Question

A supermarket has determined that daily demand for milk containers has an approximate bell shaped distribution, with a mean of 55 containers and a standard deviation of six containers. How often can we expect between 49 and 61 containers to be sold in a day?

Answer 1

Answer:

$\\ P(49$ 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:

$\\ \mu = 55$

And a standard deviation of:

$\\ \sigma = 6$

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:

$\\ z = \frac{x-\mu}{\sigma}$

Thus,

The z-scores for 49 and 61 containers are:

$\\ z_{49} = \frac{49 - 55}{6} = \frac{-6}{6} = -1$ [1]

$\\ z_{61} = \frac{61 - 55}{6} = \frac{6}{6} = 1$ [2]

Well, this is a special case when in both cases the values are one standard deviation from the mean, but in one case ($\\ z_{49} = -1$) the values are smaller than the mean and in the other case ($\\ z_{61} = 1$) the values are greater than the mean.

In other words, the cumulative probability for ($\\ z_{61} = 1$), obtained from any Table of the Normal Distribution available on the Web, is: 0.8413 (or 84.13%) and the cumulative probability for ($\\ z_{49} = -1$) 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:

$\\ P(49$ or 68.26%.

See the graph below.