Question

emmy went camping with the school outing club. they roasted marshmallows by the campfire every night for 7 nights, consuming:
81 marshmallows 92 marshmallows 84 marshmallows 82 marshmallows
31 marshmallows 15 marshmallows 98 marshmallows
what was the mean absolute deviation of the numbers of marshmallows eaten?
if the answer is a decimal, round it to the nearest tenth.
mean absolute deviation (mad): marshmallows

Explanation:

Step1: Calculate the mean

The formula for the mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\). Here \(n = 7\), and \(x_{1}=81,x_{2}=92,x_{3}=84,x_{4}=82,x_{5}=31,x_{6}=15,x_{7}=98\).
\(\sum_{i=1}^{7}x_{i}=81 + 92+84 + 82+31+15+98=483\)
\(\bar{x}=\frac{483}{7}=69\)

Step2: Calculate the absolute deviations

\(\vert81 - 69\vert=12\), \(\vert92 - 69\vert=23\), \(\vert84 - 69\vert=15\), \(\vert82 - 69\vert=13\), \(\vert31 - 69\vert=38\), \(\vert15 - 69\vert=54\), \(\vert98 - 69\vert=29\)

Step3: Calculate the mean of the absolute deviations (MAD)

The formula for MAD is \(MAD=\frac{\sum_{i = 1}^{n}\vert x_{i}-\bar{x}\vert}{n}\)
\(\sum_{i = 1}^{7}\vert x_{i}-\bar{x}\vert=12 + 23+15+13+38+54+29=184\)
\(MAD=\frac{184}{7}\approx26.3\)

Answer:

\(26.3\)