Statistics Workshop 1: Probability

Johannes M. Zanker,

Workshop Objectives: Materials:

Some Basics about Probability

knowing about probability enables you to draw general conclusions from limited observations, make quantitative predictions


In a situation where several different outcomes are possible, we define the probability of any particular outcome as fraction or proportion.
Formula:     p(A) =   ,                 p(A): probability of A


A note on random sampling:

If you draw a sample without replacement, the underlying frequencies will change! A random sample must satisfy two requirements:

Probabilities from Frequency Distributions

Because probabilities and proportions are equivalent, we can read probability almost immediately from frequency tables or frequency distributions:

Formula: p(X) =    ,            X: observed value
                                                        f: frequency of observation

                                    N: total number of observations Example:

a set of measurements, X, for instance marks between 1 and 6, can be displayed as frequency distribution table or graph, f(X)

p(X=4) = 3/10 = 0.3 = 30%

p(X>=5) = (1+1)/10 = 0.2 = 20%

  frequency histogram

Example 1


An urn contains 20 red balls, 20 green balls, and 20 yellow balls, thoroughly mixed. By making three random selections from the urn, what is the probability of selecting:

(a) three red balls, sampling with replacement
(b) three green balls, sampling without replacement
(c) three balls of any one colour, sampling without replacement
(d) three balls, each of a different colour, sampling with replacement

(a)    p(red ball first) = 20/60
        p(red ball second) = 20/60
        p(red ball third) = 20/60
        p(three reds consecutively with replacement) = 1/3 x 1/3 x 1/3 = 0.037

(b)    p(green ball first) = 20/60
        p(green ball second) = 19/59
        p(green ball third) = 18/58
        p(three greens consecutively without replacement) = 1/3 x 19/59 x 18/58 = 0.033

(c)    p(same colour second) = 19/59
        p(same colour third) = 18/58
        p(three colours the same) = 1 x 19/59 x 18/58 = 0.10

(d)    p(any colour first) = 60/60
        p(any other colour second) = 40/60
        p(third colour third) = 20/60
        p(three colours the same) = 1 x 2/3 x 1/3 = 0.22

Question 1

A box of chocolates in your dentists reception area contains 20 Mars bars and 30 Snickers. Of the 20 Mars bars, 15 are normal size and 5 are extra large. Of the 30 Snickers bars, 15 are normal size, and 15 are extra large. Suppose that you are allowed to randomly select 1 piece from this treasure box, to support your dentist's business.

(a) What is the probability of obtaining a Snickers?
(b) What is the probability of obtaining a normal size Mars bar?
(c) What is the probability of being rewarded with an extra-large chocolate?
(d) Which selection is more likely, an extra-large Mars bar or an extra-large Snickers ?
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Question 2

From observing his customers, the milkman knows that different households consume different amounts of milk. He counted 2, 5, 3, 7, 12, 5, 8, 3, 4, 1 households, which order 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 pints of milk per week, respectively. He wants to honour his best customers (those who buy the largest amount of milk) by a discount, but he does not want to loose to much of his hard earned income and wants to restrict this special offer to no more than 20% of the households.

Draw a frequency histogram from a frequency table, and work out the consumption (pints per week) which should be discounted as special offer.

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The Normal Distribution

many frequency distributions generated from large samples of biological data are bell-shaped, unimodal and symmetrical – described mathematically as normal distribution !

Formula:     Y =             (NB you don't need to know this formula)

characteristic properties of the normal distribution:


the regularity of the normal distribution allows us to

Formula:     z =     ,           X: observed value
                                                    m: mean of distribution

                                                        s: standard deviation of distribution


The Unit Normal Table (Standard Normal Table)

the exact proportions of the normal distribution corresponding to each z-score are listed in the unit normal table à basis for calculating probabilities

please note that for negative values of z the ‘body’ is on the right side and the ‘tail’ is on the left side of the mean!

Example 2

What proportion of the population would pass a standard IQ test with a score larger than 120 ?

                    z =  = 20/16 = 1.25 z = 1.25 corresponds to (0.1151+0.0968)/2 = 0.10595 in tail of the distribution

Question 3

Over the last ten years, a high school Spanish teacher has been giving the same comprehensive final examination to all students in first year Spanish. The mean on this exam is 78.4, and the standard deviation is 14.8. To pass the course, a student must score at least 55. To be placed in the top, honours section, a student must score at least 99. If a student is selected at random, use the properties of the normal distribution to determine the following probabilities:

(a) the student will pass the course
(b) the student will be placed in the honours section
(c) the student will score higher than 80
(d) the student will score lower than 85
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Question 4

The body size of adult humans is reasonably well approximated by a normal distribution with an average height of 173 cm and a standard deviation of 15 cm.

(a) Sketch this normal distribution on a metrical scale and indicate the range covered by the part of the 70 % 'most normal' (i.e. closest to the mean) height.
(b) What is the probability of randomly selecting n individual who is taller than 2 m?
(c) What is the maximum height of any individual belonging to the shortest 10% of this population?
(d) Is it more likely to encounter an individual of 150 cm or less or an individual in the range between 180 cm and 190 cm?
To see the answer click here

last update 30/10/2002
Johannes M. Zanker