Area Bounded by Curves
Note:- area bounded with y axis
A
x
Eg:-Find the area bounded by the x axis and the Curve
Area between two Curves
The area bounded by the curves
Eg:-Find the area bounded by and
Eg:-Find the area of the region of the xy plane defind by the
inequalities
Ex:-(1)Find the area of the region of the xy plane bounded by
the curve
(2)Find the area of the region of the x-y plane define by the
following inequalities.
Volume of revolution
Curve is rotated one revolution aboutx-axis formed solids volume
will be given.
Note:-
If the curve is rotated about y-axis.
Eg:-(1) is a curve given A portion bounded by y=1, y=2 x axis
and curve is
rotate 3600 about x axis. Find the volume formed,
(2)The curve defined by the inequalities is rotated completely
about the y axis. Find the volume generated.
(3)The area enclosed by the curve and the line y=3 is rotated
about the line y=3. Find the volume of the solid generated?
Ex:-(1) the region defined by there inequalities is rotated
completely
about x axis. Find the volume generated?
(2) the area bounded by these curves rotated completely about y
axis.
Find the volume generated?
(3)Find the volume generated, when the area bounded between
is
rorated about .
A Difinite Integral is,
If we cannot evaluate difinite integral with an antiderivative,
we use numerical methods like the trapezoidal rule and Simpsons
rule. These rules enable us to estimate an integrals value to as
decimal places.
The trapezoidal rule approximates streches of curves with the
line of segments.
We add this area of the trapezoids made by joining the ends of
these segments to x-axis
gives, estimated area.
In this case, the estimate area in grater than the original.
This is called over estimate.
Over estimate area = estimated area -
In this type of curves give the estimate area less than the
original area. When we consider trapezoids, we lose small area in
every trapezoids. This type is called Under estimate.
Under estimate area
formula for the trapezoid rule
n trapezoids are considered,
sub interval length
Eg:-Use the trapezian rule with five ordinates to evaluate
ordinates y values
Eg:-Use the trapezian rule with five ordinates to evaluate
ordinates y values
x00.160.320.480.64
y11.04081.17351.43331.8965
y0 y1 y2 y3 y4 (5 Ordinates)
Simpsons Rule
eg:-Use simpson role with five ordinates (4 ships) to find an
approxomate value
x0
y00.840910.84090
y0 y1 y2 y3 y4 (5 Ordinates)
(2)Estimate to 4 decimal place using five ordinates and
applying
(a)The trapezium rule
(b)Simpson rule
Using the subtitution or other wise evaluate and hence
determine the accuracy of your estimated volume.
(Take )x01
y10.94120.8 0.640.5
y0 y1 y2 y3 y4 (5 Ordinates)
using trapezium rule
Using Sinpson rule
accuracy method
Conclusion
The trapezium rule gave an estimate correct to 2 decimal place
and
Simpson rule gave an estimate correct to 4 decimal place.
\
TRIGONOMETRY
Trigonometric Ratios
Using Pythagoras thmAB2+BC2=AC2
00300450600900
sin0
1
cos10
tan01
1st Quadrant angles
2nd Quadrant angles
sin=positive(+)
cos=negative(-)
tan=negative(-)3rd Quadrant angles
sin= negative (-)
cos=negative (-)
tan= positive (+)4th Quadrant angles
sin= negative (-)
cos= positive (+)
tan= negative (-)angles are measured in degree and radians.
1800 = radians ( c )
10 = radian
similarly
Write, Reciprocal trigonometric ratios
I
II
Identities (Questions)
A. Simplify
B. Proof the following
Answers
A.
B. Proove
Find the positive angle1.
2.
3.
4.
5.
6.
7.
8.
Bearings - Introduction
1. Draw a line, and show the North direction.
2. Use clockwise sense to measure the angles from the North.
3. Write the bearing in 3 digits.
Ex:
1. Point X is 0500 bearing from Y.
2. Point B is 1500 bearing of from A.
Ex
1. Car A is 0300 bearings from Car B Draw the diagram.
Find the bearing of A from B.
2. A Car in morning 15 km North direction and morning another
bearing of 0900. Find the bearing of the Car from the initial
point.Ans
1.
Bearing of B from A is 2100
2.
Bearing is 0300 from the starting point.
Angle of elevation
An observer in the ground and the object is in above him, then
the angle between lines, horizontal and object is called angle of
elevation
is called angle of elevation.
Angle of depression
is angle of depression
Question
1. A man observer a vertical tree, distance between the man and
bottom or the tree is height of the tree is 50m. Find the angle of
elevation to the top of the tree?
2. A man sit on the top of the building and observer in the
ground object. Angle of the depression is 600 and the height of the
building is 50m. Find the distance between building and ground.
3.
Observer in at D. Angle of depression of the bottom of the
building is 450 angle of the elevation of the top the building is
600 If the height of AD=, Find the height of the building BC
Ans
Sine and Cosine RuleThe solution for an oblique triangle can be
done with the application of the Law of Sine and Law of Cosine,
simply called the Sine and Cosine Rules. An oblique triangle, as we
all know, is a triangle with no right angle. It is a triangle whose
angles are all acute or a triangle with one obtuse angle.
The two general forms of an oblique triangle are as shown:
Sine Rule (The Law of Sine)
The Sine Rule is used in the following cases:
CASE 1: Given two angles and one side (AAS or ASA)
CASE 2: Given two sides and a non-included angle (SSA)
The Sine Rule states that the sides of a triangle are
proportional to the sines of the opposite angles. In symbols,
Case 2: SSA or The Ambiguous Case
In this case, there may be two triangles, one triangle, or no
triangle with the given properties. For this reason, it is
sometimes called the ambiguous case. Thus, we need to examine the
possibility of no solution, one or two solutions.
