Question

To qualify for a police academy, candidates must score in the top 10% on a general abilities test. Assume the test scores are normally distributed and the test has a mean of 200 and a standard deviation of 20. Find the lowest possible score to qualify.

Answer 1

Answer:

$z=1.28<\frac{a-200}{20}$

And if we solve for a we got

$a=200 +1.28*20=225.6$

So the value of height that separates the bottom 90% of data from the top 10% is 225.6.  

See explanation below

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:

$X \sim N(200,20)$  

Where $\mu=200$ and $\sigma=20$

For this case we can use the z score in order to solve this problem, given by this formula:

$z=\frac{x-\mu}{\sigma}$

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.1$   (a)

$P(X<a)=0.9$   (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.  

As we can see on the figure attached the z value that satisfy the condition with 0.9 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.9 and P(z>1.28)=0.1

If we use condition (b) from previous we have this:

$P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.9$  

$P(z<\frac{a-\mu}{\sigma})=0.9$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.28<\frac{a-200}{20}$

And if we solve for a we got

$a=200 +1.28*20=225.6$

So the value of height that separates the bottom 90% of data from the top 10% is 225.6.  

Answer 2

The lowest possible score to qualify in the top 10% is approximately 226.

What is z score?

Z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:

z = (raw score - mean) / standard deviation

Given;  mean of 200 and a standard deviation of 20

P(z > c) = 10% = 0.1

1 - P(z < c) = 0.1

P(z < c) = 0.9

P(z < c) = 1.282

Hence:

1.282 = (x - 200)/20

x = 225.64

The lowest possible score to qualify in the top 10% is approximately 226.

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