Question

The percent of fat calories that a person in America consumes each day is normally distributed with a mean of about 36 and a standard deviation of 10 . Suppose that one individual is randomly chosen. Let$X=$percent of fat calories.

a. $X~$----------

b. Find the probability that the percent of fat calories a person consumes is more than 40 . Graph the situation. Shade in the area to be determined.

c. Find the maximum number for the lower quarter of percent of fat calories. Sketch the graph and write the probability statement.

Short answer

(a) The percent of fat calories that a person in America consumes each day is denoted as: $\mathrm{X}~\mathrm{N}(\mathbf{36},\mathbf{10})$.

(b) The likelihood that a person in America consumes more than 40 percent of fat calories per day is provided by $P(x>40)=0.3446$.

(c) The maximum number for the lower quarter of percent of fat calories (X) that a person in America consumes each day is $\mathbf{29}\mathbf{.}\mathbf{26}$

Step-by-step solution

Part (a) Step 1: Given information

X is a random variable with a mean of 36 and a standard deviation of 10 that represents the percent of fat calories consumed by each American each day.

Part (a) Step 2: To fill:X~--- where X is a random variable representing the percent of fat calories consumed each day by an American.

If a random variable X has a normal distribution with a mean of $\mu$and a standard deviation of s, it is denoted as $X~N(\mu ,s)$.

The random variable $X$ is denoted as $X~N(36,10)$in this case.

As a result, we infer that the random variable X representing the percent of fat calories consumed each day by an American is denoted as $X~N(36,10)$.

Part (b) Step 1: Given information

X is a random variable that represents the percent of fat calories consumed each day by an American, such that$X~N(36,10)$.

This indicates a $\mu =36$ mean and a$s=10$ standard deviation.

Part (b) Step  2:Find the likelihood that a person in America consumes more than 40 percent of their daily fat calories(X), assuming that X~N(36,10).

The probability that the percent of fat calories that a person in America consumes each day is more than 40 is given as:

$\begin{array}{rcl}P(x& >& 40)=P\left(\frac{x-\mu }{\sigma }>\frac{40-36}{10}\right)\\ & =& P(z>0.4)\end{array}$

The required probability is shown by the shaded region in the graph below. Hence, it can be written as:

$P(z>0.4)=0.5-P(0\le z\le 0.4)$$=0.5$

Therefore, we conclude that the probability that percent of fat calories that a person in America consumes each day is more than 40 is given by $P(x>40)=\mathbf{0}\mathbf{.}\mathbf{3446}$

Part (c) Step 1: Given information

X is a random variable denoting the percent of fat calories that a person in America consumes each day, such that $X~N(36,10)$.

This implies, Mean, $\mu =36$ and standard deviation, $s=10$.

Part (c) Step 2: Calculation

The probability that the percent of fat calories for an individual is less than k is given by:

$P(x<k)=P\left(\frac{x-\mu }{\sigma }<\frac{k-36}{10}\right)=P\left(z<z_1\right)=0.25$

From the normal distribution tables, we get

$z_1=-0.6745$

Replacing x with k and z with$z_1$in the equation, $x=\mu +z\sigma$, we get

$k=36+(-0.6745)\times 10=29.26$

The required probability is shown by the shaded region in the graph below.

Therefore, we conclude that the maximum number for the lower quarter of percent of fat calories $(X)$ that a person in America consumes each day is $\mathbf{29}\mathbf{.}\mathbf{26}$