Question

5 from unit 1 lesson 11 the dot - plot displays the number of marshmallows added to hot cocoa by several kids marshmallows added to hot cocoa 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

Explanation:

Step1: Find the data values and their frequencies

Let's assume from the dot - plot, the data values \(x_i\) and their frequencies \(f_i\) are:
For \(x_1 = 3\), \(f_1=1\); for \(x_2 = 4\), \(f_2 = 1\); for \(x_3=5\), \(f_3 = 7\); for \(x_4 = 6\), \(f_4=1\); for \(x_5 = 7\), \(f_5 = 1\). The total number of data points \(n=f_1 + f_2+f_3+f_4+f_5=1 + 1+7+1+1=11\).

Step2: Calculate the mean \(\bar{x}\)

\(\bar{x}=\frac{\sum_{i = 1}^{5}f_ix_i}{n}=\frac{3\times1 + 4\times1+5\times7+6\times1+7\times1}{11}=\frac{3 + 4+35+6+7}{11}=\frac{55}{11}=5\).

Step3: Calculate the absolute deviations \(|x_i-\bar{x}|\) and their sum

For \(x_1 = 3\), \(|x_1-\bar{x}|=|3 - 5| = 2\); for \(x_2 = 4\), \(|x_2-\bar{x}|=|4 - 5| = 1\); for \(x_3=5\), \(|x_3-\bar{x}|=|5 - 5| = 0\); for \(x_4 = 6\), \(|x_4-\bar{x}|=|6 - 5| = 1\); for \(x_5 = 7\), \(|x_5-\bar{x}|=|7 - 5| = 2\).
The sum of the absolute - deviations weighted by their frequencies is \(S=\sum_{i = 1}^{5}f_i|x_i-\bar{x}|=1\times2+1\times1 + 7\times0+1\times1+1\times2=2 + 1+0+1+2 = 6\).

Step4: Calculate the MAD

\(MAD=\frac{S}{n}=\frac{6}{11}\approx0.6\).

Answer:

A. 0.6 marshmallows