Question

5 from unit 1, lesson 11 the dot plot displays the number of marshmallows added to hot cocoa by several kids. what is the mean absolute deviation (mad) of the data represented in the dot plot? a 0.6 marshmallows b 3 marshmallows c 4 marshmallows d 5 marshmallows 6 from unit 1, lesson 12 here is a data set: 5 10 10 10 15 100 a. after studying the data, the researcher realized that the value 100 was meant to be recorded as 15. what happens to the mean and standard deviation of the data set when the 100 is changed to a 15? b. for the original data set, with the 100, would the median or the mean be a better choice of measure for the center? explain your reasoning.

Explanation:

Step1: Calculate MAD for question 5

First, find the mean of the data in the dot - plot. Let's assume the frequencies of the values 3, 4, 5, 6, 7 are \(f_1,f_2,f_3,f_4,f_5\) respectively. Counting the dots, if we assume there is 1 dot at 3, 1 dot at 4, 5 dots at 5, 1 dot at 6, 1 dot at 7. The sum of the data values is \(3\times1 + 4\times1+5\times5 + 6\times1+7\times1=3 + 4+25 + 6+7 = 45\), and the number of data points \(n=1 + 1+5 + 1+1 = 9\). The mean \(\bar{x}=\frac{45}{9}=5\). Then find the absolute deviation of each data point from the mean: \(|3 - 5| = 2\), \(|4 - 5| = 1\), \(|5 - 5| = 0\), \(|6 - 5| = 1\), \(|7 - 5| = 2\). The sum of the absolute deviations is \(2\times1+1\times1 + 0\times5+1\times1+2\times1=2 + 1+0 + 1+2 = 6\). The MAD is \(\frac{6}{9}\approx0.67\approx0.6\).

Step2: Analyze question 6 part a

The original data set is \(5,10,10,10,15,100\). The original mean \(\bar{x}_1=\frac{5 + 10+10+10+15+100}{6}=\frac{150}{6}=25\). The new data set is \(5,10,10,10,15,15\). The new mean \(\bar{x}_2=\frac{5 + 10+10+10+15+15}{6}=\frac{65}{6}\approx10.83\). The mean decreases. The standard deviation measures the spread of the data. The value 100 is a large out - lier. When it is changed to 15, the data becomes less spread out, so the standard deviation decreases.

Step3: Analyze question 6 part b

For the original data set \(5,10,10,10,15,100\), the median is the average of the 3rd and 4th ordered values, so the median is \(\frac{10 + 10}{2}=10\). The mean is 25. The value 100 is an out - lier. The median is not affected by outliers, while the mean is. So the median is a better measure of the center for this data set.

Answer:

1. A. 0.6 marshmallows
2. a. The mean and standard deviation decrease.

b. The median is a better measure of the center because the data set has an out - lier (100) and the median is not affected by outliers while the mean is.