Cosine Rule (The Law of Cosine)The Cosine Rule is used in the
following cases:
1. Given two sides and an included angle (SAS)
2. Given three sides (SSS)
The Cosine Rule states that the square of the length of any side
of a triangle equals the sum of the squares of the length of the
other sides minus twice their product multiplied by the cosine of
their included angle. In symbols:
Find the all the missing sides in this triangle, then work out
its area
1)
Fill in all the missing angles in these triangles
2a)
b)
3a) Find the missing length x b) Find the missing angle x
4)
5)
Graphical Method of Solution of a Linear Programming Problem
So far we have learnt how to construct a mathematical model for
a linear programming problem. If we can find the values of the
decision variables x1, x2, x3, ..... xn, which can optimize
(maximize or minimize) the objective function Z, then we say that
these values of xi are the optimal solution of the Linear Program
(LP).
a) The determination of the solution space that defines the
feasible solution. Note that the set of values of the variable x1,
x2, x3,....xn which satisfy all the constraints and also the
non-negative conditions is called the feasible solution of the
LP.
b) The determination of the optimal solution from the feasible
region.
a) To determine the feasible solution of an LP, we have the
following steps.
Step 1: Since the two decision variable x and y are
non-negative, consider only the first quadrant of xy-coordinate
plane
Draw the line ax + by = c (1)
For each constraint,
the line (1) divides the first quadrant in to two regions say R1
and R2, suppose (x1, 0) is a point in R1. If this point satisfies
the in equation ax + by c or ( c), then shade the region R1. If
(x1, 0) does not satisfy the inequality, shade the region R2.
Step 3: Corresponding to each constant, we obtain a shaded
region. The intersection of all these shaded regions is the
feasible region or feasible solution of the LP.
Let us find the feasible solution for the problem of a
decorative item dealer whose LPP is to maximize profit
function.
Z = 50x + 18y (1)
Subject to the constraints
Step 1: Since x0, y0, we consider only the first quadrant of the
xy - plane
Step 2: We draw straight lines for the equation
2x+ y = 100 (2)
x + y = 80
To determine two points on the straight line 2x + y = 100
Put y = 0, 2x = 100
x = 50
(50, 0) is a point on the line (2)
put x = 0 in (2), y =100
(0, 100) is the other point on the line (2)
Plotting these two points on the graph paper draw the line which
represent the line 2x + y =100.
This line divides the 1st quadrant into two regions, say R1 and
R2. Choose a point say (1, 0) in R1. (1, 0) satisfy the inequality
2x + y 100. Therefore R1 is the required region for the constraint
2x + y 100.
Similarly draw the straight line x + y = 80 by joining the point
(0, 80) and (80, 0). Find the required region say R1', for the
constraint x + y 80.
The intersection of both the region R1 and R1' is the feasible
solution of the LPP. Therefore every point in the shaded region
OABC is a feasible solution of the LPP, since this point satisfies
all the constraints including the non-negative constraints.
b) There are two techniques to find the optimal solution of an
LPP.
Corner Point Method
The optimal solution to a LPP, if it exists, occurs at the
corners of the feasible region.
The method includes the following steps
Step 1: Find the feasible region of the LLP.
Step 2: Find the co-ordinates of each vertex of the feasible
region.
These co-ordinates can be obtained from the graph or by solving
the equation of the lines.
Step 3: At each vertex (corner point) compute the value of the
objective function.
Step 4: Identify the corner point at which the value of the
objective function is maximum (or minimum depending on the LP)
The co-ordinates of this vertex is the optimal solution and the
value of Z is the optimal value
Example: Find the optimal solution in the above problem of
decorative item dealer whose objective function is Z = 50x +
18y.
In the graph, the corners of the feasible region are
O (0, 0), A (0, 80), B(20, 60), C(50, 0)
At (0, 0) Z = 0
At (0, 80) Z = 50 (0) + 18(80)
= 1440
At (20, 60), Z = 50 (20) +18 (60)
= 1000 + 1080 = Rs.2080
At (50, 0) Z = 50 (50 )+ 18 (0)
= 2500.
Since our object is to maximize Z and Z has maximum at (50, 0)
the optimal solution is x = 50 and y = 0.
The optimal value is 2500.
If an LPP has many constraints, then it may be long and tedious
to find all the corners of the feasible region. There is another
alternate and more general method to find the optimal solution of
an LP, known as 'ISO profit or ISO cost method'
ISO- PROFIT (OR ISO-COST)
Method of Solving Linear Programming Problems
Suppose the LPP is to
Optimize Z = ax + by subject to the constraints
This method of optimization involves the following method.
Step 1: Draw the half planes of all the constraints
Step 2: Shade the intersection of all the half planes which is
the feasible region.
Step 3: Since the objective function is Z = ax + by, draw a
dotted line for the equation ax + by = k, where k is any constant.
Sometimes it is convenient to take k as the LCM of a and b.
Step 4: To maximise Z draw a line parallel to ax + by = k and
farthest from the origin. This line should contain at least one
point of the feasible region. Find the coordinates of this point by
solving the equations of the lines on which it lies.
To minimise Z draw a line parallel to ax + by = k and nearest to
the origin. This line should contain at least one point of the
feasible region. Find the co-ordinates of this point by solving the
equation of the line on which it lies.
Step 5: If (x1, y1) is the point found in step 4, then
x = x1, y = y1, is the optimal solution of the LPP and
Z = ax1 + by1 is the optimal value.
The above method of solving an LPP is more clear with the
following example.
Example: Solve the following LPP graphically using ISO- profit
method.
maximize Z =100 + 100y.
Subject to the constraints
Suggested answer:
since x0, y0, consider only the first quadrant of the plane
graph the following straight lines on a graph paper
10x + 5y = 80 or 2x+y =16
6x + 6y = 66 or x +y =11
4x+ 8y = 24 or x+ 2y = 6
5x + 6y = 90
Identify all the half planes of the constraints. The
intersection of all these half planes is the feasible region as
shown in the figure.
Give a constant value 600 to Z in the objective function, then
we have an equation of the line
120x + 100y = 600 (1)
or 6x + 5y = 30 (Dividing both sides by 20)
P1Q1 is the line corresponding to the equation 6x + 5y = 30. We
give a constant 1200 to Z then the P2Q2 represents the line.
120x + 100y = 1200
6x + 5y = 60
P2Q2 is a line parallel to P1Q1 and has one point 'M' which
belongs to feasible region and farthest from the origin. If we take
any line P3Q3 parallel to P2Q2 away from the origin, it does not
touch any point of the feasible region.
