Answer:
or 68.26%.
Step-by-step explanation:
The daily demand for milk containers has a Normal (or Gaussian) distribution, and we can use values from the <em>cumulative distribution function</em> and z-scores to solve the question.
We know from the question that the <em>mean</em> of the distribution is:

And a <em>standard deviation</em> of:

The z-scores permit calculates the probabilities for any case whose values have a <em>Normal </em>o <em>Gaussian</em> distribution. Then, for this, we need to calculate the z-scores for 49 containers and 61 containers to establish the corresponding probabilities, as well as the differences between these two values to determine the probability between them.
These z-scores are given by:

Thus,
The z-scores for 49 and 61 containers are:
[1]
[2]
Well, this is a special case when <em>in both cases</em> the values are <em>one standard deviation</em> from the mean, but in one case (
) the values are smaller than the mean and in the other case (
) the values are greater than the mean.
In other words, the cumulative probability for (
), obtained from any Table of the Normal Distribution available on the Web, is: 0.8413 (or 84.13%) and the cumulative probability for (
) is: 1 - 0.8413 = 0.1587 (or 15.87%), because of the symmetry of the Normal Distribution.
Then, the probability of expecting to sell between 49 and 61 containers in a day is the difference of both obtained probabilities:
or 68.26%.
See the graph below.