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

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

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

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) = (1)/(n)\sum |x - \mu|`

For Educational Documentary

First, calculate the mean

`\mu = (\sum x)/(n)`

So:

`\mu_A = (26 + 28 + 30 + 18+27 + 18 + 20 +31+17+17)/(10)`

`\mu_A = (232)/(10)`

`\mu_A = 23.2`

The mean absolute deviation is then calculated as:

`m(x)_A = (1)/(n)\sum |x - \mu_A|`

`m(x)_A = (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 = (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 = (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 = (1)/(10)*52`

`m(x)_A = 5.2`

For Prime time Drama

First, calculate the mean

`\mu = (\sum x)/(n)`

So:

`\mu_B = (39+39+35+29+40+27+41+29+32+30)/(10)`

`\mu_B = (341)/(10)`

`\mu_B = 34.1`

The mean absolute deviation is then calculated as:

`m(x)_B = (1)/(n)\sum |x - \mu_B|`

`m(x)_B = (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 = (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 = (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 = (1)/(10)*47`

`m(x)_B = 4.7`