Question

1. healthy fast food? here is a dotplot of the amount of fat (to the nearest gram) in 12 different hamburgers served at a fast-food restaurant: dotplot image (a) the distribution of fat content has a mean of \\(\bar{x} = 22.83\\) grams and a standard deviation of \\(s_x = 9.06\\) grams. interpret the standard deviation. (b) suppose the restaurant replaces the burger that has 22 grams of fat with a new burger that has 35 grams of fat. how would this affect the mean and the standard deviation? justify your answer.

Explanation:

Part (a)

Step 1: Recall the meaning of standard deviation

The standard deviation measures the typical distance of the data points from the mean. For a sample, the formula for the sample standard deviation \( s_x \) is \( s_x=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}} \), where \( x_i \) are the data points, \( \bar{x} \) is the mean, and \( n \) is the sample size.

Step 2: Interpret the given standard deviation

Given that the mean fat content \( \bar{x}=22.83 \) grams and the sample standard deviation \( s_x = 9.06 \) grams. This means that, on average, the fat content of each hamburger in the sample deviates from the mean fat content of \( 22.83 \) grams by approximately \( 9.06 \) grams. In other words, the fat content of the hamburgers is, on average, about \( 9.06 \) grams away from the mean fat content of \( 22.83 \) grams.

Step 1: Effect on the mean

The mean \( \bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n} \). Initially, we have \( n = 12 \) data points. Let the sum of the original fat contents be \( S \). So, \( \bar{x}_{\text{original}}=\frac{S}{12}=22.83 \), which means \( S=12\times22.83 = 273.96 \).

We are replacing a burger with fat content \( x_{\text{old}} = 22 \) grams with a new burger with fat content \( x_{\text{new}}=35 \) grams. The new sum \( S_{\text{new}}=S - 22+35=S + 13 \).

The new mean \( \bar{x}_{\text{new}}=\frac{S_{\text{new}}}{12}=\frac{S + 13}{12}=\frac{S}{12}+\frac{13}{12}=\bar{x}_{\text{original}}+ 1.083\approx22.83 + 1.08=23.91 \) (approximate). So, the mean will increase because we are replacing a value (22) that is less than the original mean (22.83) with a value (35) that is greater than the original mean.

Step 2: Effect on the standard deviation

The standard deviation measures the spread of the data from the mean. The formula for the sample standard deviation is \( s_x=\sqrt{\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}} \).

The original data point was \( x = 22 \), and the original mean was \( \bar{x}=22.83 \). The deviation of this point from the mean was \( 22 - 22.83=- 0.83 \), and the squared deviation was \( (-0.83)^2 = 0.6889 \).

The new data point is \( x = 35 \), and the new mean is \( \bar{x}_{\text{new}}\approx23.91 \). The deviation of this new point from the new mean is \( 35 - 23.91 = 11.09 \), and the squared deviation is \( (11.09)^2\approx123.0 \).

Since we are replacing a data point with a squared deviation of approximately \( 0.6889 \) with a data point with a much larger squared deviation (approximately \( 123.0 \)), the sum of squared deviations \( \sum_{i = 1}^{n}(x_i-\bar{x})^2 \) will increase. When we take the square root of the ratio of this new sum (divided by \( n - 1 \)), the standard deviation will also increase. In other words, the new data point (35) is further from the new mean than the old data point (22) was from the old mean, so the spread of the data (as measured by the standard deviation) will increase.

Answer:

On average, the fat content of each hamburger in the sample deviates from the mean fat content of \( 22.83 \) grams by approximately \( 9.06 \) grams.

Part (b)