Hope I'm not too late! This explanation should get you through 1,2,4, and 5. For 3 you'll need a little bit of probability/combinatorics.
Approach to 3:
First, find the probability that any one test is between 65/75 (find the z scores, calculate the probability by use of calculator or table). Then we need to figure out a binomial distribution for the 3/5 tests part where p = the probability and n = 5 (tests): f(3; 5, p)
means compute the probability of 3 successes in 5 test-takings where the probability of a success is p.
A few more common-sense definitions:
A normal distribution/bell curve emerges when you have lots of collections of objects (think standardized testing, IQs, height/weights); most people will average around the middle, and you will have a smaller (and smaller) percentage of the population towards either extreme of the trait. We call this middle the mean
, which is expressed by the greek letter mu (μ)
When describing a normal distribution, a useful variable is the standard deviation
, (σ or s, though there is a slight difference between these). This basically marks how widely scattered around the mean the traits are. The graph sentythee provided above is a useful depictor of standard deviation: approximately 68% of a population will fall within one standard deviation from the mean (μ-σ to μ+σ), 95% within 2 standard deviations (μ-2σ to μ+2σ), and 99.7% within 3 standard deviations (μ-3σ to μ+3σ). This 68-95-99.7 rule
is actually worth memorizing, as it can save you a lot of time on tests.
For example, if I had a sample survey with a mean weight of 150 pounds and a standard distribution of 15, ASSUMING A NORMAL DISTRIBUTION, this would mean that 68% of the survey would lie between 135 and 165 pounds, 95% between 120 and 180 pounds, and 99.7% between 95 and 195 pounds.
Given a normal distribution, it's useful to think about where a particular individual lies within that distribution. For example, if you take a standardized test, it's helpful to see your percentile, or where you lie relative to the other test takers. We do this by measuring an individual's z score
, which is a fancy statistics way of talking about their individual percentile. the z score expresses how many standard deviations a person is from the mean.
The z score is computed via the following formula:
z = (x - μ)/σ
x is the individual's particular statistic for the trait.
For example, given the previous normal weight distribution (μ=150, σ=15), we would expect an individual with weight 165 (translation: x = 165) to be exactly 1 standard deviation above the mean. Therefore their z score should equal 1:
z = (165-150)/15 = (15/15) = 1.
An individual with weight 135 should yield a z score of -1, meaning they are 1 standard deviation below the mean.
An individual with a weight of 153 should yield a z score of
z = (153-150)/15 = .2
meaning he/she is .2 standard deviations below the mean.
When we are calculating percentages/percentiles/probabilities of particular individuals, we use something called the standard normal distribution. This is a normal distribution with μ=0 and σ=1. It can be proved that the area under this curve is 1 (in mathematical terms, that this is a NORMALIZED function). This proves very useful:
For example, take the slightly heavyset person with the weight of 165 (x = 165 ... z = 1) in the previous example. Some useful questions people might ask are:
- What percentile is this person's weight (eg. what percentage of people have a lower weight than this person?)
- What percentage of people have MORE EXTREME weights than this person (eg. what % of people have a higher weight than this person [z=1], AND what % of people have a lower weight than a similarly slightly lightweight person [z = -1]
- What percentage of people have LESS EXTREME weights than this person (eg. what % of people have weights with z scores less than 1 and more than -1)?
Notice all of these questions are PERCENT questions.
FACT: all probabilities in a given situation must sum to 1.
FACT: the standard normal distribution's area sums to 1.
Interjecting a little calculus, if we shade the proper area under a curve, this means we can find the probability associated with the events.
Ignore the quartz part, i just poached this image off somewhere online.
Notice how the shaded area is bounded by the lines x (or z) = -1 and x (or z) = 1. Another way of phrasing this question would be question #3 above (people with z scores with a magnitude <1). In the standard normal distribution, we would expect this shaded area under this curve to be .68 (see the 68-95-99.7 rule above). In answer to the question, we might say that
"There is a (.68 probability/68% chance) that a person will (have a weight LESS extreme than 165/have a z score less extreme than 1/have a weight within 1 standard deviation of the mean)"
Any of these following are correct ways of expressing this information.
Also, notice how if we looked at the white area under the curve, (the part NOT shaded), we would expect this area to be 1-.68 = .32. This area denotes the parts with z scores MORE extreme than z=1 (question #2). To answer this question, we might say the following:
"There is a (.32 probability/32% chance) that a person will (have a weight MORE extreme than 165/have a z score MORE extreme than 1/have a weight greater than 1 standard deviation from the mean)"
Any of these are correct ways of expressing this information.
Lastly, note that if we took the line x = 1 and shaded to the left of it:
It would denote the percentile, the % of individuals with z < 1 (% of individuals with weight < 165).
Notice how we can again use the 68-95-99.7 rule and the symmetry of the standard normal distribution to solve:
- 68% of individuals are within 1 standard deviation of the mean (|z| < 1)
- 100%-68% = 32% of individuals are greater than 1 standard deviation from the mean (|z| >1). NOTICE how this area applies to both ends(tails) of the spectrum. To get the area at one end, we just divide by 2:
- 32/2 = 16% of individuals should have a z > 1, and 16% of individuals should have a z < -1.
If we want to find % z < 1, we add the middle chunk (|z| < 1, 68%) and the left tail (z < -1, 16%).
68+16 = 84%, so we can say that (answer to question #1)
"An individual with a weight of 165 would be in the 84th percentile (84% of people would weigh less than him)."
A survey was taken of people's weights, and the statistician computed the average of the weights to be equal to 150 pounds. The standard deviation associated with this is 15 pounds. Without using a calculator or table,
a) What is the z score associated with a person who weighs 120 pounds?
b) What percentile weight would this person fall into?
c) What percentage of people have a weight greater than 120 pounds?
d) What percentage of people have a MORE EXTREME weight than 120 pounds?
e) Estimate the percent of people who have a weight LESS THAN 165 pounds and MORE THAN 105 pounds.