The co-ordinates of the point M can be obtained by solving the
equation 2x + y = 16
x + y =11 which give
x = 5 and y = 6
The optimal solution for the objective function is x = 5 and y =
6
The optimal value of Z
120 (5) + 100 (6) = 600 + 600
= 1200
Statistics and probability
Definitions:
Event Any collection of results or outcomes of a procedure
Simple Event An outcome or event that cannot be broken down into
simpler components
Sample Space The collection of all simple events that could
result from a procedure.
Complement The complement of an event A consists of all outcomes
where A does not occur.
Notations:
P this means probability
A, B, C, these stand for events
Ac, this means the complement of A
P(A) this means the probability that A occurs
S this is used to denote the sample space
Example:
Q: Raj rolls a 6-sided die. Then the sample space of this is as
followsA: S = {1, 2, 3, 4, 5, 6}Q: Sue measures how many coin flips
it takes to get 3 heads. What is the sample space of this
procedure?
A: S = {3, 4, 5,} = {all integers > 2}Q: Fred sees what
proportion of cars on his block are SUVs. What is the sample space
of this procedure?
A: S = {any real number between 0 and 1} = [0,1]
Since events come in a lot of different ways, there are 3
general approaches to finding the probabilities for events. The
method that is most useful depends on the situation.
Approach 1: Relative Frequency Approximation
For procedures that can be repeated over and over again, we can
estimate the probability of an event A by using the following:
From theoretical arguments (see Law of Large Numbers, p.141), it
turns out that this value p will get closer to P(A) as the number
of trials gets larger.Approach 2: Classical Approach
For procedures with equally likely outcomes (e.g. rolling a die,
flipping a coin, etc.), we can find P(A) directly, by
computing:
Approach 3: Subjective Probability
For procedures that cannot be repeated, and do not have equally
likely outcomes, the true probability of an event is usually not
able to be determined. In situations like this, we can estimate the
probability using our knowledge and experience of the subject. For
instance, we could ask What is the probability that the Columbus
Blue Jackets will win the Stanley Cup this year? No one knows the
true probability, but people who know a lot about hockey could give
a ballpark figure.Examples:
A situation where Approach 1 is used is in baseball. If we want
to know the probability that a player will get a hit when they go
up to bat, we cannot use Approach #2 because the outcomes are not
equally likely. We could use Approach #3, but that would be
subjective. However, by dividing the number of hits by the number
of at-bats gives the batting average, which is an estimate of the
true probability of getting a hit.
A situation where Approach 2 is used is something like rolling a
die. Each face is equally likely to turn up, so we can find
probabilities using this approach. Lets say A is the event of
rolling an even number. What is P(A)?
Another Example
Q: Joe flips one coin 3 times and records the 3 outcomes. What
is the sample space?
A: S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}
Q: What is the probability of getting 1 or 2 heads?
A: Since all outcomes are equally likely (we assume the coin is
fair), we can use Approach #2.
Q: What is the probability of the complement of the previous
event? (i.e. P(not 1 or 2 heads))
A: There are 2 ways this can happen (HHH or TTT), so it will
be
Section 4-3: Addition RuleCompound Event An event that is
comprised of two or more simple events.
Generally, compound events are written in terms of their simple
events.
For example, the event It will rain or snow today could be
written as A or B, where A is the event that it rains today and B
is the event that it snows today. So this event would happen if it
rained today, snowed today, or both.
Another type of event is of the form A and B, in which the event
only occurs if both A and B occur.
For example, the event It is at least 70 degrees and sunny
outside could be written as A and B, where A is the event that it
is at least 70 degrees outside, and B is the event that it is
sunny.The Formal Addition RuleTo see how this rule is derived, lets
examine a Venn Diagram. The area within each circle corresponds to
the probability of that event occurring. Where the two circles
overlap (dark grey), both A and B occur. However the area, say, in
circle A that does not overlap B (grey) would be when A occurs but
B does not. The area outside of both circles (light grey)
corresponds to neither A nor B occurring.
How would we find P(A or B) then? We want the area within the
two circles (the grey and dark grey areas) because thats where A
happens, B happens, or they both happen. What we could do is add
together the area in circle A, and the area of circle B. The
problem is that we could the overlapping area (dark grey) twice.
That means we need to subtract it. Using the fact that the area in
circle A is P(A), the area in circle B is P(B), and the overlap is
P(A and B), we get the formal addition rule:
P(A or B) = P(A) + P(B) P(A and B)
This rule works for any events A and B. Anytime you know three
of the four quantities in the equation, you can solve for the
fourth.
Disjoint Events
Disjoint Events are events that cannot both happen at the same
time. For example, let A be the event that a traffic light is
green, and B be the event that the traffic light is red. The event
A and B cannot happen, because traffic lights are never green and
red at the same time. If two events are disjoint, then P(A and B) =
0. Disjoint events are also often called mutually exclusive
events.Addition Rule for Disjoint Events
Using the fact that P(A and B) = 0 for disjoint events, we can
rewrite the formal addition rule as:P(A or B) = P(A) + P(B)
Complementary EventsRecall from the previous section that for an
event A, its complement Ac is the event that A does not occur.
Since Ac only happens when A does not, and vice versa, P(A and Ac)
= 0. In other words, A and Ac are disjoint. Therefore, P(A or Ac) =
P(A) + P(Ac), by the addition rule for disjoint events. But what is
P(A or Ac)? This means the probability that either A happens, or A
does not happen. This probability is 1, since something has to
happen, whether it is A or not. Therefore, we have a trio of
equivalent formulas:
P(A) + P(Ac) = 1
P(Ac) = 1 P(A)
P(A) = 1 P(Ac)
Examples: For the following questions, imagine we are drawing
one card from a deck of 52 cards.
Q: What is the probability of drawing a queen?
A: Using Approach 2 from the previous section, and letting A be
the event in question,
Q: What is the probability of drawing a diamond?
A: Using Approach 2 from the previous section, and letting B be
the event in question,
Q: What is the probability of drawing a queen of diamonds?
A: This is the event A and B, and we can use Approach 2
again:
Q: What is the probability of drawing a queen or a diamond?
