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Sections in Text: 6.3, 6.4, 7.
One of the main reasons we spend so much time with the Normal Distribution, and why it models so
many situations in nature is because both we, and nature, tend to add quantities together. Hereโs an
intuitive example that we will present now and explore more later.
Suppose you are waiting on line at the supermarket and start to keep track of how long it takes an
individual to be โchecked outโ once you reach the cashier. Since not all orders are the same, you will
notice that there is a distribution of times spent with the cashier. You might find that people take on average 4 minutes but that when you construct a histogram you get something like this:
Notice that you would โbalanceโ the histogram at the average value of 4 minutes. There is nothing โsymmetric and mound shapedโ about this distribution. However this histogram is for the amount of time taken by a single individual. Usually there is a line when I go to the supermarket. The total time I invest in line is the total time taken by the people ahead of me plus the time taken by me. If there is only one person ahead of me I would expect to see a histogram like this:
(^00 5 10 15 20 25 30 )
100
200
300
400
500
600 Histogram of "Cashier Service Times" at a Supermarket
Time Spent in Minutes
Number of Customers
(^00 5 10 15 20 25 30 35 40 )
50
100
150
200
250
300
350 Histogram of "Total Time Spent" at Cashier at a Supermarket
Number of Customers
Time Spent in Minutes
Notice that you would โbalanceโ the histogram at the average value of 8 minutes. Suppose there are 10 people in line. What sort of distribution emerges when we process all 10?
Notice that the figure is becoming more โsymmetric and mound shapedโ. Weโll say more about this later.
Letโs start with an example to review the Normal Distribution and then look at the connection between the normal and the binomial. Very often psychologists will give people โpen and paperโ surveys to measure some characteristic. As an example, if you want to tell if someone tends to be more optimistic or pessimistic you might give them the โLife Orientation Testโ (Scheier, Carver, & Bridges, 1994). Here is the test:
LOT-R
Please be as honest and accurate as you can throughout. Try not to let your response to one
statement influence your responses to other statements. There are no "correct" or "incorrect"
answers. Answer according to your own feelings, rather than how you think "most people" would
answer.
A = I agree a lot B = I agree a little C = I neither agree nor disagree D = I DISagree a little E = I DISagree a lot
_1. In uncertain times, I usually expect the best. [2. It's easy for me to relax.]
- If something can go wrong for me, it will.
- I'm always optimistic about my future. [5. I enjoy my friends a lot.] [6. It's important for me to keep busy.]
- I hardly ever expect things to go my way. [8. I don't get upset too easily.]_
(^010 20 30 40 50 60 70 80 )
50
100
150
200
250
Time Spent in Minutes
Number of Customers
Histogram of "Total Time Spent" at Cashier (10 Customers)
๏ท What score would someone have to obtain to be within the top 10% of the population?
Remember that we have two related formulas:
Depending upon whether we have a score and want a probability or have a probability and want a score. If we want the data point which separates the top 10% from the bottom 90% we first see what ๐ง value corresponds to an area of 0.9000. From my table I see that ๐ด = 0.8997 corresponds to ๐ง = 1.28_. Close enough. Then the score cutting off the top 10% is_ ๐ฅ = ๐ + ๐ง๐ = 15 + 1.28 โ 3 = 18.84. Is this consistent with your graph?
Hereโs an important example. Recall from our previous lectures that a binomial random variable (discrete) may be fairly well approximated by the normal distribution (continuous). As a rule of thumb, the approximation is good when np > 5 and n(1 โ p) = nq > 5. That is, when the average number of successes and also the average number of failures are both at least as high as 5.
