Question

A small company repairs home computers. In the past 200 days, it has been quite busy, taking in and repairing 2 computers on 52 of those days; 3 computers on 84 of those days; 4 computers on 40 of those days; and 5 computers on 24 of those days. Build a probability distribution for the number of computers repaired on any given day, show it is a valid distribution, and calculate its mean and variance.

Answer 1

The probability distribution is shown in the table below

The value of
`x`: 2, 3, 4, and 5 shows the number of computers repaired and
`P(X=x)` shows the probability

Mean = (2×0.26×) + (3×0.42) + (4×0.2) + (5×0.12)
Mean = 0.52+1.26+0.80+0.6
Mean = 3.18

Variance = [(2²×0.26)+(3²×0.42)+(4²×0.2)+(5²×0.12)] - (3.18)²
Variance = 0.9076