Respuesta :
Answer:
a) P(X ≤ 2) = 0.677
b) P(X ≥ 5) = 0.043
c) P(1 ≤ X ≤ 4) = 0.835
d) P(x=0) = 0.072
e) expected value: 2 boards
standard deviation: 1.34 boards
Step-by-step explanation:
We can model this with a binomial random variable, with sample size n=20 and probability of success p=0.1.
The probability that k number of defective boards in the sample is: [tex]P(x=k)=\dbinom{n}{k}p^k(1-p)^{n-k}=\dbinom{20}{k}\cdot0.1^k\cdot0.9^{20-k}[/tex]
a) We have to calculate the probability that 2 or less number of defective boards. This can be calculated as:
[tex]P(x\leq2)=P(x=0)+P(x=1)+P(x=2)\\\\\\P(x=0)=\dbinom{20}{0}\cdot0.1^{0}\cdot0.9^{20}=1\cdot1\cdot0.122=0.122\\\\\\P(x=1)=\dbinom{20}{1}\cdot0.1^{1}\cdot0.9^{19}=20\cdot0.1\cdot0.135=0.270\\\\\\P(x=2)=\dbinom{20}{2}\cdot0.1^{2}\cdot0.9^{18}=190\cdot0.01\cdot0.15=0.285\\\\\\\\P(x\leq2)=0.122+0.270+0.285\\\\\\P(x\leq2)=0.677[/tex]
The probability that 2 or less number of defective boards is 0.677.
b) We have to calculate the probability that 5 or more number of defective boards. This can be calculated as:
[tex]P(x\geq5)=1-[P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)]\\\\\\P(x=0)=\dbinom{20}{0}\cdot0.1^{0}\cdot0.9^{20}=1\cdot1\cdot0.122=0.122\\\\\\P(x=1)=\dbinom{20}{1}\cdot0.1^{1}\cdot0.9^{19}=20\cdot0.1\cdot0.135=0.270\\\\\\P(x=2)=\dbinom{20}{2}\cdot0.1^{2}\cdot0.9^{18}=190\cdot0.01\cdot0.15=0.285\\\\\\P(x=3)=\dbinom{20}{3}\cdot0.1^{3}\cdot0.9^{17}=1140\cdot0.001\cdot0.167=0.190\\\\\\P(x=4)=\dbinom{20}{4}\cdot0.1^{4}\cdot0.9^{16}=4845\cdot0.0001\cdot0.185=0.090\\\\\\\\\\\\[/tex]
[tex]P(x\geq5)=1-[0.122+0.270+0.285+0.190+0.090]\\\\P(x\geq5)=1-0.957=0.043[/tex]
The probability that 5 or more number of defective boards is 0.043.
c) We have to calculate P(1 ≤ X ≤ 4)
[tex]P(1\leq X \leq4)=P(x=1)+P(x=2)+P(x=3)+P(x=4)\\\\\\P(x=1)=\dbinom{20}{1}\cdot0.1^{1}\cdot0.9^{19}=20\cdot0.1\cdot0.135=0.270\\\\\\P(x=2)=\dbinom{20}{2}\cdot0.1^{2}\cdot0.9^{18}=190\cdot0.01\cdot0.15=0.285\\\\\\P(x=3)=\dbinom{20}{3}\cdot0.1^{3}\cdot0.9^{17}=1140\cdot0.001\cdot0.167=0.190\\\\\\P(x=4)=\dbinom{20}{4}\cdot0.1^{4}\cdot0.9^{16}=4845\cdot0.0001\cdot0.185=0.090\\\\\\\\P(1\leq X \leq4)=0.270+0.285+0.190+0.090=0.835[/tex]
d) The probability that none of the 25 boards is defective is P(x=0), but now calculated for a sample size n=25.
[tex]P(x=0)=\dbinom{25}{0}\cdot0.1^{0}\cdot0.9^{25}=1\cdot1\cdot0.072=0.072\\\\\\[/tex]
e) The expected value and standard deviation for X (n=20, p=0.1) are:
[tex]\mu=np=20\cdot0.1=2\\\\\sigma=\sqrt{np(1-p)}=\sqrt{20\cdot0.1\cdot0.9}=\sqrt{1.8}=1.34[/tex]
