Question

customers at a sandwich shop are given a pull - tab card with every purchase, and every card is a winner. nick has received 10 cards, with his winnings of: $6.00 $3.00 $7.70 $4.50 $6.90 $9.30 $3.70 $8.40 $3.40 $4.10 what was the mean absolute deviation of the amounts won? if the answer is a decimal, round it to the nearest ten cents. mean absolute deviation (mad): $

Explanation:

Step1: Calculate the mean

First, sum up all the winnings: $6.00 + 3.00+7.70 + 4.50+6.90 + 9.30+3.70 + 8.40+3.40 + 4.10=57$
The number of data - points $n = 10$.
The mean $\bar{x}=\frac{57}{10}=5.7$

Step2: Calculate the absolute deviations

For $x_1 = 6.00$, the absolute deviation is $|6.00 - 5.7|=0.3$
For $x_2 = 3.00$, the absolute deviation is $|3.00 - 5.7| = 2.7$
For $x_3 = 7.70$, the absolute deviation is $|7.70 - 5.7|=2$
For $x_4 = 4.50$, the absolute deviation is $|4.50 - 5.7| = 1.2$
For $x_5 = 6.90$, the absolute deviation is $|6.90 - 5.7|=1.2$
For $x_6 = 9.30$, the absolute deviation is $|9.30 - 5.7| = 3.6$
For $x_7 = 3.70$, the absolute deviation is $|3.70 - 5.7| = 2$
For $x_8 = 8.40$, the absolute deviation is $|8.40 - 5.7|=2.7$
For $x_9 = 3.40$, the absolute deviation is $|3.40 - 5.7| = 2.3$
For $x_{10}=4.10$, the absolute deviation is $|4.10 - 5.7| = 1.6$

Step3: Calculate the sum of absolute deviations

$0.3+2.7 + 2+1.2+1.2+3.6+2+2.7+2.3+1.6 = 19.3$

Step4: Calculate the mean absolute deviation

The mean absolute deviation $MAD=\frac{19.3}{10}=1.93$

Answer:

$1.90$