Respuesta :
Answer:
a. P(x≤9)=0.9999
b. P(x=6)=0.0430
c. P(x≥6)=0.0611
Step-by-step explanation:
The question is incomplete:
a.At most 9 will come to a complete stop?
b.Exactly 6 will come to a complete stop?
c.At least 6 will come to a complete stop?
d.How many of the next 20 drivers do you expect to come to a complete stop?
The amount of drivers from the sample that will come to a complete stop can be modeled by a binomial random variable with n=15 and p=0.2.
The probability that exactly k drivers from the sample come to a complete stop is:
[tex]P(x=k) = \dbinom{n}{k} p^{k}q^{n-k}[/tex]
a. We have to calculate the probability that at most 9 come to a complete stop:
[tex]P(x\leq9)=\sum_{k=0}^9P(x=k)\\\\\\P(x=0) = \dbinom{15}{0} p^{0}q^{15}=1*1*0.0352=0.0352\\\\\\P(x=1) = \dbinom{15}{1} p^{1}q^{14}=15*0.2*0.044=0.1319\\\\\\P(x=2) = \dbinom{15}{2} p^{2}q^{13}=105*0.04*0.055=0.2309\\\\\\P(x=3) = \dbinom{15}{3} p^{3}q^{12}=455*0.008*0.0687=0.2501\\\\\\P(x=4) = \dbinom{15}{4} p^{4}q^{11}=1365*0.0016*0.0859=0.1876\\\\\\P(x=5) = \dbinom{15}{5} p^{5}q^{10}=3003*0.0003*0.1074=0.1032\\\\\\P(x=6) = \dbinom{15}{6} p^{6}q^{9}=5005*0.0001*0.1342=0.043\\\\\\[/tex]
[tex]P(x=7) = \dbinom{15}{7} p^{7}q^{8}=6435*0*0.1678=0.0138\\\\\\P(x=8) = \dbinom{15}{8} p^{8}q^{7}=6435*0*0.2097=0.0035\\\\\\P(x=9) = \dbinom{15}{9} p^{9}q^{6}=5005*0*0.2621=0.0007\\\\\\P(x\leq9)=0.0352+0.1319+0.2309+0.2501+0.1876+0.1032+0.043+0.0138+0.0035+0.0007\\\\P(x\leq9)=0.9999[/tex]
b. We have to calculate the probability that exactly 6 will come to a complete stop:
[tex]P(x=6) = \dbinom{15}{6} p^{6}q^{9}=5005*0.0001*0.1342=0.043\\\\\\[/tex]
c. We have to calculate the probability that at least 6 will come to a complete stop:
[tex]P(x\geq6)=\sum_{k=6}^{15}P(x=k)\\\\\\P(x=6) = \dbinom{15}{6} p^{6}q^{9}=5005*0.0001*0.1342=0.043\\\\\\P(x=7) = \dbinom{15}{7} p^{7}q^{8}=6435*0*0.1678=0.0138\\\\\\P(x=8) = \dbinom{15}{8} p^{8}q^{7}=6435*0*0.2097=0.0035\\\\\\P(x=9) = \dbinom{15}{9} p^{9}q^{6}=5005*0*0.2621=0.0007\\\\\\P(x=10) = \dbinom{15}{10} p^{10}q^{5}=3003*0*0.3277=0.0001\\\\\\P(x=11) = \dbinom{15}{11} p^{11}q^{4}=1365*0*0.4096=0\\\\\\P(x=12) = \dbinom{15}{12} p^{12}q^{3}=455*0*0.512=0\\\\\\[/tex]
[tex]P(x=13) = \dbinom{15}{13} p^{13}q^{2}=105*0*0.64=0\\\\\\P(x=14) = \dbinom{15}{14} p^{14}q^{1}=15*0*0.8=0\\\\\\P(x=15) = \dbinom{15}{15} p^{15}q^{0}=1*0*1=0\\\\\\P(x\geq6)=0.043+0.0138+0.0035+0.0007+0.0001+0+0+0+0\\\\P(x\geq6)=0.0611[/tex]
