Answer:
The probability that three or fewer were walk-ins is P=0.254.
If this outcome occured, it can be produced by pure chance, as it nos a very unlikely event, although the expected number of walkins is 4.5.
Step-by-step explanation:
This problem can be modeled as a binomial experiment, with p=0.5 and n=9.
p is the probability that the customers are walk-ins.
n is the sample size.
We have to calculate the probabilty that three or fewer are walk-ins.
[tex]P(x\leq3)=\sum_{k=0}^3P(x=k)[/tex]
[tex]\\ P(x=k)=\binom{n}{k}p^k(1-p)^{n-k}\\\\\\P(x=0) = \binom{9}{0} p^{0}q^{9}=1*1*0.002=0.002\\\\P(x=1) = \binom{9}{1} p^{1}q^{8}=9*0.5*0.0039=0.0176\\\\P(x=2) = \binom{9}{2} p^{2}q^{7}=36*0.25*0.0078=0.0703\\\\P(x=3) = \binom{9}{3} p^{3}q^{6}=84*0.125*0.0156=0.1641\\\\[/tex]
[tex]P(x\leq3)=\sum_{k=0}^3P(x=k)=0.0020+0.0176+0.0703+0.1641\\\\P(x\leq3)=0.254[/tex]
