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Inverse Problems: Percentiles and Quantiles

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E.g. Karen scores in the 99th percentile for the Bursary English exam in 1999. ... the students who sat the 1999 Bursary English exam had lower marks than Karen. ... – PowerPoint PPT presentation

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Title: Inverse Problems: Percentiles and Quantiles


1
Inverse Problems Percentiles and Quantiles
  • If you fall into the 98th percentile in a
    population, this means that 98 of the population
    falls below you.
  • E.g. Karen scores in the 99th percentile for the
    Bursary English exam in 1999. This means that 99
    of all the students who sat the 1999 Bursary
    English exam had lower marks than Karen.
  • The lower quartile is the 25th percentile
  • The median is the 50th percentile
  • The upper quartile is the 75th percentile

2
Quantiles
  • A quantile is the same as a percentile but is
    expressed as a proportion. E.g. the 75th
    percentile would be the 0.75 quantile.
  • More formally, if 0 ? p ? 1, then the p-quantile
    qp is defined to be the value such that Pr(X ?
    qp) p.
  • This is just a complicated way of saying that the
    lower tail probability for some value qp is p
  • Check the following with Minitab or Excel. For
    the Normal distribution the
  • 0.95 quantile is approximately 1.6449
  • 0.975 quantile is approximately 1.96
  • 0.99 quantile is approximately 2.326

3
Using a computer program to find to find
percentiles/quantiles for the Normal distribution
  • Last time we gave the computer an x value, a mean
    and a standard deviation and it gave us back a
    lower tail probability
  • This time, we provide the computer with a
    probability, a mean and a standard deviation, and
    we want an x value
  • Most software packages return x values for lower
    tail probabilities. That is, given p, the
    computer returns an x such that Pr(X ? x) p.
  • This means that we must rephrase our questions in
    terms of lower tail probabilities before we can
    answer them
  • This of course is where the most skill is involved

4
  • Assume that we have phrased our question as a
    lower tail probability.
  • In Minitab
  • Enter the probabilities for which you require an
    x into a column (say C1)
  • Select Calc gt Probability Distributions gt Normal
  • Click on the Inverse Cumulative Probability radio
    button
  • Enter the mean into the mean box and the standard
    deviation into the standard deviation box
  • Enter the column in which your probabilities are
    stored (C1) into the input box and click on OK.
  • In Excel
  • Enter the probability into a cell, say A1
  • Click on an adjacent cell (say B1) and then click
    on the paste function button fx
  • Select the Statistical from the function
    category and NORMINV from the function name and
    click on OK
  • Enter the address of the probability (A1) into
    the probability box, the mean into the mean box,
    and the standard deviation into the std_deviation
    box and click on okay
  • Use Fill Down if you have more than one
    probability

5
Example
  • Suppose that IQ scores are Normally distributed
    with a mean of 100 and a standard deviation of
    15. Use the Minitab output to answer the
    following questions
  • Find the IQ score of the bottom 20 of the people
    in the population (i.e. find the 20th percentile)
  • What IQ score is exceed by only the top 10?
  • Find the interquartile range for IQ scores.

Normal with mean 100.000 and standard deviation
15.0000 P( X lt x) x P( X lt x)
x 0.1000 80.7767 0.7500
110.1173 0.2000 87.3757 0.8000
112.6243 0.2500 89.8827 0.9000
119.2233
6
(a)
(b)
(c)
7
Exercise
  • Assume that the natural gestation period for
    human births is approximately Normally
    distributed with a mean 266 days and a standard
    deviation of about 16 days. Use the Minitab
    output to answer the following questions
  • What is the maximum gestation for the bottom 15
    of births?
  • What is the range of gestation period for the
    central 60 of births?
  • What is the maximum gestation period exceed by
    90 of births?

Normal with mean 100.000 and standard deviation
15.0000 P( X lt x) x P( X lt x)
x 0.1000 80.7767 0.6000
103.8002 0.1500 84.4535 0.8000
112.6243 0.2000 87.3757 0.9000
119.2233
8
(a)
(b)
(c)
9
Working in standard units
  • Sometimes it is useful to measure distance in
    terms on the number of standard deviations from
    the mean.
  • The z-score measures the distance in terms on the
    number of standard deviations from the mean
  • To find the z-score for a particular value x, we
    subtract the mean and divide by the standard
    deviation, i.e.
  • If x is smaller than the mean then the z-score is
    negative
  • If x is larger than the mean then the z-score is
    positive
  • Example If
    find the z-score for
  • (a) x 28 (b) x 16 (c) x 30 (d) x 13

10
Example
  • A student sat 3 exams. Her results are Maths 72,
    English 68, Art History 64. Assume that the
    results of each of these exams are approximately
    Normally distributed as follows
  • Maths m 63 s 11
  • English m 58 s 13
  • Art History m 54 s 10
  • Assume that large numbers of students sat each of
    the three exams. In which subject is this student
    the strongest in terms of ranking with those who
    sat the same exam as her?

11
The Standard Normal distribution
  • If X is Normally distributed with a mean of m
    and a standard deviation of s, i.e.
    then the random variable corresponding to
    the z-score
  • is Normal with mean 0 and standard deviation 1
  • The Normal distribution with m 0 and s 1 is
    called the standard Normal distribution and we
    usually write
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