Answer:
a) 2 red lights
b) SD = 1.26
c) mean = 10, SD = 1.26
Step-by-step explanation:
a) The number of red lights she expects to hit by day can be gotten by calculating the mean of the distribution.
\(E(X) = \sum xP(x)\)
\(E(X) = (0*0.05) + (1*0.25) + (2*0.35) + (3*0.15) + (4*0.15) + (5*0.15)\\E(X) = 2.25\)
Since the number of lights cannot be a decimal, she expects to hit 2 lights each day
b)
Variance, \(V(X) = \sum(x- \mu)^{2} P(x)\)
\(V(X) = [(0-2.25)^{2}*0.05] + [(1-2.25)^{2}*0.25] + [(2-2.25)^{2}*0.35] + [(3-2.25)^{2}*0.15] + [(4-2.25)^{2}*0.15] + [(5-2.25)^{2}*0.05]\)
V(X) = 0.253 + 0.391 + 0.022 + 0.084 + 0.459 + 0.378
V(X) = 1.587
Standard Deviation, \(SD = \sqrt{V(X)}\)
\(SD = \sqrt{1.587}\)
SD = 1.26
c) In a 5 day work week, the commuter is expected to hit an average of 5 2 red lights, i.e. mean = number of red lights hit per day number of days
mean = 2 * 5
mean = 10
The standard deviation will not change, SD = 1.26