A: We could could up the total number of cards that fit this
bill (13 diamonds one of which is a queen and the other 3 queens =
16
possible cards out of 52), or we can just use the formal
addition rule:
Q: What is the probability of drawing a number card? (Aces
included)
A: Lets call this event C. There are 4 of each number 1 to 10,
for a total of 40 out of 52 cards.
Q: What is the probability of drawing a number card or a
queen?
A: Now we can use the addition rule for disjoint events, since A
and C cant happen at the same time.
Section 4-4: Multiplication Rule: The BasicsNow that we can find
A or B probabilities, we focus on how to find A and B
probabilities. Intuitively, for A and B to happen, we need two
things to take place:
i. A needs to happen
ii. Given that A happened, B needs to happen
This leads us to
The Formal Multiplication RuleP(A and B) = P(A)P(B|A)
Here, P(B|A) is what is called a conditional probability. It
stands for the probability that B happens given A already happened.
(The vertical bar means given) Since A and B is the same as B and
A, we can also write the formula as:P(A and B) = P(B)P(A|B)
Which form you use depends on what information is available.
Example: Students can take a standardized test at three test
centers A, B, and C. Suppose that after the most recent test, 500
students went to A, 200 went to B, and 300 went to C. Furthermore,
the proportion of students who passed the exam were 50%, 80%, and
75%, respectively.
Q: What is the probability that a randomly selected student took
the test at center B?
A: There are a total of 1000 students, and 200 went to B. Thus
P(B) = 200/1000 = 0.20.
Q: What is the probability that a student who took the test at B
passed the exam?
A: Now, we want to find the probability of passing given that
the student took the test at B. We are told in the problem that 80%
of
the students at center B passed. Thus we have: P(Pass | B) =
0.80.
Q: What is the probability that a student both took the test at
B and passed?
A: Using the Multiplication Rule, P(Pass and B) = P(B)P(Pass |
B) = (0.20)(0.80) = 0.16.
Independent Events
Two events are called independent if the occurrence of one does
not affect the chances of the other one occurring. Statistically,
what this means is that A and B independent ( P(A | B) = P(A). In
other words, the probability of A happening given that B happened
is just the same as if we didnt know whether B happened (because
the occurrence of B has no effect on the occurrence of A).
Note: if A and B are disjoint, then we know that only one can
occur. Thus, knowing that B happened tells you that A definitely
did not happen, and we have P(A | B) P(A). Thus disjoint events are
never independent events.Multiplication Rule for Independent
Events
Using the fact that P(A | B) = P(A) for independent events, we
see that the formal multiplication rule turns into:P(A and B) =
P(A)P(B)
The Law of Total Probability
This rule is very intuitive, and is useful for finding
probabilities of events. To explain it, we refer to the test center
example above.Q: What is the probability of a randomly selected
student passing the exam?
A: From the information above, we can find out (similar to the
previous example) that:
P(A) = 0.50
P(B) = 0.20
P(C) = 0.30
P(Pass | A) = 0.50 P(Pass | B) = 0.80 P(Pass | C) = 0.75
We want to find P(Pass). What are the possible scenarios where a
student passes the exam? They could take the test at A and
pass, they could take it at B and pass, or they could take it at
C and pass.
So P(Pass) = P(A and Pass OR B and Pass OR C and Pass). But each
of those 3 scenarios are disjoint, because a student cant take
the test at more than one center. Therefore, by the addition
rule we can add these probabilities as follows:
P(Pass) = P(A and Pass) + P(B and Pass) + P(C and Pass)
Then, by the multiplication rule, we can find all of these
probabilities:
P(A and Pass) = P(A)P(Pass | A) = (0.50)(0.50) = 0.25
P(B and Pass) = 0.16
P(C and Pass) = P(C)P(Pass | C) = (0.30)(0.75) = 0.225
Thus P(Pass) = 0.25 + 0.16 + 0.225 = 0.635
In general, if you have disjoint events B1, B2, , Bn that
represent every possible outcome of a procedure, then you can
write:P(A) = P(A and B1) + P(A and B2) + + P(A and Bn) =
The most common way to use this rule is if you have two events A
and B, then:P(A) = P(A and B) + P(A and Bc)
Examples:
A telemarketing company makes phone calls to potential customers
all across the U.S. For each call, the probability of the customer
answering the phone is 0.20. For the next couple of questions,
assume calls are independent of each other.Q: Lets say the company
makes 10 phone calls. What is the probability that all of them are
answered?
A: P(10 calls answered) = P(1st call answered AND 2nd call
answered AND AND 10th call answered)
= P(1st call answered)P(2nd call answered) P(10th call answered)
( (Mult. Rule for Independent Events)
= (0.20)(0.20) (0.20) = 0.2010 = 0.0000001024
Not very likely, is it?Q: Lets say the company makes 2 phone
calls. What is the probability that exactly one of them is
answered?
A: First, note that from the Complement Rule, the probability
that a call is not answered is 1 0.20 = 0.80. Thus:
P(1 call answered) = P(1st call answered and 2nd call not
answered OR 1st call not answered and 2nd call answered)
= P(1st call answered and 2nd call not answered) + P(1st call
not answered and 2nd call answered) ( (Add. Rule)
= P(1st call answered)P(2nd call not answered) + P(1st call not
answered)P(2nd call answered) ( (Mult. Rule)
= (0.20)(0.80) + (0.80)(0.20) = 0.16 + 0.16 = 0.32
Q: Now suppose that if a customer answers the phone, their
chance of buying the product is 0.10. (Note that if they do not
answer the
phone, their chance of buying it is 0). What is the overall
chance of a telemarketer selling the product when they call a
home?
A: In the question, we are told P(Buying | Call Answered) = 0.10
and P(Buying | Not Answered) = 0. We want to find P(Buying).
From the Law of Total Probability,
P(Buying) = P(Buying and Call Answered) + P(Buying and Not
Answered)
= P(Call Answered)P(Buying | Call Answered) + P(Not
Answered)P(Buying | Not Answered) ( (Mult. Rule)
= (0.20)(0.10) + (0.80)(0) = 0.02.
Note: If this seemed complicated, try just replaced Buying with
A, Call Answered with B, and Not Answered with Bc. Then
the calculations above follow directly from the Law of Total
Probability written before.I. DESCRIPTIVE STATISTICS
There are certain statistics that can help us condense large
amounts of data into a few numbers that are easier to comprehend.
