1. Find the ratio of the width to the length of the rectangular swimming pool sketched below.

Leakage from underground gasoline tanks at service stations can damage the environment. It is estimated that 25% of these tanks

Gnesinka [82]

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.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

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 $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

It is estimated that 25% of these tanks leak.

This means that $p = 0.25$

15 tanks chosen at random

This means that $n = 15$

a.What is the expected number of leaking tanks in such samples of 15?

$E(X) = np = 15*0.25 = 3.75$

b.What is the probability that fewer than 3 tanks will be found to be leaking?

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{15,0}.(0.25)^{0}.(0.75)^{15} = 0.0134$

$P(X = 1) = C_{15,1}.(0.25)^{1}.(0.75)^{14} = 0.0668$

$P(X = 2) = C_{15,2}.(0.25)^{2}.(0.75)^{13} = 0.1559$

$P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0134 + 0.0668 + 0.1559 = 0.2361$

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

$\mu = E(X) = np = 2000*0.25 = 500$