Answer:
a) 3.75
b) 23.61% probability that fewer than 3 tanks will be found to be leaking
c) 0% the probability that at least 600 of these tanks are leaking
Step-by-step explanation:
For each tank there are only two possible outcomes. EIther they leak, or they do not. The probability of a tank leaking is independent of other tanks. So we use the binomial probability distribution to solve this question.
Binomial probability distribution
The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
In which
is the number of different combinations of x objects from a set of n elements, given by the following formula.
![C_{n,x} = \frac{n!}{x!(n-x)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
And p is the probability of X happening.
The expected value of the binomial distribution is:
![E(X) = np](https://tex.z-dn.net/?f=E%28X%29%20%3D%20np)
The standard deviation of the binomial distribution is:
![\sqrt{V(X)} = \sqrt{np(1-p)}](https://tex.z-dn.net/?f=%5Csqrt%7BV%28X%29%7D%20%3D%20%5Csqrt%7Bnp%281-p%29%7D)
To solve question c), i am going to approximate the binomial distribution to the normal.
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
When we are approximating a binomial distribution to a normal one, we have that
,
.
It is estimated that 25% of these tanks leak.
This means that ![p = 0.25](https://tex.z-dn.net/?f=p%20%3D%200.25)
15 tanks chosen at random
This means that ![n = 15](https://tex.z-dn.net/?f=n%20%3D%2015)
a.What is the expected number of leaking tanks in such samples of 15?
![E(X) = np = 15*0.25 = 3.75](https://tex.z-dn.net/?f=E%28X%29%20%3D%20np%20%3D%2015%2A0.25%20%3D%203.75)
b.What is the probability that fewer than 3 tanks will be found to be leaking?
![P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)](https://tex.z-dn.net/?f=P%28X%20%3C%203%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%202%29)
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 0) = C_{15,0}.(0.25)^{0}.(0.75)^{15} = 0.0134](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20C_%7B15%2C0%7D.%280.25%29%5E%7B0%7D.%280.75%29%5E%7B15%7D%20%3D%200.0134)
![P(X = 1) = C_{15,1}.(0.25)^{1}.(0.75)^{14} = 0.0668](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20C_%7B15%2C1%7D.%280.25%29%5E%7B1%7D.%280.75%29%5E%7B14%7D%20%3D%200.0668)
![P(X = 2) = C_{15,2}.(0.25)^{2}.(0.75)^{13} = 0.1559](https://tex.z-dn.net/?f=P%28X%20%3D%202%29%20%3D%20C_%7B15%2C2%7D.%280.25%29%5E%7B2%7D.%280.75%29%5E%7B13%7D%20%3D%200.1559)
![P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0134 + 0.0668 + 0.1559 = 0.2361](https://tex.z-dn.net/?f=P%28X%20%3C%203%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%2B%20P%28X%20%3D%202%29%20%3D%200.0134%20%2B%200.0668%20%2B%200.1559%20%3D%200.2361)
23.61% probability that fewer than 3 tanks will be found to be leaking
c.Now you do a larger study, examining a random sample of 2000 tanks nationally. What is the probability that at least 600 of these tanks are leaking?
Now we have n = 2000. So
![\mu = E(X) = np = 2000*0.25 = 500](https://tex.z-dn.net/?f=%5Cmu%20%3D%20E%28X%29%20%3D%20np%20%3D%202000%2A0.25%20%3D%20500)
![\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{2000*0.25*0.75} = 19.36](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%7BV%28X%29%7D%20%3D%20%5Csqrt%7Bnp%281-p%29%7D%20%3D%20%5Csqrt%7B2000%2A0.25%2A0.75%7D%20%3D%2019.36)
This probability is 1 subtracted by the pvalue of Z when X = 600. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{600 - 500}{19.36}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B600%20-%20500%7D%7B19.36%7D)
![Z = 5.16](https://tex.z-dn.net/?f=Z%20%3D%205.16)
has a pvalue of 0.
0% the probability that at least 600 of these tanks are leaking