To find the probability that the thickness of the walls of a randomly selected can is less than 0.118 mn, we can use the properties of the normal distribution.
Given:
- Mean thickness (\( \mu \)) = 0.1 mn
- Standard deviation (\( \sigma \)) = 0.002 mn
First, we need to calculate the z-score for a thickness of 0.118 mn. The z-score is calculated using the formula:
\[ z = \frac{X - \mu}{\sigma} \]
Where:
- \( X \) is the value we are interested in (0.118 mn)
- \( \mu \) is the mean (0.1 mn)
- \( \sigma \) is the standard deviation (0.002 mn)
Plugging in the values, we get:
\[ z = \frac{0.118 - 0.1}{0.002} = \frac{0.018}{0.002} = 9 \]
Next, we look up the z-score of 9 on the standard normal distribution table. However, a z-score of 9 is extremely high, and standard normal distribution tables typically do not cover such extreme values. In practice, a z-score this high implies a probability that is essentially 1, since almost all values under the normal distribution curve fall well below this z-score.
Thus, the probability of selecting a can with a wall thickness less than 0.118 mn is practically 100%.
Therefore, the probability that the thickness of the walls of a randomly selected can will be less than 0.118 mn is 100 percent.