-score for 12.19 oz.:
\[ z_2 = \dfrac{12.19 - 12}{0.15} = 1.27 \]
Now, we look up the probabilities associated with these z-scores in the standard normal distribution table.
The probability of a can having less than 11.90 oz. of soda is the same as the probability of a z-score less than -0.67, which is approximately 0.2514.
The probability of a can having less than 12.19 oz. of soda is the same as the probability of a z-score less than 1.27, which is approximately 0.8980.
Therefore, the probability that a can chosen at random will have between 11.90 and 12.19 oz. of soda is:
\[ P(11.90 < X < 12.19) = 0.8980 - 0.2514 = 0.6466 \]
So, the probability is approximately 0.6466 or 64.66%.