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USING CLASSPAD
Consider the shaded area in the diagram below.
Find the area between the function , the normal
to the curve at, the axis and the axis.
NB The normal is the line perpendicular to the tangent.
USING CLASSPAD 2
MATHEMATICS METHODS UNIT 4
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The syllabus contains:
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Random Sampling -
(7 hours)
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Sample proportion -
(4 hours)
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Confidence Intervals for Proportion
(9 hours)
RANDOM NUMBERS 1
Mathematics / Year 8 / Statistics and Probability / Data representation and
interpretation / ACMSP293
Content Description
Explore the variation of means and proportions of random samples drawn from the
same population
Elaborations
using sample properties to predict characteristics of the population
RANDOM NUMBERS 2
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What would be the result
if you purchased 5
“Boost” bars
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Use a simulation to repeat
the purchase 30 times
RANDOM SAMPLING 1
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On classpad
enter
rand (
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What happened
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Define the rand ( function
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If you found 100 random numbers
what would the histogram look like ?
RANDOM SAMPLING 2
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Expected histogram if selecting 100 pseudo random numbers.
RANDOM SAMPLING 3
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Enter 100 pseudo numbers into the calculator
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Draw a histogram of the result
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How did it compare with the expected histogram
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Recalculate the result
RANDOM SAMPLING 4
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Selected 10 pseudo numbers, added them
together and repeated this experiment
30 times, what would you expect the
histogram to look like ?

Central Limit Theorem
RANDOM SAMPLING 5
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On the classpad enter rand ()  8
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What happened ?
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On the classpad enter int (rand ()  8 + 1)
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What happened ?
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Draw a histogram obtained if a single die is rolled 120 times
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How many sixes would you expect if you rolled a single die 120 times ?
SAMPLE PROPORTION
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A sample proportion is the ratio of the number of times a property occurs (e.g
success) in a sample, divided by the number in the sample.
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The sample proportion is denoted by 𝑝 (read as 𝑝 hat)
𝑝 =
●
𝑥
𝑛
(where 𝑥 = number of successes and 𝑛 = sample size)
Example
From a sample of 60 students, 3 had red hair. What is the sample proportion of
red hair for the students ?
𝑝 =
𝑥
𝑛
=
3
= 0.05
60
SAMPLE PROPORTION 2
Sampling Distributions
Example 1
Select a smartie from a box containing 10 smarties, 3 of which are blue. Replace
the smartie and repeat the experiment 10 times. Record the number of blue
smarties for the 10 experiments.
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This is a Bernoulli trial.
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Record the theoretical probabilities on the table.
X
0
1
2
3
4
5
6
7
8
9
10
𝑝
0
10
1
10
2
10
3
10
4
10
5
10
6
10
7
10
8
10
9
10
10
10
P(𝑃 = 𝑝)
SAMPLING PROPORTION 3
Sampling Binomial Distribution
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To find the theoretical probabilities
P (x=n) = 10Cn (0.3)n (0.7)10-n
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Method 1 – Using main
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Method 2 – Using Statistics
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Method 3 – Using graph & table
SAMPLING PROPORTION 4
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Lets do example 1 as simulation
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Repeat the simulation 30 times.
CONFIDENCE INTERVALS 1
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This 𝑝 is one value of a population of all of the possible values in the 𝑃
population – the proportion of blue smarties in every 380 g bag.
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So what is the actual proportion of blue smarties ?
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For large random samples the Central Limit Theorem tells us that the sample
estimate 𝑝 will lie on a approximate normal distribution of the estimates from all
possible samples.
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So we know how many standard deviations from the centre of the normal
distribution encompass 90% of the distribution.
CONFIDENCE INTERVAL 2
Sampling for unknown Population Proportions
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We will consider the proportion of blue smarties in the population of all smarties
produced.
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The 380 g bag contained 352 smarties, of which
43 were blue
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For this sample
𝑝 =
𝑥
𝑛
=
43
352
=
0.12
CONFIDENCE INTERVAL 3
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We can now find the 90% confidence interval
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( 𝑝 )  1.64 
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0.092 ≤ 𝑝 ≤ 0.148
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This answer can be found using the classpad
𝑝
1−𝑝
𝑛
( to 3 decimal places )
CONFIDENCE INTERVALS 4
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What does this mean
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Our result of 𝑝  0.12 could be the worst case scenario or the best case
scenario.
0.92
1.2
1.48
CONFIDENCE INTERVAL 5
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The margin of error = zc 
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Error = 0.028
estimate – error
𝑝
1−𝑝
𝑛
estimate
estimate + error
Confidence interval
0.92
●
1.2
1.48
This provides an interval estimate for the true population value and
we are 90% confident that the population proportion 𝑝 =0.12 lies in
this interval.