Question

Assume that a randomly selected subject is given a bone density test. Those test scores are normally distributed with a mean of 0 and a standard deviation of 1. Find the probability that a given score is less than negative 0.93 and draw a sketch of the region.

Answer 1

Answer:

$\\ P(z$

Step-by-step explanation:

We can solve this question taking into account the concepts of the standard normal distribution and the z-scores. We can say that the former is a special normal distribution from which we can obtain probabilities for all normally distributed data. Likewise, the z-scores are "transformed" values for raw scores, and we can use them to find probabilities using the standard normal distribution.

The question tells us something important "Those test scores are normally distributed with a mean of 0 and a standard deviation of 1".

Thus, it refers to the standard normal distribution. In this distribution, every given score is a z-score. A z-score tells us the distance from the mean in standard deviation units. Moreover, a negative score means that the value is below the mean, whereas a positive score tells us that the value is above the mean in the distribution.

Well, we have to find the probability that "a given score is less than negative 0.93", that is z = -0.93.

With the help of a standard normal table, we can find this probability using z = -0.93 as an entry to the table to find it. However, in most tables of this kind, we have positive values for z-scores. Fortunately, the standard normal distribution, as well as the normal distribution, is symmetrical and we can verify that:

$\\ P(z$

Then, the probability is as follows:

$\\ P(z$

$\\ P(z$

$\\ P(z$

That is, "the probability that a given score is less than negative 0.93" is 0.17619 or $\\ P(z$.

We can see that the graph below represents this region. It is below the mean (at the left hand from the population mean, at 0.93 standard deviation units, and it represents an area that is, approximately, 17.619% of the total area of the standard distribution).