Question

the graph shows the distribution of the amount of chicken (in ounces) that adults eat in one sitting. the distribution is approximately normal, with a mean of 8 ounces and a standard deviation of 1.2 ounces. what percentage of adults eat between 5.6 and 8 ounces of chicken in one sitting? 2.5% 34% 47.5% 95%

Explanation:

Step1: Calculate the z - score for \(x = 5.6\)

The formula for the z - score is \(z=\frac{x-\mu}{\sigma}\), where \(\mu = 8\) (mean) and \(\sigma=1.2\) (standard deviation).
For \(x = 5.6\), \(z=\frac{5.6 - 8}{1.2}=\frac{- 2.4}{1.2}=-2\)

Step2: Use the empirical rule for the normal distribution

The empirical rule states that for a normal distribution:

• Approximately \(68\%\) of the data lies within \(\mu\pm\sigma\) (\(z=\pm1\))
• Approximately \(95\%\) of the data lies within \(\mu\pm2\sigma\) (\(z = \pm2\))
• Approximately \(99.7\%\) of the data lies within \(\mu\pm3\sigma\) (\(z=\pm3\))

The mean \(\mu = 8\) (\(z = 0\)). We want to find the percentage between \(z=-2\) and \(z = 0\).
Since the normal distribution is symmetric about the mean, the percentage of data between \(z=-2\) and \(z = 0\) is half of the percentage of data between \(z=-2\) and \(z = 2\).

The percentage of data between \(z=-2\) and \(z = 2\) is \(95\%\). So the percentage of data between \(z=-2\) and \(z = 0\) is \(\frac{95\%}{2}=47.5\%\)

Answer:

\(47.5\%\)