Answer:
[tex] \\ P(49<x<61) = 0.8413 - 0.1587 = 0.6826 [/tex] 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 cumulative distribution function and z-scores to solve the question.
We know from the question that the mean of the distribution is:
[tex] \\ \mu = 55 [/tex]
And a standard deviation of:
[tex] \\ \sigma = 6 [/tex]
The z-scores permit calculates the probabilities for any case whose values have a Normal o Gaussian 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:
[tex] \\ z = \frac{x-\mu}{\sigma} [/tex]
Thus,
The z-scores for 49 and 61 containers are:
[tex] \\ z_{49} = \frac{49 - 55}{6} = \frac{-6}{6} = -1 [/tex] [1]
[tex] \\ z_{61} = \frac{61 - 55}{6} = \frac{6}{6} = 1 [/tex] [2]
Well, this is a special case when in both cases the values are one standard deviation from the mean, but in one case ([tex] \\ z_{49} = -1 [/tex]) the values are smaller than the mean and in the other case ([tex] \\ z_{61} = 1 [/tex]) the values are greater than the mean.
In other words, the cumulative probability for ([tex] \\ z_{61} = 1 [/tex]), obtained from any Table of the Normal Distribution available on the Web, is: 0.8413 (or 84.13%) and the cumulative probability for ([tex] \\ z_{49} = -1 [/tex]) 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:
[tex] \\ P(49<x<61) = 0.8413 - 0.1587 = 0.6826 [/tex] or 68.26%.
See the graph below.
