Respuesta :
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.
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.
[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]
And p is the probability of X happening.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
It is estimated that 25% of these tanks leak.
This means that [tex]p = 0.25[/tex]
15 tanks chosen at random
This means that [tex]n = 15[/tex]
a.What is the expected number of leaking tanks in such samples of 15?
[tex]E(X) = np = 15*0.25 = 3.75[/tex]
b.What is the probability that fewer than 3 tanks will be found to be leaking?
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)[/tex]
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{15,0}.(0.25)^{0}.(0.75)^{15} = 0.0134[/tex]
[tex]P(X = 1) = C_{15,1}.(0.25)^{1}.(0.75)^{14} = 0.0668[/tex]
[tex]P(X = 2) = C_{15,2}.(0.25)^{2}.(0.75)^{13} = 0.1559[/tex]
[tex]P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0134 + 0.0668 + 0.1559 = 0.2361[/tex]
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
[tex]\mu = E(X) = np = 2000*0.25 = 500[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{2000*0.25*0.75} = 19.36[/tex]
This probability is 1 subtracted by the pvalue of Z when X = 600. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{600 - 500}{19.36}[/tex]
[tex]Z = 5.16[/tex]
[tex]Z = 5.16[/tex] has a pvalue of 0.
0% the probability that at least 600 of these tanks are leaking
