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 9.2 and 11.6 ounces of chicken in one sitting? 15.85% 34% 49.85% 99.7%

Explanation:

Step1: Calculate z - scores

The formula for the z - score is \(z=\frac{x - \mu}{\sigma}\), where \(\mu = 8\) (mean) and \(\sigma=1.2\) (standard deviation).
For \(x = 9.2\): \(z_1=\frac{9.2 - 8}{1.2}=\frac{1.2}{1.2}=1\)
For \(x = 11.6\): \(z_2=\frac{11.6 - 8}{1.2}=\frac{3.6}{1.2}=3\)

Step2: Use the empirical rule (68 - 95 - 99.7 rule)

The empirical rule states that for a normal distribution:

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

The area to the left of \(z = 1\) is \(0.8413\) (from the standard normal table: \(P(Z<1)=0.8413\))
The area to the left of \(z = 3\) is \(0.9987\) (from the standard normal table: \(P(Z < 3)=0.9987\))

Step3: Find the probability between \(z = 1\) and \(z = 3\)

\(P(1<Z<3)=P(Z < 3)-P(Z < 1)\)
\(P(1<Z<3)=0.9987 - 0.8413=0.1584\approx15.85\%\)

Answer:

15.85%