As a reminder, consider the following graphs:
Note that in the first figure we have ๐๐ฃ๐๐๐๐๐ ๐๐ข๐๐๐๐ ๐๐ ๐ ๐ข๐๐๐๐ ๐ ๐๐ = ๐ โ ๐ = 5 โ 0.2 = 1 < 5 and the match isnโt very good. By the time we get to the last figure we have ๐๐ฃ๐๐๐๐๐ ๐๐ข๐๐๐๐ ๐๐ ๐ ๐ข๐๐๐๐ ๐ ๐๐ = ๐ โ ๐ = 50 โ 0.2 = 10 โซ 5 as well as ๐๐ฃ๐๐๐๐๐ ๐๐ข๐๐๐๐ ๐๐ ๐๐๐๐๐ข๐๐๐ = ๐ โ 1 โ ๐ = 50 โ 0.8 = 40 โซ 5 and our approximation is quite good.
0 2 4 6
0
n=5 p=0.
0 5 10
0
n=10 p=0.
0 10 20 30
0
n=30 p=0.
0 20 40
0
n=50 p=0.
binomial normal
binomial normal
binomial normal
binomial normal
Since most people donโt enjoy the binomial formula, letโs think about how we might use the normal approximation to approximate binomial probabilities. Below is a graph of the probability distribution of a binomial random variable describing a situation with n = 12 independent trials, each with probability of success p = 0.55, i.e. tossing a biased coin 12 times and counting the number of heads. The graph of the probability density function of a normal distribution has been superimposed.
We can use our table to get the probability of 10 successes: I read off the number ๐๐๐๐ = 0.034. This looks about right on the figure below. Just note for what follows that the height of the spike equals the binomial probability.
Now, how did I fit a normal distribution? What mean did I use? What standard deviation?
If you think about it, we would like the center or mean of our binomial distribution to match that of the normal distribution, so I set the mean of the normal distribution equal to ๐ = ๐๐. It is the same for the
standard deviations- we match the spread of the normal to that of the binomial: ๐ = ๐ ๐ 1 โ ๐
This gives me ๐ = ๐๐ = 12 โ 0.55 = 6.6 and ๐ = ๐ ๐ 1 โ ๐ = 12 โ .55 โ .45 = 1.7234.
As always, look at these numbers and make sure you see that 6.6 is right in the middle and if we look โthree up and three downโ, i.e. from 6.6 โ 3 โ 1.7234 = 1.4298 and 6.6 + 3 โ 1.7234 = 11.7702 we have captured just about all of the area under our curve.
Now how do we use our table? There are 3 steps here:
๏ท Notice that the โheight of the spikeโ pretty much touches the normal curve, so if we can figure out the height on the normal curve we know the binomial probability. ๏ท Since the formula for the normal curve is more than we want we need to use thez-table somehow. The z-table gives us areas whereas we want a height. Hereโs a clever idea: the area of a rectangle is equal to its base times its height, so if we set up a rectangle with a base of 1 we get area = 1*height or (ignoring units) area = height. ๏ท Since the area under the curve is well approximated by the area of a rectangle (see graph) we can say that the โheight of the spikeโ at 10 is approximately equal to the area under the curve from 9.5 to 10.5.
(^0) -1 0 1 2 3 4 5 6 7 8 9 10 11 12 13
n=12 p=0. binomial normal
- Binomial to Normal Approximation: Since we want the probabilities from 40 to 60, we break our
interval up into pieces. The first goes from 39.5 to 40.5 in order to represent 40, the second goes from 40.5 to 41.5 in order to represent 41, all the way up to 59.5 to 60.5 to represent 60. Now find the area under the normal curve from 39.5 to 60.5 when n=100 and p=0.5. We have
๐ = ๐๐ = 100 โ 0.5 = 50 , and ๐ = ๐ ๐ 1 โ ๐ = 100 .5. 5 = 5. Do you see that the curve below is centered at 50 and if you look three up and three down (between 45 and 65) you have just about all your probability? Hereโs the calculation:
probability (^) 0. 9642
Pretty close!
And hereโs another presentation: Calculate via Excel and using the normal approximation to the binomial distribution the probability that, if you roll a fair die 100 times you will obtain between 10 and 20 (inclusive) โSIXESโ.
0 10 20 30 40 50 60 70 80 90 100 0
n=100 p=0. binomial normal
