Question

The average cholesterol content of a certain brand of eggs is 215 milligrams, and the standard deviation is 15 milligrams. Assume the variable is normally distributed. If a single egg is randomly selected, what is the probability that the egg will be with cholesterol content greater than 220 milligrams

Answer 1

Answer:

"The probability that the egg will be with cholesterol content greater than 220 milligrams" is 0.37070 (37.070% or simply 37%)

Step-by-step explanation:

We have here a normally distributed random variable. The parameters that characterize this distribution is the mean, $\\ \mu$, and the standard deviation, $\\ \sigma$.

In this question, we have that:

• $\\ \mu = 215$ milligrams.
• $\\ \sigma = 15$ milligrams.

And we want to know the probability that a randomly selected single egg "will be with cholesterol content greater than 220 milligrams."

To answer the latter, we need to use the following key concepts:

• Z-scores.
• The cumulative standard normal distribution, and
• The [cumulative] standard normal table.

The z-scores are standardized values and represent the distance (for the raw score) from the mean in standard deviations units. A positive z-score indicates that it is above $\\ \mu$ and, conversely, a negative result that the value is below it.

The cumulative distribution function generates the values for the cumulative standard normal distribution displayed in the standard normal table.

The standard normal distribution is employed to find probabilities for any normally distributed data and we only need to calculate the corresponding z-score (or standardized value). This distribution has a $\\ \mu = 0$ and $\\ \sigma = 1$.

As we can see, all of these concepts are intertwined, and each of them is important because:

1. To find the corresponding probability, we first need to obtain the z-score.
2. After this, we can consult the standard normal table, whose values are tabulated from the cumulative standard normal distribution, to find the requested probability.

Finding the probability

We can get the z-score using the formula:

$\\ z = \frac{x - \mu}{\sigma}$ [1]

Where x is the raw value we want to standardize using the previous formula, and, in this case is 220 milligrams, $\\ x = 220$ milligrams.

Thus (without using units):

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

$\\ z = \frac{220 - 215}{15}$

$\\ z = \frac{5}{15}$

$\\ z = \frac{5}{5} * \frac{1}{3}$

$\\ z = 1 * \frac{1}{3}$

$\\ z = 0.3333333...$

To consult the standard normal table, we only need $\\ z = 0.33$, because it only has values for two decimal digits. As a result, the value will be a little inexact (but near to the true value) compared to that obtained using statistical software (or maybe a more precise table).

With this value (which is positive and, therefore, above the mean), we need to carefully see the first column of the mentioned table to find z = 0.3. Then, in the first row, we only need to select that column for which we can add the next digit, in this case, 3 (it appears as +0.03). That is, we are finding the probability for $\\ z = 0.33$.

Then, the cumulative probability for $\\ z = 0.33$ is:

$\\ P(x$

However, the question is asking for "cholesterol content greater than 220 milligrams" or

$\\ P(x>220) = P(z>0.33)$

Since

$\\ P(x220) = 1$

Which is the same for a standardized value:

$\\ P(z0.33) = 1$

Then

$\\ P(z>0.33) = 1 - P(z$

Therefore

$\\ P(x>220) = P(z>0.33) = 1 - P(z$

$\\ P(x>220) = 1 - P(z$

$\\ P(x>220) = 1 - 0.62930$

$\\ P(x>220) = 0.37070$

Thus, "the probability that the egg will be with cholesterol content greater than 220 milligrams" is 0.37070 (37.070% or simply 37%).

The graph below shows a shaded area that corresponds to the found probability.