These statistics are called descriptive statistics and they are
used to convey summary information about any set of numbers or
data. By using descriptive statistics we can get a quick estimate
of what our data (e.g., students' Mathematics scores; people's
lifes expectation)II. MEASURES OF CENTRAL TENDENCYMeasures of
central tendency are statistics that identify the center of a
distribution of scores. The most common measures of central
tendency are the mode, the median, and the mean. To make
calculations easier, instead of discussing full data sets .A. The
modeThe mode is a statistic that identifies the most frequently
occurring score in a distribution. For example, in the following
distribution, the mode is 6:
4 6 7 8 6 3 5 9 6
Mode = 6 (it occurs 3 times)
B. The medianThe median is a statistic that identifies the
middle score in a distribution. For example, in the distribution
above, the median is also 6. To determine this, you must first
rearrange the numbers and order them from lowest to highest:
3 4 5 6 6 6 7 8 9
Median = 6
In a distribution with an odd number of scores, such as the one
above, the median is simply the middle score. In this case it is 6
(a number which has 4 scores to the left of it and 4 scores to the
right). In a distribution with an even number of scores, the median
is found by taking the average of the two middle scores. For
example, in the following distribution, the median is 5.5:
2 3 4 5 6 7 8 9
Median = 6 + 5 = 5.5
2
C. The meanThe mean, more commonly known as the average, is the
most common measure of central tendency in statistics. It is
determined by dividing the sum of the scores (N) in a distribution
by the total number of scores (N) in that distribution. For
example, in the distribution below, the mean is 6:
4 6 7 8 6 3 5 9 6
Mean = N = 4+6+7+8+6+3+5+9+6 = 6
N 9
In the above distribution, the mode, median, and mean were all
the same value (6). The number six thus represents the center of
that distribution no matter how we measure it. However, the mode,
median, and mean are rarely the same for a given distribution. Here
are some examples for you to work out on your own. You can look up
the answers on the last page of this handout.
ModeMedianMeanData Set 1:4 6 7 2 7 9 3 7__________________
Data Set 2:3 8 5 7 4 4 7 4 __________________
Data Set 3:8 9 7 8 1 5 6 8__________________
III. MEASURES OF DISPERSION
A. VarianceMeasures of central tendency provide useful
information, but they do not always accurately represent the entire
distribution. In addition to a measure of central tendency, it is
usually helpful to know something about the extent to which the
scores in a given distribution differ (or deviate) from the mean.
Another way to state this issue is to ask the question, "How much
variation is there in a given set of scores?" Obviously, if all the
scores in a set of data are the same (e.g., if every professor goes
off on 6 boring tangents/irrelevant anecdotes during lecture) then
there is no variation. This rarely happens in research (or in real
life), however; people vary in their characteristics, behaviors,
attitudes, and emotions. If everyone was the same, then the science
of Psychology could proceed by merely studying one person. How
dull. Variety, as they say, is the spice of life.
The concept of "variation" in a set of data is illustrated
easily by the following example. Given your tremendous popularity
you receive invitations to 3 parties that all fall on the same
night. All you are told is that the mean age of the 8 guests at
each party will be 22 years old. Given this information, which
party do you wish to attend? What other information might be
helpful in making your decision?
The ages of those attending the 3 different parties are as
listed in the following table:
PersonParty 1Party 2Party 3
#12034 2
#22393
#322102
#424413
#522384
#62473
#7203178
#821681
Mean:222222
Party 1 is going to be a night of who knows what with some young
adults. Party 2 is going to be a dinner party with parents and
their children. Party 3 is going to be Little Billy's third
birthday party (hosted by his Grandma and Grandpa Jones).
Therefore, although the mean age for each party is the same, these
three data sets are very different from one another. Determining
the mode and median for each party would give us a more complete
picture, but it would be very helpful to have some statistic that
tells us how much the ages of the guests at each party deviated or
"varied" from the mean. Other than merely forming a subjective
impression of our data, how do we determine this "variability" in
scores?
The thought that immediately comes to mind is: " Let's simply
add up how much each score differs from the mean." To do this, we
could take the mean, which is 22, and then subtract the mean from
each individual score. So, for example, for Person 1 at Party 1 we
would have 20 - 22 = -2. For Person 2 and Party 1 we would have 23
- 22 = 1.
At this point, it will facilitate your understanding of this
concept if you do the following. For Party 1, continue with these
calculations and subtract the mean from each of the 8 scores. Then
total up these "deviation scores," being sure to pay attention to
the plus and minus signs. After you get your total, then perform
the same calculations for Party 2 and Party 3. Move on to the
paragraph below after you have finished these calculations.
Now you see the problem. For any set of data, if we simply
subtract the mean from each individual score and add these
deviation scores up, we will get a total of zero. Therefore, the
average deviation will also come out to be zero (i.e., the sum of
deviation scores divided by the number of scores, or 0/8 = 0 in our
example.) Thus, we cannot determine the "average deviation" in this
way. Astoundingly, to the great benefit of humankind, there are two
statistics that provide very meaningful information about
variability: the variance and the standard deviation. These are
called measures of dispersion because they quantify the degree to
which a set of scores, overall, differs from the mean.
The word "variance" -- because it sounds foreign, mystical, or
technical -- sometimes produces a fear/anxiety response in
students; some students sweat profusely, while others experience
increased heart rate, nausea, or light-headedness. In extreme cases
some students have heard voices telling them "Drop this course!
Become an English major!" In case you are having any of the above
reactions to the sight or sound of the word "variance," then do the
following two things: 1) remember that Statistics are our friends,
and 2) think of the word "variance" as a synonym for the word
"variation."
Variance is defined as the average of the squared deviations
about the mean. This is represented mathematically by the following
equation, where X represents a single score within a distribution
and N represents the total number of scores:
Variance = (X-Mean)2
N
The variance is arrived at by performing the following
computations in the listed order:
Step 1. find the mean (sum of the scores divided by the number
of scores)
Step 2.compute the deviation scores (the difference between an
individual score and the mean)
Step 3.square each of the deviation scores
Step 4.divide the sum of the deviation scores by the number of
scores
For example, the variance for Party 1 would be computed as
follows:
PersonAgeMeanDeviation Score
(Age-Mean)Deviation Squared
#12022-24
#2232211
#3222200
#4242224
#5222200
#6242224
#72022-24
#82122-11
Average:2222018/8 = 2.25
The answer to step 4 above, then, is 18 divided by 8, or 2.25.
