Respuesta :
Answer:
The probability of founding exactly one defective item in the sample is P=0.275.
The mean and variance of defective components in the sample are:
[tex]\mu=0.375\\\\\sigma^2=0.347[/tex]
Step-by-step explanation:
In the case we have a lot with 3 defectives components, the proportion of defectives is:
[tex]p=3/40=0.075[/tex]
a) The number of defectives components in the 5-components sample will follow a binomial distribution B(5,0.075).
The probability of having one defective in the sample is:
[tex]P(k=5)=\binom{5}{1}p^1(1-p)^4=5*0.075*0.925^4=0.275[/tex]
b) The mean and variance of defective components in the sample is:
[tex]\mu=np=5*0.075=0.375\\\\\sigma^2=npq=5*0.075*0.925=0.347[/tex]
The Chebyschev's inequality established:
[tex]P(|X-\mu|\geq k\sigma)\leq \frac{1}{k^2}[/tex]
