In a video for every 6 Enemy is defeated you earn three points if you do feeder 22 and I miss how Point would you have earned

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

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

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{2000*0.25*0.75} = 19.36$

This probability is 1 subtracted by the pvalue of Z when X = 600. So

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

$Z = \frac{600 - 500}{19.36}$

$Z = 5.16$

$Z = 5.16$ has a pvalue of 0.

0% the probability that at least 600 of these tanks are leaking

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A scientist need 10 L of a 20% acid solution for an experiment where she had only a 5% solution and a 40% solution to the neares

andrezito [222]

Answer:

Amount of 5% solution = 5.7 Litres

Amount of 40% solution = 4.3 Litres

Step-by-step explanation:

First, of all, let x be the amount of 5% solution used

Since, the scientist needs 10 L of solution, then;

Let 10 - x be the amount of 40% solution used

Now, we are told that she needs 10 L of a 20% acid solution.

This means that;

5%x + 40%(10 - x) = 20%(10).

Simplifying this gives;

0.05x + 0.40(10 - x) = 0.20 × 10

Expanding the bracket gives;

0.05x + 4 - 0.4x = 2

0.4x - 0.05x = 4 - 2

0.35x = 2

x = 2/0.35

x ≈ 5.7 litres

Thus, she will need to mix 5.7 liters of the 5% solution.

Since 10 - x be the amount of 40% solution used , then;

the amount of 40% solution = 10 - 5.7 = 4.3 litres

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