This is the variance of ages at Party 1.
B. Standard DeviationThe standard deviation is the most commonly
reported measure of dispersion. It is simply the square root of the
variance. In the case of Party 1, the standard deviation is the
square root of 2.25, or 1.5. Why take the square root of the
variance? Well, remember that when we computed the variance we had
to square each deviation score before adding these scores up. (As
we saw earlier, if you add the deviation scores without squaring
them, then you will always get a total of zero.) Therefore, because
we squared the deviation scores to get the variance, we now take
the square root of the variance in order to convert our numbers
back to their original units of measurement. Thus, for Party 1 the
mean is 22 years, the variance is 2.25, and the standard deviation
is 1.5 years. At this point I strongly encouraged you to calculate
the variance and standard deviation for Party 2 and Party 3 to
check your understanding of these statistics. The correct answers
are on the last page of the handout.)C. The RangeThe range is a
relatively crude measure of dispersion. It represents the highest
score in a distribution minus the lowest score. It is a crude
measure of dispersion because the composition of other scores in
the distribution (other than the high and low scores) have no
effect on the range. For example, each of the 3 distributions below
has a range of 8, despite the fact that the distributions are quite
different. Each distribution, however, would have a different value
for the variance (and hence, for the standard deviation).
10 7 6 5 4 3 2 Range = 10 - 2 = 8.
10 10 10 9 9 9 2Range = 10 - 2 = 8
10 4 4 3 2 2 2
Range = 10 - 2 = 8.
Variance and the standard deviation are more sensitive measures
of dispersion, because they are influenced by each particular score
in the distribution. Change even one score, and you'll change the
values of the variance and standard deviation.
D. Additional Comments: This Handout Versus The TextYou should
be aware of the following differences between the formula used in
this handout versus the one used in your textbook. On page 423 of
your textbook, a formula is given for calculating the standard
deviation. This formula calculates the standard deviation by
dividing by n-1. The examples in this handout divide by n. Remember
that n corresponds to the number of people attending each party
(8). Why the difference and which divisor should you use? The
answer depends on what you want to use the standard deviation for.
Consider the following two examples.
Example 1. Assume that I wish to know how much variability in
age there is for students in this class. To gather this
information, I have the 25 students in this class complete an
anonymous questionnaire on which they indicate their age. If I am
only concerned with the variability of age for this specific group
of people (and I am not interested in trying to generalize my
findings to other students), then I would divide by n. In this
instance, the standard deviation I am calculating would be
considered a descriptive statistic, since I only want to describe
the variability of this particular set of numbers. (Aside: Computer
programs and calculators can both calculate this number for us very
quickly so why do I bother to make you calculate it by hand? No, I
dont enjoy inflicting pain and anguish on my students (Ok maybe I
do just a little!). The real reason for asking you to become
familiar with this way of calculating the standard deviation is
that it is useful in helping you gain a better conceptual
understanding of measures of dispersion.)
Example 2. Alternatively, assume that I wish to know the
variability in age for all students at Shoreline. In general, what
kind of spread is there in age for students attending Shoreline? It
would be possible for me to contact every single student at
Shoreline, ask how old he or she is, and calculate the standard
deviation of this rather large set of numbers. But this would be
very time consuming (and besides, Im also very lazy!). What I could
do instead is to determine the variability in age of a small sample
of Shoreline students and then use this number to make an educated
guess about the variability in age of the entire population of
Shoreline students. So assume that I randomly sample 25 Shoreline
students about their ages. Since I am using my sample to try to
infer the variability in age for the whole population, when I
calculate the standard deviation in this example, I divide by n-1
instead of n. In this second case, the standard deviation I am
calculating would be considered an inferential statistic, since I
am using the standard deviation of my sample to make an inference
about the overall standard deviation of the entire population of
students. For the purposes of this class, when you are asked to
calculate the standard deviation, please divide by n.IV.
Z-SCORES
Imagine that you and a friend are somewhat competitive and you
want to compare how you each did on your last psychology exam.
Unfortunately, you are taking different psychology classes (you are
in Psychology 209 while your friend is in Psychology 204). What
information would you need to find out who did better? You would
probably start by comparing the points you each got correct on your
respective exams. Lets say that you got 40 points correct and your
friend got 20 points correct. Looks like you win, right? Not
necessarily. You also need to know how many total points were on
each exam. If your test had 50 possible points, and your friends
test had 25, you are still tied (40/50 = 20/25 = 80%). The next
thing you might want to know is what were the respective class
means on each of the tests. If I tell you that the mean for your
test was 30 (out of 50) while the mean for your friends test was 15
(out of 25), can you now tell who did better? It seems like you may
have done better since you scored 10 points higher than the mean
but your friend only scored 5 points higher than the mean. But
remember your test had twice as many total points on it as your
friends test (50 vs. 25), so the difference of each of your scores
from the means still seems roughly equivalent.
As you can see, making a comparison in this situation is a bit
difficult. But thanks to the standard deviation, there is still a
way to make a comparison. If you also know the standard deviation
of the scores in each of the classes, this can allow you and your
friend settle your dispute. If I tell you that the standard
deviation for both classes is 10 points, who did better on their
test relative to the rest of their class? Fortunately for you, it
looks like you are the winner. Your scores is a full standard
deviation above the mean ((40-30)/10 = 1) but your friend was only
a half of a standard deviation above the mean in his or her class
((20-15)/10 = 0.5). Thus, using standard deviations, we can see
that you outscored a larger percentage of your class than your
friend did of his/her class.
What we have just done in this example is to calculate something
called a z-score. A z-score counts the difference between an
individual score and the mean in terms of standard deviations. For
instance, we could say that you scored 10 points above the mean on
your test, or equivalently, that you scored 1.0 standard deviation
above the mean. This 1.0 represents your z-score for this test.
