Answer :
Answer:
a) Binomial distribution, with n=20 and p=0.10.
b) P(x>1) = 0.6082
c) P(3≤X≤5) = 0.3118
d) E(X) = 2
e) σ=1.34
Step-by-step explanation:
a) As we have a constant "defective" rate for each unit, and we take a random sample of fixed size, the appropiate distribution to model this variable X is the binomial distribution.
The parameters of the binomial distribution for X are n=20 and p=0.10.
[tex]X\sim B(0.10,20)[/tex]
b) The probability of k defective surge protectors is calculated as:
[tex]P(x=k) = \binom{n}{k} p^{k}q^{n-k}[/tex]
In this case, we want to know the probability that more than one unit is defective: P(x>1). This can be calculated as:
[tex]P(x>1)=1-(P(0)+P(1))\\\\\\P(x=0) = \binom{20}{0} p^{0}q^{20}=1*1*0.1216=0.1216\\\\P(x=1) = \binom{20}{1} p^{1}q^{19}=20*0.1*0.1351=0.2702\\\\\\ P(x>1)=1-(0.1216+0.2702)=1-0.3918=0.6082[/tex]
c) We have to calculate the probability that the number of defective surge protectors is between three and five: P(3≤X≤5).
[tex]P(3\leq X\leq 5)=P(3)+P(4)+P(5)\\\\\\P(x=3) = \binom{20}{3} p^{3}q^{17}=1140*0.001*0.1668=0.1901\\\\P(x=4) = \binom{20}{4} p^{4}q^{16}=4845*0.0001*0.1853=0.0898\\\\P(x=5) = \binom{20}{5} p^{5}q^{15}=15504*0*0.2059=0.0319\\\\\\P(3\leq X\leq 5)=P(3)+P(4)+P(5)=0.1901+0.0898+0.0319=0.3118[/tex]
d) The expected number of defective surge protectors can be calculated from the mean of the binomial distribution:
[tex]E(X)=\mu_B=np=20*0.10=2[/tex]
e) The standard deviation of this binomial distribution is:
[tex]\sigma=\sqrt{np(1-p)}=\sqrt{20*0.1*0.9}=\sqrt{1.8}=1.34[/tex]