Answer:
- 61 and 62 inches, or
- 60 and 63 inches.
Step-by-step explanation:
The given requirements on mean and mean absolute deviation let us write some equations for the possible student heights.
Setup
Let x and y represent the absolute deviations from the mean of the remaining two students. We can describe the mean of the heights by ...
(55 +66 +(61+x) +(61-y))/4 = 61 ⇒ 243 +x -y = 244
And the mean absolute deviation is required to be ...
(|55 -61| +|66 -61| +x +y)/4 < 4 ⇒ 11 +x +y < 16
Solution
We can solve the first equation for x and substitute into the second equation:
x = 1 +y . . . . . . . add y-243 to the first equation
11 +(1 +y) +y < 16 . . . . . substitute for x in the second equation
2y < 4 . . . . . . . . . . . simplify, subtract 12
y < 2 . . . . . . . . . . divide by 2
The definitions of x and y require they both be non-negative. Then integer values of y may be ...
y ∈ {0, 1}
The corresponding values of x are ...
x ∈ {1, 2}
Remembering that the shorter student is 61-y, and the taller is 61+x, we find the two other students may have heights of ...
- 61 and 62 inches, or
- 60 and 63 inches.