Similarly, the z-score for your friend was 0.5 standard deviation
above the mean. He or she scored one half of a standard deviation
better than the mean. It turns out that z-scores are a convenient
way to compare scores from different groups that have used
different numerical scales, as was the case in our example above
(the scale used on your test was from 0-50 while the scale used in
your friends class was from 0-25). By converting our numbers to
z-scores, this allows us to make meaningful comparisons between the
numbers in each of these groups, something we couldnt do initially.
Z-scores also come in handy when trying to understand our next
topic, the normal distribution.
V. THE NORMAL DISTRIBUTION
One of the most important concepts in statistics is a curve
called the "normal distribution" or "normal curve." The normal
distribution is an essential part of inferential statistics. While
we will not be covering inferential statistics in this course, I
would still like to introduce you to the normal curve and some of
its unique properties. By learning a little about the normal curve
now, you will be ahead of the game when you take your required
statistics course.
The normal distribution is a bell shaped curved that is
symmetric (that is, on either side of the mean it looks the same).
A typical normal distribution is shown in Figure 1. The x-axis
represents the all possible values or scores on some characteristic
that we are interested in (e.g., SAT scores, reaction times, etc).
The y-axis represents the frequency or percentage of each score on
the x-axis occurs. Because it is symmetric, the mean, median, and
mode for the normal distribution are all equal to one another. The
main reason that the normal curve is useful is because there are
many variables out in the real world which distribute themselves
normally (i.e. they approximate a normal distribution). For
instance, IQ is one variable that is normally distributed; if we
were to look at the distribution of IQs for all adults, this
distribution would look pretty much like the theoretical normal
curve pictured below.
Figure 1.
Another useful feature of the normal distribution concerns the
standard deviation. Specifically, in any normally distributed set
of numbers, the standard deviation can be used to divide the
distribution into segments which contain fixed percentages of
scores. For instance, we know that in a normal distribution, 34% of
the scores fall between the mean and one standard deviation above
the mean (dont worry about how this percentage is calculated; it is
simply a product of the mathematics of the normal distribution).
Since the normal curve is symmetric, we also know that 34% of the
scores will fall between the mean and one standard deviation below
the mean. We can put these two pieces of information together to
deduce that 68% of the scores in a normal distribution fall within
one (plus or minus) standard deviation of the mean. Similar
percentages also exist for segments further away from the mean and
these are also shown in Figure 1.
To see why are these percentages useful, consider the following
example. Assume that you took an IQ test and received a score of
130. Looking at this score, you assume that you did well but, being
the competitive person you are, you want to know precisely how many
people did as well or better than you. If we tell you that the
average IQ score is 100 and that the standard deviation is 15 (both
of which are true), you could determine, using the percentages
listed in Figure 1, that only 2.5% of the people scored 130 or
higher (Wow! Youre pretty smart!). Alternatively, you could say
that you scored better than 98.5% of the people. In either case,
the percentages generated by the normal distribution allow you to
more accurately determine how well you did.
Another reason that the normal curve is useful concerns
inferential statistics, a topic that we will discuss briefly later
in this course. Inferential statistics will be covered in greater
detail when you take statistics. I am briefly covering this
information now in hopes that it will make it easier to assimilate
when you come across it again in the future.Metric and Imperial
unitsConversion Tables
The following tables show some of the more common measures, and
the conversion between larger and smaller units. The left hand
tables show the units for the metric system while the right hand
tables show for the Imperial measurement systems.
Metric lengthImperial Length
10 millimeters=1 centimeter12 inches=1 foot
10 centimeters =1 decimeter3 feet=1 yard
10 decimeters=1 meter22 yards=1 chain
10 meters=1 decameter10 chains=1 furlong
10 decameters=1 hectometer8 furlongs=1 mile (5280 feet)
10 hectometers=1 kilometer (1000 meters)
Metric areaImperial area
100 square mm=1 square centimeter144 square inches=1 square
foot
10000 square cm =1 square meter9 square feet=1 square yard
100 square m=1 are4840 square yards=1 acre
100 ares=1 hectare640 acres=1 square mile
100 hectares=1 square kilometer
=1000000 square meters
Metric massImperial weight
1000 grams=1 kilogram16 ounces=1 pound
1000 kilograms=1 ton14 pounds=1 stone (UK)
8 stones (UK)=1 hundredweight (UK)
=112 pounds (UK)
100 pounds=1 hundredweight (USA)
20 hundredweight (UK)=1 ton (UK)
=2240 pounds
20 hundredweight (USA)=1 ton (USA)
=2000 pounds
Metric capacityImperial liquid capacity
10 mililiters=1 centilitre2 teaspoons=1 dessertspoon
10 centiliters=1 decilitre3 teaspoons=1 tablespoon
10 deciliters=1 litre2 tablespoons=1 fluid ounce
1000 liters=1 cubic meter5 fluid ounces=1gill
2 gills=1 cup
2 cups=1 pint
=20 fluid ounces
2 pints=1 quart
4 quarts=1 gallon
USA liquid capacity
3 teaspoons=1 tablespoon
2 tablespoons=1 fluid ounce
4 fluid ounces=1 gill
2 gills=1 cup
2 cups=1 pint
16 fluid ounces
2 pints=1 quart
4 quarts=1 gallon
Representing Units
Length
The standard unit of length in the metric system is the meter.
Other units of length and their equivalents in meters are as
follows: 1 millimeter = 0.001 meter1 centimeter = 0.01 meter1
decimeter = 0.1 meter1 kilometer = 1000 meters
We symbolize these lengths as follows:1 millimeter = 1 mm1
centimeter = 1 cm1 meter = 1 m1 decimeter = 1 dm1 kilometer = 1
km
For reference, 1 meter is a little longer than 1 yard or 3 feet.
It is about half the height of a very tall adult. A centimeter is
nearly the diameter of a dime, a little less than half an inch. A
millimeter is about the thickness of a dime.
Volume
The standard unit of volume in the metric system is the liter.
