Question

Brian recorded the number of commercials played during two types of television shows: an educational documentary and a primetime drama. Both types of shows lasted 1 hour, including commercials. He recorded the number of commercials for 10 episodes of each type of show.

Answer 1

Answer:

$m(x)_A = 5.2$ --- Mean absolute deviation of educational documentary

$m(x)_B = 4.7$ --- Mean absolute deviation of prime time drama

Step-by-step explanation:

Given

$n = 10$

See attachment for table

Required

Determine the mean absolute deviation of each

Mean absolute deviation m(x) is calculated as:

$m(x) = \frac{1}{n}\sum |x - \mu|$

For Educational Documentary

First, calculate the mean

$\mu = \frac{\sum x}{n}$

So:

$\mu_A = \frac{26 + 28 + 30 + 18+27 + 18 + 20 +31+17+17}{10}$

$\mu_A = \frac{232}{10}$

$\mu_A = 23.2$

The mean absolute deviation is then calculated as:

$m(x)_A = \frac{1}{n}\sum |x - \mu_A|$

$m(x)_A = \frac{1}{10}(|26 -23.2| +|28 -23.2| +|30 -23.2| +|18 -23.2| +|27 -23.2| +|18 -23.2| +|20 -23.2| +|31 -23.2| +|17 -23.2| +|17 -23.2|)$

$m(x)_A = \frac{1}{10}(|2.8| +|4.8| +|6.8| +|-5.2| +|3.8| +|-5.2| +|-3.2| +|7.8| +|-6.2| +|-6.2|)$

$m(x)_A = \frac{1}{10}(2.8 +4.8 +6.8 +5.2 +3.8 +5.2 +3.2 +7.8 +6.2 +6.2)$

$m(x)_A = \frac{1}{10}*52$

$m(x)_A = 5.2$

For Prime time Drama

First, calculate the mean

$\mu = \frac{\sum x}{n}$

So:

$\mu_B = \frac{39+39+35+29+40+27+41+29+32+30}{10}$

$\mu_B = \frac{341}{10}$

$\mu_B = 34.1$

The mean absolute deviation is then calculated as:

$m(x)_B = \frac{1}{n}\sum |x - \mu_B|$

$m(x)_B = \frac{1}{10}(|39-34.1|+|39-34.1|+|35-34.1|+|29-34.1|+|40-34.1|+|27-34.1|+|41-34.1|+|29-34.1|+|32-34.1|+|30-34.1|)$

$m(x)_B = \frac{1}{10}(|4.9|+|4.9|+|0.9|+|-5.1|+|5.9|+|-7.1|+|6.9|+|-5.1|+|-2.1|+|-4.1|)$

$m(x)_B = \frac{1}{10}(4.9+4.9+0.9+5.1+5.9+7.1+6.9+5.1+2.1+4.1)$

$m(x)_B = \frac{1}{10}*47$

$m(x)_B = 4.7$