The weight of a can made from a machine is normally distributed with a mean of 25 grams and a
standard deviation of 0.4 grams. What percentage of the cans from this machine would not meet the
minimum required weight of 24 grams? Round to the nearest hundredth of a percent
Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score. Multiply by 100 to get the percentage.
To find the percentage of cans that would not meet the minimum required weight of 24 grams, we need to calculate the probability that a randomly selected can weighs less than 24 grams.
To do this, we can use the formula for the standard normal distribution, also known as the Z-score formula:
Z = (X - μ) / σ
Where:
- Z is the standard score
- X is the value we want to find the probability for (24 grams in this case)
- μ is the mean of the distribution (25 grams in this case)
- σ is the standard deviation of the distribution (0.4 grams in this case)
Substituting the values into the formula:
Z = (24 - 25) / 0.4 = -2.5
Now, we need to find the cumulative probability for this Z-score using a z-table or statistical software. The cumulative probability represents the area under the normal distribution curve up to a certain Z-score.
Looking up the Z-score of -2.5 in a standard normal distribution table, we find that the cumulative probability is approximately 0.00621.
This means that approximately 0.621% of the cans from this machine would weigh less than 24 grams. Rounding to the nearest hundredth of a percent, we get 0.62%.