- #1
brad sue
- 281
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Hi,
I need have this problem check because I have a problem at the last question:
the probability is 0.6 that a well driller will find a well of water at a depth less than 100 feet in a certain area. Wells are to be drilled for six new homeowners. Assue that finding a well of water at a depth of less than 100 feet is independent from drilling to drilling and that the probability is 0.6 on every drilling. If X is the number of wells of water found. Find:
1-P(X>4)
My solution: P(X>4)=1-B(4;6,0.6)
2-P(X=4)
P(X=4)=b(4;6,0.6)
3-mean
mean=n*p=3.6
4-variance
Var^2=n*p*(1-p)=1.44
5-P(X=mean)
Here since mean =3.6, I have problem to find P(X=mean)
Thank you
B.
I need have this problem check because I have a problem at the last question:
the probability is 0.6 that a well driller will find a well of water at a depth less than 100 feet in a certain area. Wells are to be drilled for six new homeowners. Assue that finding a well of water at a depth of less than 100 feet is independent from drilling to drilling and that the probability is 0.6 on every drilling. If X is the number of wells of water found. Find:
1-P(X>4)
My solution: P(X>4)=1-B(4;6,0.6)
2-P(X=4)
P(X=4)=b(4;6,0.6)
3-mean
mean=n*p=3.6
4-variance
Var^2=n*p*(1-p)=1.44
5-P(X=mean)
Here since mean =3.6, I have problem to find P(X=mean)
Thank you
B.