It is conjectured that an impurity exists in 30% of all drinking wells in a certain rural community. In order to gain some insight on this problem, it is determined that some tests should be made. It is too expensive to test all of the many wells in the area, so 10 were randomly selected for testing. a. Using the binomial distribution, what is the probability that exactly three wells have the impurity assuming that the conjecture is correct? b. What is the probability that more than three wells are impure?

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Jordan Mitchell · Accepted Answer

Step 1  Given:  p=0.30n=10  a) The probability that exactly three wells have the impurity is obtained as below:  P(X=x)=nCxpx(1−p)n−x  P(X=3)=10C3×0.303(1−0.30)10−3  =120×0.303×0.707  =0.2668  Thus, the probability that exactly three wells have the impurity is 0.2668.  Step 2  b) The probability that more than three wells are impure is obtained as below:  P(X&amp;gt;3)=1−P(X≤3)  =1−{P(X=0)+P(X=1)+P(X=2)+P(X=3)}  =1−{{10}C00.3000.7010+10C10.3010.709+10C20.3020.708+10C30.3030.707}  =1−(0.0282+0.1211+0.2335+0.2668)  =1−0.6496  =0.3504  Thus, the probability that more than three wells are impure is 0.3504.

Bertha Jordan · Accepted Answer

Step 1  a) b(x; 10, 0.3)  =P(X=3)  =B(3; 10, 0.3)−B(2; 10, 0.3)  =0.6496−0.3828=0.2668  b) P(X&amp;gt;3)=1−B(3; 10, 0.3)  =1−0.6496=0.3504

nick1337 · Accepted Answer

This is the well-known procedure for a hypothesis test. You are trying to determine if the value X=6 lies in the critical region (usually defined as the upper or lower 5% of the distribution). You can determine this to be the case in this example if p(X≥6&amp;lt;0.05) which is true. We conclude therefore that there is evidence that the percentage of polluted wells is greater than 30% at the &#039;&#039;5% level&quot; of significance. The individual probability p(X=6)is of no importance. However the occurrence of 6 polluted wells in the sample can be regarded as very unlikely if the proportion is only 30%

It is conjectured that an impurity exists in 30% of all drin

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