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normal distribution is a bell shaped curve where 50% of the values are below the mean and 50% of the values are above the mean.
in a normal distribution, the mean, the mode, and the median all fall at the 50% mark.
here's a reference that describes a normal distribution.
the reference describes what a normal distribution and what a standard deviation is (spread of data around the mean).
it also describes what a z-score is and what calculating a value between 2 values on the graph means.
here's your question again:
X = the average income
X has a normal distribution with μ = 55000 and σ = 9000
We need to find P(18000 < X < 72000).
Since X has a normal distribution with μ = 55000 and σ = 9000, Z = (X-μ)/σ has a standard normal distribution, therefore 97.06% of household incomes is this state likely to be between 18,000 to $72,000/year.
What is the P(18000< X <72000) stating?
My answer to your questions and the problem:
p(18000 < X < 72000) is stating the probability that you pick a person out of the population at random and their income is between 18000 and 72000 dollars a year.
if the normal distribution is accurate, this should happen 97% of the time or thereabout. what that means is, if you pick a person out of the population at random 1000 times, that person will have an income between 18000 and 72000 roughly 970 times out of the 1000. if you pick a person out of the population at random 10,000 times, that person will have an income between 18000 and 72000 roughly 9,700 times.
probabilities and frequencies go together.
if you observe something happening 9000 times out of every 10,000 times, then the probability of you observing something happening each time you look is considered to be 9000/10000 = .9, or 90%.
now you have a normal distribution and the mean is 55000.
that's the average.
add all all the incomes of the people in the population and divide that by the number of people and you get an average of 55,000 per person. if you multiply the number of people in the population by 55,000, they you will get the total income of the population, as you should, since that's how you derived the average in the first place.
the standard deviation is a measure of the spread of the data. you measure the difference between the mean and each person's age and you square it. then you add up all the squared differences and divide that by the number of people. then you take the square root of that. that's the standard deviation. it's kind of like saying what is the average difference between each person's score and the mean. if all of the people had incomes of 55,000 then the standard deviation would be equal to 0. apparently, with this population, the average spread is around 9000. that's 9000 dollars above 55,000 and 9000 below 55,000.
the z-score is telling you how many times a score is above or below the mean.
the z-score is calculated in the following manner.
z-score = (data minus mean)/standard deviation.
for example, suppose you sample a person and his income is 70,000. since the mean is 55,000, then his z-score is (70,000 - 55,000)/9,000 which is equal to 15,000 / 9,000 which is equal to 1.6666667.
what this is saying is that the person's income is 1.666667 times the standard deviation above the mean.
note that, if a person has an income of 55,000 which is exactly the mean, then that person's z-score will be (55,000 - 55,000)/9,000 = 0.
with a z-score, the mean is always 0 and every score is that many standard deviations above or below that mean.
a person who makes 18,000 dollars will have a z-score of (18,000 - 55,000)/9000 which equals -37,000/9,000 = -4.11111 standard deviations from the mean.
that's pretty far from the mean and only an extremely small percentage of the people in the normal distribution would have incomes that small. in fact, the number calculated is .00002 of the population in a normal distribution would have an income that small. that's so close to 0 that we would consider it 0 for all practical purposes.
bear in mind this is a population with a normal distribution. in real life, the percentage of people with an income of 18,000 or less might be larger.
back to z-scores:
a person with an income of 72,000 will have a z-score of (72,000 - 55,000) / 9,000 = 1.8888888888. that translates to 1.8888888 times the standard deviation from the mean, only this time in a positive direction. according to the charts i'm using, that comes out to be .029454 of the population have incomes higher than 72,000 per year. subtract that from 1 and you get .970546 of the people have incomes less than 72,000 per year.
in your problem you are asked to find the percentage of people who have incomes between 18,000 and 72,000. since roughly 0% have incomes less than 18,000 per year and roughly 2.94% have incomes above 72,000 per year, you add up those percentages to get 2.94% have incomes either below 18,000 or above 72,000 which means that 100 - 2.96 = 79.6% of the people have incomes between 18,000 and 72,000 per year.
if you need instructions on how to read a z-score, check out the tutorials i have on the subject.
i also have a tutorial on a nifty little calculator that can halp you find the answers to your problems easily.
this calculator is only to be used to check your work because it will not teach you how to use the charts, which will you will need to know in order to pass a test on how to read them.
here are the tutorials.
email me at email@example.com if you have further questions regarding this material.