Question

Water samples from a particular site demonstrate a mean coliform level of 10 organisms per liter with a standard deviation of 2. Values vary according to a Normal distribution. What is the 90th percentile on this distribution? Round to two decimal places.

Answer 1

Answer:

The 90th percentile on this distribution is 12.57 organisms per liter.

Step-by-step explanation:

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

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

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This is the percentile that the score is in the distribution.

In this problem, we have that:

Water samples from a particular site demonstrate a mean coliform level of 10 organisms per liter with a standard deviation of 2, so $\mu = 10, \sigma = 2$.

What is the 90th percentile on this distribution?

Z has a pvalue of 0.90 when it is between $Z = 1.28$ and $Z = 1.29$, so we use $Z = 1.285$

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

$1.285 = \frac{X - 10}{2}$

$X - 10 = 2.57$

$X = 12.57$

The 90th percentile on this distribution is 12.57 organisms per liter.

Answer 2

Answer:

A. The 90th percentile is 12.56

Step-by-step explanation:

A.  The 90th percentile is a possible value of x, it´s a reference value which tells us that any x below that percentile has a probability to happen below 90%.

The 90th percentile it´s a value of x that its probability is as close from below to 90% (i.e. P(z≤y)≈0.9 where y is the percentile)

We find the 90th percentile in a standard normal distribution and then, with the mean and the STD of the actual distribution, we "transform" the 90th percentile of the standard normal distribution to the one that we are looking for as follows.

The standard normal distribution has many tools  that you are able to use to find probabilities in it. My favorite is a table (that is attached) that has so many useful values of the distribution (standard normal) in it. The table works like this:

P(z≤a.bc) = The value in the coordinates (a.b,c)

where a is the whole part of the number, b its first decimal value, and c is its second decimal value

We are looking for some "y" that give us P(z≤y)≤≈0.9. in the chart we find that y=1.28 because:

P(z≤1.28)=0.89997≈0.9

Now we proceed to "transform" that 1.28 in the value that we are looking for, there´s a process called "Normalization" that takes us from a particular normal distribution of any mean and STD to a standard normal distribution with mean 0 and STD 1 and it works like this:

P(x<a)=P( (x-μ)/σ ≤ (a-μ)/σ )=P(z≤b)

Where μ is the mean of our particular normal distribution and σ is the STD of that distribution. Now, we need some process that does the same but backward because we want to get back to our normal distribution, so we kinda isolate the x in this expression:

z=(x-μ)/σ

x=σz+μ

So, with z=1.28, we find x:

x=1.28*2+10

x=12.56

With this value we have that

P(x≤12.56) = P( (x-10)/2 ≤ (12.56-10)/2) = P(z≤1.28)

P(x≤12.56) = 0.8997 ≈0.9

Due to this, x=12.56 is our 90th percentile.