Question

On a normal distribution of IQ test scores, with a mean of 100 and a standard deviation of 15 points, a score of 85 places you approximately in what percentile of the population?

Answer 1

A score of 85 on an IQ test with a mean of 100 and a standard deviation of 15 places a person in the 16th percentile, meaning they score better than approximately 16% of the population.

Step-by-step explanation:

On a normal distribution of IQ test scores, with a mean of 100 and a standard deviation of 15 points, a score of 85 is one standard deviation below the mean. This typically places an individual within the 16th percentile, meaning they score better than approximately 16% of the population. The bell curve of a normal distribution indicates that about 68% of values (IQ scores, in this case) fall within one standard deviation of the mean. Therefore, to find out what percentage falls below a score of 85, we can subtract half of this 68% from 100%.

Step 1: Identify the mean and standard deviation. In this case, the mean is 100 and the standard deviation is 15.

Step 2: Determine how many standard deviations your score is from the mean. A score of 85 is one standard deviation below the mean.

Step 3: Refer to the empirical rule or a z-table. The empirical rule says that 68% of data fall within one standard deviation of the mean. Therefore, 100% - 34% (half of 68%) equals 66%. We must subtract another 2% that the empirical rule allocates for the tail beyond one standard deviation, giving us approximately the 16th percentile.

A score of 85 places you in the 16th percentile of the IQ distribution.

Answer 2

`P(X<85)=P((X-\mu)/(\sigma)<(85-\mu)/(\sigma))=P(Z<(85-100)/(15))=P(Z<-1)`

And we can find this probability using the normal standard table or excel and we got:

`P(Z<-1)=0.1587`

So then we can conclude that the value of 85 is approximately the 16 percentile for this case.

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(100,15)`

Where
`\mu=100` and
`\sigma=15`

We are interested on this probability

`P(X<85)`

And the best way to solve this problem is using the normal standard distribution and the z score given by:

`z=(x-\mu)/(\sigma)`

If we apply this formula to our probability we got this:

`P(X<85)=P((X-\mu)/(\sigma)<(85-\mu)/(\sigma))=P(Z<(85-100)/(15))=P(Z<-1)`

And we can find this probability using the normal standard table or excel and we got:

`P(Z<-1)=0.1587`

So then we can conclude that the value of 85 is approximately the 16 percentile for this case.