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:

Step-by-step explanation:
In the case we have a lot with 3 defectives components, the proportion of defectives is:

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:

b) The mean and variance of defective components in the sample is:

The Chebyschev's inequality established:

A= l * W (area = length times width)
<span>Answer:
3 + 4 + 4 + 2s = 172121391827
f + r + f + d+ = +_+_+_71230973210123
f - 4 = 9
s - 222 = 9009
r + 555 = 1231212312123</span>
In order to answer this we need the table, however to do this all you would do would be to multiply the p value in the table with the corresponding r value in the table and then multiply this by 5, then order them, smallest to greatest.
M=2,000÷((1−(1+0.072÷12)^(−12
×2))÷(0.072÷12))
M=89.73