One liter is equal to 1000 cubic centimeters in volume. Other units
of volume and their equivalents in liters are as follows: 1
milliliter = 0.001 liter1 centiliter = 0.01 liter1 deciliter = 0.1
liter1 kiloliter = 1000 liters
From these units, we see that 1000 milliliters equal 1 liter; so
1 milliliter equals 1 cubic centimeter in volume. We symbolize
these volumes as follows:1 milliliter = 1 ml1 centiliter = 1 cl1
deciliter = 1 dl1 liter = 1 l1 kiloliter = 1 kl
For reference, 1 liter is a little more than 1 quart. One
teaspoon equals about 5 milliliters.
Mass
The standard unit of mass in the metric system is the gram.
Other units of mass and their equivalents in grams are as follows:1
milligram = 0.001 gram1 centigram = 0.01 gram1 decigram = 0.1 gram1
kilogram = 1000 grams
We symbolize these masses as follows:1 milligram = 1 mg1
centigram = 1 cg1 decigram = 1 dg1 gram = 1 g1 kilogram = 1 kg
For reference, 1 gram is about the mass of a paper clip. One
kilogram is about the mass of a liter of water.
Time
The following conversions are useful when working with time:1
minute = 60 seconds1 hour = 60 minutes = 3600 seconds1 day = 24
hours1 week = 7 days1 = 365 1/4 days (for the Earth to travel once
around the sThis gives us a total of 52 complete 7 day weeks in
each calendar year, with 1 day left over (or 2 in a leCoordinate
Geometry
I. Distance between two points.
Look at horizontal and vertical distances.
Ex: Distance from ((3, (2) to (5, 4)
= 8 = 6
(AC)2 = 82 + 62
(AC)2 = 100
AC = 10
Distance from (x1, y1) to (x2, y2)
d2 = (x2 ( x1)2 + (y2 ( y1)2
d =
I. Midpoint.
( middle of a segment.
( find the midpoint for the vertical and
the horizontal segment.
( to find the middle use and
. The point (1, 2) is the
midpoint.
Median ( a segment from a vertex of a triangle to the midpoint
of the opposite side.
Mid(A, C) = = (3, 1)
=
I. Slope of a line or segment.
Slope (m) = how steep a line is.
=
= for points (x1, y1) and (x2, y2).
Examples:
1. Find slopes for each of the following.
a.
b.
c.
m =
m =
m =
***The slope of a horizontal line (parallel to the xaxis) is
0.***The slope of a vertical line (parallel to the yaxis) is
undefined.
***Lines rising to the right have a positive slope.***Lines
falling to the right have a negative slope.I. Slope yintercept form
of a line.
yintercept
( point where a relation intersects the yaxis. x = 0
xintercept
( point where the relation intersects the xaxis. y = 0
1. Find the yintercept and slope for the following.
a. 3x + 2y ( 12 = 0
b. y = (5x + 4
two points are (0, 6) and (2, 3) two points are (0, 4) and (3,
(11)
m =
m =
m = and yintercept is (0, 6) m = (5 and yintercept is (0, 4)
From above y = mx + b; m = slope b = yintercept
2. Determine equations of the following lines given slope and
point.
a. m = (2; (0, (3)
b. m = 3; yintercept is 4c. m = ; (0, (6
y = (2x ( 3
y = 3x + 4
y = (
d. m = (5 and same yintercept as 3x ( 4y + 32 = 0
yint is 8y = (5x + 8
3. Determine equations of the following.
a.
m =
b. m = , b =
y =
y =
4. The equation of a line is y = 2x + b. Find b if the line
passes through the point:
a. (2, (6)
b. ((1, (3)
(6 = 2(2) + b
(3 = 2((1) + b
(10 = b
(1 = bI. Parallel and perpendicular lines.
Parallel lines have
Perpendicular lines have
the same slope.
slopes that are negative
reciprocals of each other.
Examples.
1. Find slopes of parallel and perpendicular lines to:
Parallel
Perpendicular
a. m =
b. y =
c. 4x + 5y ( 11 = 0
2. a. Given A (0, 2), B ((3, (4), C (2, (4)b. Given A (2, 3), B
(6, 5), C ((1, 4)
and D ((8, 1), show that . and D (3, 6), is?
Ye
3. Which of the following lines are: i. parallel? ii.
perpendicular?
a. 3x + 4y (24 = 0b. 4x + 3y ( 16 = 0c. 3x ( 4y + 10 = 0d. 6x +
8y + 15 = 0
m =
m =
m =
m =
i. a & d are parallel
ii. b & c are perpendicular
4. Find y in A (3, 8), B ((1, (2), C (4, 1) D (2, y) such
that:
a.
b.
20 = (4y + 4
5y ( 5 = 4
16 = (4y
5y = 9
(4 = y
y =
5. Given , find k.
a.
b.
5k = 60
(2k = 15
k = 12
k = (7.5
6. For what value of k are the lines 3x ( 4y + 4 = 0 and kx + 3y
+ 10 = 0 perpendicular?
m = m =
(3k = (12
k = 4
7. Find the equation of the line through A ((1, 5) which is
perpendicular to 3x ( 2y ( 12 = 0.
(2y = (3x + 12
m =
y =
3y ( 15 = (2x ( 2
m =
2x + 3y ( 13 = 0
8. Find the equation of a line which is parallel to 3x + 2y + 8
= 0 and has a yint of (2.
2y = (3x ( 8
y = mx + b
y =
y =
x0
A
B
C
Hypotenuse
Adjacent
Opposite
EMBED Equation.3
EMBED Equation.3
a
x
450
a
600
a
a
600
a
2a
2a
300 300
1st
4th
3rd
2nd
1st
4th
3rd
2nd
1st
4th
3rd
2nd
1st
4th
3rd
2nd
10 = QUOTE
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
EMBED Equation.3
450
450
300
300
EMBED Equation.DSMT4
300
300
600
600
EMBED Equation.DSMT4
300
300
600
600
EMBED Equation.DSMT4
450
450
600
600
EMBED Equation.DSMT4
500
N
Y
X
500
1500
N
B
A
N
300
300
N
B
A
1800
15km
EMBED Equation.DSMT4
900
Y
N
X
Z
ground
observer
h
object
h
horizontal
observer
object
d
EMBED Equation.DSMT4
man
50m
tree
50
=600
observer
object
x
=600
B
A
observer
D
building
y
x
C
Negative reciprocals
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