Answer
Mean =μ = $∑xP(x)$ = 3.05 units
Thus, it is expected that 3.05 units of remote starting systems are installed on a given day.
Standard Deviation
= $\sqrt∑x^{2}P(x) -μ^{2}$
= $\sqrt 10.75-3.05^{2}$
=1.2031 units of remote starting system
If k = 2, at least 75% of the number of remote starting systems installed on a given day lie between $μ-2σ$ and $μ+2σ$.
μ =3.05, σ = 1.2031
$μ-2σ$ = 3.05-2(1.2031) = 0.6438
$μ+2σ$ = 3.05+2(1.2031) = 5.4562
Using Chebyshev's theorem, we can state that at least 75% of the remote starting systems are expected to install 0.6438 to 5.4562 units on a given day.
Work Step by Step
See above.
