Probability Distribution: - a table or a graph that displays the theoretical probability for each outcome of an experiment.
- P (any particular outcome) is between 0 and 1
- the sum of all the probabilities is always 1.
Binomial Distribution:
Is a distribution in which the probabilities are symmetrical on the left and right side.
Uniform Probability Distribution: - a probability distribution where the probability of one outcome is the same as all the others.
Doing it on your Calculator:
Binompdf (Binomial Probability Distribution: displays a binomial probability distribution when the number of trials and the theoretical probability of the favourable outcome are specified.
Step 1 Press 2nd VARS
Step 2 Press 0
binompdf ( number or Trails, Theoretical Probability of favorable outcome)
Example 1: Using your graphing calculator, determine the probabilities of having any number of girls in a family of 5 children.
-Number of trails = 5 (5 children)
-Probability of favorable outcome= ½
1. binompdf (5, ½)
2. Store answer in L2 ( Ans ->L2)
3. Enter 0 to 5 in L1 (STAT Editor)
4. Adjust WINDOW Settings
5. Select Historgram in STAT PLOT and graph (2ND Y=, Plot1 on, type: histogram, Xlist L1, Freq: L2)
6. Graph
Answer: (0.03125, 0.15625, 0.3125, 0.3125,0.15625,0.03125)
Example 2: When a soccer player takes a shot on net they score 47% of the time. Create a binomial distribution using your graphing calculator to represent the next 10 shots on net.
-Number of trails = 10(10 children shots)
-Probability of favorable outcome=0.47
(Turn STAT plot on)
1. binompdf (3, 0.47)
2. Store answer in L2 ( Ans ->L2)
3. Enter 0 to 3 in L1 (STAT Editor)
4. Adjust WINDOW Settings
5. Select Historgram in STAT PLOT and graph (2ND Y=, Plot1 on, type: histogram, Xlist L1, Freq: L2)
6. Graph
Answer: (0.148877, 0.396,0.351, 0.1038)
For Binomial Distributions
Mean(µ or X): - the average or the middle of a given set of data.To calculate the mean you total up all of your values and divide by the number of values.
There are 2 ways to calculate mean:
1) Calculator
step #1 Enter numbers into L1
step #2 Press STAT then CALC then 1
setp #3 Read x=
2) Use the formula
µ=np
n= number of trials
p= probability of a favorable outcome
Standard Deviation (σ): - the measure of how far apart are the data spread out from the mean. Standard deviation is called a measure of dispersion. A small standard deviation means that the data is clustered fairly close to the mean. A large standard deviation means that the data is spread apart quite far from the mean.
Dashed line has a larger standard of deviation
Solid line has a smaller standard deviation
There are 2 ways to calculate Standard deviation:
1) Calculator
step #1 Enter numbers into L1
step #2 Press STAT then CALC then 1
setp #3 Read σx=
2) Use the formula
σ =√(np(1-p))
n= number of trials
p= probability of a favorable outcome
Frequency Distribution: - a Histogram (bar graphs with no gaps) OR a Curve showing the frequency of occurrence over the range of values of a data set.
e.x
3-3: The Normal Distribution
-Anormaldistributionofdataiswherethemean,medianandmodeare thesameandinthemiddle;suchthatthedataisdistributedsymmetricallyaboutthemiddle;andthathasagraphthatappearsasacontinuous
bell shapedcurve.
Normaldistributionsaretheresultofusingverylargesamplesofcertainkindsofdata,
chosenrandomly.
Normal Distribution (Bell Curve): a probability distribution that has been normalized for standard use and exhibits the following characteristics.
a. The distribution has a mean (µ) and a standard deviation (σ).
b.The curve is symmetrical about the mean.
c.Most of the data (99.7%) is within ±3 standard deviation of the mean.
d.The area under the curve represents probability. The total area under the entire curve is 1 or 100%.
e The normal curve extends indefinitely in both directions (The curve gets really close to the x-axis, but never touches it).
f. All of the data is represented by the area under the curve.
g. The mean, median and mode lie in the centre of the data
h. 50% of your data lies to the left of the mean and 50% of the data lies to the right of the mean.
To plot a standard normal curve using TI-83:
Press Y=
Press 2nd Vars
Press 1
Type in Y= normalpdf(X,0,1)
Standardized Normal Curve If you have data that follows a normal distribution then you can “normalize it”. This means that you turn it into the standard normal distribution curve (with a mean of 0 and a standard deviation of 1). This is useful as you can use the standard normal curve to figure out probability percentages.
To do this we need to use the following formula:
Raw-Scores (X): - the scores as they appear on the original data list.
z-score (z): - the number of standard deviation a particular score is away from the mean in a normal distribution.
µ-the mean of your original data
σ-the standard deviation of your original data
The standard normal distribution is sometimes called the z distribution. A z score always reflects the number of standard deviations above or below the mean a particular score is.
For instance, if a person scored a 70 on a test with a mean of 50 and a standard deviation of 10, then they scored 2 standard deviations above the mean. Converting the test scores to z scores, an X of 70 would be:
So, a z score of 2 means the original score was 2 standard deviations above the mean. Note that the z distribution will only be a normal distribution if the original distribution (X) is normal.
Determining Probability form Z-scores
Once we calculate Z scores this will allow us to determine the probability of that event occurring.
Ex The heights of students are normally distributed with a mean of 145 cm and
a standard deviation of 15 cm. What percentages of students have a height greater than 160 cm?
Step 1 calculate their Z-score
Z= (160-145)/15
Z= 1
That means that our score is 1 point off of the mean, using our standard deviation picture above only 15.87% of a normal distribution lie above 1 standard deviation form the mean. So our answer is 15.87.
If we calculate a Z score that is not a whole number we can use the following table to determine the probabilities.
