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Copyright 2009

A113 Mathematics

Problem 14: Salary

6th

Presentation

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Recap

Mean is the average of a set of numerical values.

Standard deviation

A measure of dispersion of data (how spread outyour data is) from the mean

A smaller standard deviation value will indicate thatmost of the data points are nearer to the mean.

Variance is the square of standard deviation.

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Frequency table

The frequency table shows all possible outcomes of an eventand the number of times each outcome occurs.

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Salary Range ($) Frequency Relative Frequency

1000

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0

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Relative frequency histogram

The distribution of the individual entries is as shown below:

Salary Range ($)

R

elativeFrequency

Mean = 1494

Standard deviation = 289.6

Variance = 83849

Distribution of the individual entries of salary

The distribution showsthat all outcomes have

about the same chanceof occurrence for ourdata.

The mean of thedistribution isapproximately $1494.

The variance andstandard deviation are83849 and 289.6respectively.

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Sample size

Sample size is the number of observations thatconstitute the sample.

For example, when we are calculating the averageof 2 entries, the number of observations is 2.Hence, the sample size will be 2.

Likewise, if we are calculating the average of 30entries, then the sample size will be 30.

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Distribution of the averages of two entries of salary The relative frequencyis the lowest at the twotails of the distribution,and it increases as wemove towards the mean.

The mean of thedistribution isapproximately $1494,i.e. the mean of thedistribution of the

individual entries.

The variance isreduced by a factor of 2,which is the number ofentries in each sample.

Mean = 1501

Number ofentries in a

sample

Standard deviation = 202.3

Variance = 40941( 83849 / 2 )

The distribution of the average of 2 entries is as shown below:

Average of 2 entries

RelativeFrequency

Salary Range ($)

Variance ofindividual

entriesdistribution

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0.05

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Distribution of the averages of ten entries of salary

The distribution of the average of 10 entries is as shown below:

The spread of the

distribution is furtherreduced with the highestfrequency occurring atthe mean.

The mean of thedistribution isapproximately $1494.

The variance is reducedproportionally by thenumber of entries in eachsample.

Standard deviation = 94.0

Variance = 8834

( 83849 / 10 )

Average of 10 entries

RelativeFrequency

Salary Range ($)

Mean = 1499

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Distribution of the averages of thirty entries of salary

The probability ofobtaining an averagethat is approximately

$1494 increases withthe number of entriesin each sample.

The mean isapproximately $1494

and the spread of thedistribution becomeseven smaller.

Mean = 1500

Standard deviation = 52.9

Variance = 2796

( 83849 / 30 )

Average of 30 entries

The distribution of the average of 30 entries is as shown below:

RelativeFrequency

Salary Range ($)

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Central Limit Theorem

The Central Limit Theorem states that if the samplesize is large ( 30),the shape of the histogram of thesample means will resemble a bell-shaped curve,also known as the Normal distribution curve.

NormalDistribution Curve

E.g.:

RelativeFrequency

Salary Range ($)

$1501 $1499 $1500

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Normal distribution

The Normal curve is symmetrical about its mean. It is described by its mean and its standard deviation

(or variance).

The area under the Normal distribution curve represents theprobability of an event occurring where the total area is 1.

Mean

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Normal distribution

The height of a population is normally distributed. It can bemodeled by a Normal distribution curve.

For the example above, there is a 10% chance for a randomlychosen individual to have a height exceeding 172 cm.

Mean172 cm163 cm

90%

10%

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Estimating average

From the discussions made so far, the distribution of theaverage of a reasonably large sample size approaches aNormal distribution and the spread of the data will be verynarrow.

As the sample size gets larger, the average obtained from thissample is less likely to deviate too far from the actual averageof the population.

Thus, to estimate the average of a population, we can use theaverage computed from a randomly selected large sample( 30) of the population.

In conclusion, adopting Peters suggestion is an efficient and

effective way of solving Serenes problem.

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Learning points

Understand the relationship between frequency andrelative frequency

Understand the relationship between relative frequencyand probability

Understand that for large sample size ( 30), the Normal

distribution curve is a fairly accurate representation ofthe distribution of the sample means

Understand that the variance of the distribution ofsample means is inversely proportional to the samplesize

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Discussion

Suppose the distribution of the individual entries is as follows:

Suggest how the distribution of average might look like whensample size is large (30).

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0.25

Mean = 1202.5Standard deviation = 145.7

Variance = 21230

Distribution of the individual entries of salary

Salary Range ($)

RelativeFrequency