Question

question 3 of 6 a study investigated about 3000 meals ordered from chipotle restaurants using the online site chubhub. researchers analyzed the sodium content (in milligrams) for each order based on chipotle’s published nutrition information. the distribution of sodium content is approximately normal with mean 1993 mg and standard deviation 591 mg. about what percentage of the meals ordered at chipotle contained between 1200 mg and 1800 mg of sodium? (round to 2 decimal places) %

Explanation:

Step1: Identify the distribution and parameters

The sodium content is normally distributed with mean $\mu = 1993$ mg and standard deviation $\sigma = 591$ mg. We need to find the percentage of meals with sodium content between $x_1 = 1200$ mg and $x_2 = 1800$ mg.

Step2: Calculate the z - scores

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$.

For $x = 1200$:
$z_1=\frac{1200 - 1993}{591}=\frac{- 793}{591}\approx - 1.34$

For $x = 1800$:
$z_2=\frac{1800 - 1993}{591}=\frac{-193}{591}\approx - 0.33$

Step3: Find the area between the z - scores

We use the standard normal distribution table (or z - table) to find the area to the left of $z_1$ and $z_2$.

The area to the left of $z=-1.34$ (denoted as $P(Z < - 1.34)$) is approximately $0.0901$.

The area to the left of $z = - 0.33$ (denoted as $P(Z < - 0.33)$) is approximately $0.3707$.

The area between $z_1$ and $z_2$ is $P(-1.34

$=0.3707 - 0.0901=0.2806$

Answer:

28.06%