Suppose that the weight of tomato juice in mechanically filled cans follows a normal distribution with a mean ƒÝ of 464 grams and a standard deviation ƒã of 8 grams. What proportion of cans contain between 454 and 474 grams of tomato juice?
the beauty of this applet is that you can enter the data values as is, without changing them to z-scores.
http://davidmlane.com/hyperstat/z_table.html
Since these kind of questions require the use of tables anyway, I don't see the harm of using the above program directly.
To find the proportion of cans that contain between 454 and 474 grams of tomato juice, we need to calculate the z-scores for these two values and then use the standard normal distribution table.
The formula for calculating the z-score is:
z = (x - μ) / σ
Where:
x is the value you want to find the proportion for (454 and 474 in this case)
μ is the mean of the distribution (464 grams)
σ is the standard deviation of the distribution (8 grams)
For 454 grams:
z1 = (454 - 464) / 8 = -1.25
For 474 grams:
z2 = (474 - 464) / 8 = 1.25
Now, we can use the standard normal distribution table to find the proportion of cans between these z-scores.
The table provides the proportion to the left of the given z-score. Since we want the proportion between the two z-scores, we subtract the cumulative probability for the smaller z-score from the cumulative probability for the larger z-score.
P(454 < x < 474) = P(z1 < z < z2)
From the standard normal distribution table, we find:
P(z < -1.25) = 0.1056
P(z < 1.25) = 0.8944
P(454 < x < 474) = P(z < 1.25) - P(z < -1.25)
= 0.8944 - 0.1056
= 0.7888
Therefore, approximately 0.7888 or 78.88% of the cans contain between 454 and 474 grams of tomato juice.