g. a second stage smog device has been installed on 50 different industry firms in los angeles. an inspector plans to visit 10 r
1 answer:
The probability that 5 or less of the inspected firms that will result smog volation is 0.944.
Given that among 50 different industrial firms, 15 violate smog regulations and the sample size be 10.
We are required to find the probability that 5 or less of the inspected firms that will result smog violation.
Probability is basically the calculation of likeliness of happening an event among all the events possible. It lies between 0 and 1.
Probability=Number of items/ Total items.
Probability that the firm violates the regulations if 1 industry is chosen from 50 industrial firms=15/50=0.3
Probability that 5 or less of the inspected firms will result somg violations among 10 industrial firms=
=0.028+10*0.3*0.04+45*0.09*0.057+120*0.027*0.082+
210*0.0081*0.117+252*0.00243*0.16807
=0.028+0.12+0.23+0.265+0.199+0.102
=0.944
Hence the probability that 5 or less of the inspected firms that will result smog volation is 0.944.
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Answer:
Step-by-step explanation:
Hello!
The variable of interest in this case is:
X: length of time before a major repair is required on a new car. (months)
This variable has a normal distribution with mean μ= 36 months and a standard deviation σ= 9 months.
If he wants to set a guarantee period in which only 5% of the sold cars fell, the objective is to find the value of X that has below 5% of the cars that need major repairs before the end of the guarantee period.
Check the attachment, the curve represents the distribution of the population of the time it takes before a new car needs major repairs, I've marked approximately where this value of X should be.
Symbolically:
P(X≤a)=0.05
To reach the proper value of "a" you have to work using the standard normal distribution because is a tabulated distribution and you can extrapolate it to any normal distribution.
So the first step is to look in the table of the Z distribution for the value that accumulates 0.05 of probability since it is such a low probability, you have to look for the value in the left side of the table (negative values of Z). You look for 0.05 in the body of the table and then the margins for the corresponding value (see second attachment)
Then the value that accumulates 0.05 of probability is -1.64, now you have to reverse the standardization to reach the asked value of X
Z= (a- μ)/σ
(Z*σ)= a - μ
a=(Z*σ)+ μ
a=(-1.64*9)+36
a= 21.24 months.
The guarantee period should be 21.24 months so that only 5% of the sold cars will need major repairs before it wears of.
I hope it helps!