NOTICE THAT THIS CHART GIVES YOU THE PROBABILITIES BELOW THAT Z Score.
If you look for our z score 1 you find 0.8413, this is the probability that it will fall below to find the probability that it will fall above you have to go 1-0,08413 = 0.1587 or 15.9%
You can also calculate this using your calculator
To plot a standard normal curve using TI-83:
Press 2nd Vars
Press 2
Normalcdf(Xlower, Xupper,µ, σ) - use to convert Raw-Score directly to probability with
Normalcdf(Zlower, Zupper)- use to convert z-Score to probability
(- if Xloweror Zlower is at the very left edge of the curve and is not obvious, use −1 × 1099 on calculator- if Xor Z is at the very right edge of the curve and is not obvious, use
1 × 1099 )
Ex The heights of students are normally distributed with a mean of 145 cm and
a standard deviation of 15 cm. What percentages of students have a height greater than 160 cm?
Normalcdf(160, 1X1099,145, 15)
=0.15865
To move from Probabilities to Raw scores
Press 2nd Vars
Press 3
invNorm (area left of boundary, µ, σ) : - use to convert Area under the curve back to Raw-Score
invNorm (area left of boundary) : - use to convert Area under the curve (Probability) back to z-Score
Ex our probability was 0.15865 in the above question but this represents the area above the Z score
To find the area to the left of the z-score we have to go 1-0.15865 =0.84135
Thus invNorm (0.84135, 145, 15) =160
Graphs:
ShadeNorm (Xlower, Xupper, µ, σ) : - use to convert Raw-Score directly to probability with graphics (this will show you what the graph looks like).
ShadeNorm (Zlower, Zupper) : - use to convert z-Score to probability with graphics (this will show you what the graph looks like).
The Normal Approximation to a Binomial Distribution
Binomial Distribution: - a histogram that shows the probabilities of an experiment repeated many times (only success or failure – desirable or undesirable outcomes).
When the conditions np > 5 and n(1 −p) > 5 are met, we can use the normal approximation for the binomial distribution. ONLY IN THAT CASE, the mean and the standard deviation can be calculated by:
(standard deviation) σ =√(np(1-p)) (mean) µ=np
where n = number of trials and p= probability of favourable outcome
Using the bell curve to approximate a binomial distribution really depends on the number
of trials, n. When n is small, there are very few bars on the binomial distribution and the
bell curve does not fit the graph well. However, when n is large, the bell curve fits the
binomial distribution much better. Therefore, we can use the area under normal bell curve
to approximate the cumulative sum of the binomial probabilities.
Example 1: A multiple-choice test has 10 questions. Each question has 4 possible choices.
a. Determine whether the conditions for normal approximation are met.
b. Graph the resulting binomial distribution.
c. Find the probability that a student will score exactly 6 out of 10 on the test.
d. Calculate the probability that a student will at least pass the test.
a) n=10
p= (1/4)=0.25
np= 10x0.25 =2.5 (less then 5) n(1-p)= 10 x (1-0.25) = 7.5 (greater then 5)
Therefore we can use The Normal Approximation to a Binomial Distribution
b)
(Turn STAT plot on)
1. binompdf (10, 0.25)
2. Store answer in L2 ( Ans ->L2)
3. Enter 0 to 3 in L1 (STAT Editor)
4. Adjust WINDOW Settings
5. Select Historgram in STAT PLOT and graph (2ND Y=, Plot1 on, type: histogram, Xlist L1, Freq: L2)
6. Graph
c)
1. binompdf (10, 0.25, 6)
=0.016222
d)
a pass is 5/10
binompdf (10, 0.25, 5) + binompdf (10, 0.25, 6) +binompdf (10, 0.25, 7) +binompdf (10, 0.25, 8) +binompdf (10, 0.25, 9) +binompdf (10, 0.25, 10)
or
sum binompdf (10, 0.25, {5,6,7,8,9,10})
=0.07813
Example 2: A multiple-choice test has 30 questions. Each question has 4 possible choices.
a. Determine whether the conditions for normal approximation are met.
b. Find the mean and standard deviation
c. Graph the resulting binomial distribution..
d. Find the probability that a student will score exactly 17 out of 30 on the test.
e. Calculate the probability that a student will at least pass the test.
a)
n = 30 questions p = 1/4= 0.25 (probability of guessing a question correct)
np = 30 × 0.25 n(1 −p) = 30 × (1 − 0.25)
np = 7.5 (greater than 5) = 30 × 0.75
n(1 − p) = 22.5 (greater than 5)
b)
µ= np σ = √np x n(1 −p)
=(30)(0.25) σ = √(7.5)(1-0.25)
=7.5 = 2.372
c)
(Turn STAT plot on)
1. binompdf (30, 0.25)
2. Store answer in L2 ( Ans ->L2)
3. Enter 0 to 3 in L1 (STAT Editor)
4. Adjust WINDOW Settings
5. Select Historgram in STAT PLOT and graph (2ND Y=, Plot1 on, type: histogram, Xlist L1, Freq: L2)
6. Graph
d)
binompdf (30, 0.25, 17)
e)
sum binompdf (30, 0.25, {15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30} )
Confidence Intervals: - the level of assurance from a statistical report.
-symmetrical area around the mean.
- confidence intervals (CI) present a range of values around the sample estimate (you may
be familiar with this concept when hearing about poll results and the ‘margin or error’)
the 95 % CI is the range of values with which there is a 95% probability that the true value lies
Example 1: Given that µ = 72.5, σ = 5.24 and n = 130, draw a 95% confidence interval curve and determine the margin of error and the percent margin of error.
Xlower = invNorm (0.025, 72.5, 5.24) Xlower = 62.2
Xupper = invNorm (0.975, 72.5, 5.24) Xupper = 82.8