6.11 – The Normal Distribution

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6.11 The Normal Distribution

• IB Math SL/HL Y1Y2 - Santowski

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(A) Random Variables

•  Now we wish to combine some basic statistics
  with some basic probability ? we are interested
  in the numbers that are associated with
  situations resulting from elements of chance i.e.
  in the values of random variables
• We also wish to know the probabilities with which
  these random variables take in the range of their
  possible values ? i.e. their probability
  distributions

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(A) Random Variables

• So 2 definitions need to be clarified
• (i) a discrete random variable is a variable
  quantity which occurs randomly in a given
  experiment and which can assume certain, well
  defined values, usually integral ? examples
  number of bicycles sold in a week, number of
  defective light bulbs in a shipment
• discrete random variables involve a count
•  
• (ii) a continuous random variable is a variable
  quantity which occurs randomly in a given
  experiment and which can assume all possible
  values within a specified range ? examples the
  heights of men in a basketball league, the volume
  of rainwater in a water tank in a month
• continuous random variables involve a measure

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(B) CLASSWORK

• CLASSWORK (to review the distinction between the
  2 types of random variables)
• Math SL text, pg 710, Chap29A, Q1,2,3
• Math HL Text, p 728, Chap 30A, Q1,2,3

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(C) The Normal Distribution

• - data obtained by direct measurement (i.e.
  population heights) is usually continuous rather
  than discrete (all heights are possible, not just
  whole numbers
• - continuous data also has statistical
  distributions and many physical quantities are
  usually distributed symmetrically and unimodally
  about the mean ? statisticians observe this bell
  shaped curve so often that its model is known as
  the normal distribution

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(C) The Normal Distribution

• the graph of the normal distribution is also
  referred to as the standard normal curve and one
  defining equation for the curve for our purposes
  is
• where z refers to a concept called the z score
  which takes into account the mean and standard
  deviation of a set of data

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(C) The Normal Distribution

• we can graph the normal distribution as follows,
  where the x-axis is the number of standard
  deviations, ?, from the mean/median, ? (the idea
  behind our z score)
• the total area under the curve is 1 unit (aising
  from the fact that the total probability of all
  outcomes of an event can be at most 1 or 100)
• With our z-score, we set the mean, ? , to be 0
  and each 1 unit of the x-axis is 1 standard
  deviation, ?.

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(C) The Normal Distribution

• to find the area under the curve between any two
  given z-scores, we can rely on graphs
• the area under the curve between our two given
  z-scores means the proportion of values between
  our two z-scores
• so if we write p(-2 lt z lt 1) 0.81859, we mean
  that the proportion of data values that are
  between 2 standard deviation units below the mean
  and 1 unit above is 0.81859, or as a percentage
  81.859 of our data, or the probability that our
  data values lie between 2 SDs below and 1 SD
  unit above the mean is 0.0.81859 ? we can
  illustrate this on a normal distribution graph as
  follows

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(C) The Normal Distribution Tables of z scores

• We can work out the previous example without a
  graph and shading areas under a graph, by simply
  using prepared tables
• SL Math text, p735 and HL Math text, p772
• So to determine the p(-2 lt z lt 1), we check the
  table and see that a z value of 2.00 corresponds
  to a value of 0.0228 ? this means that the area
  shaded under the curve, starting from 2.00 all
  the way left to -? is 0.0288 (or 2.88 of the
  data is more than 2 SD units below the mean)
• Likewise, we check the table for our z value of
  1.00 and see the value of 0.8413 ? this means
  that the area shaded under the curve, starting
  from 1.00 all the way left to -? is 0.8413 (or
  84.13 of the data is less than 1 SD units above
  the mean)
• So what do we do with the 2 numbers? Well, we
  have accounted for some of the data twice ? the
  data more than 2 SD units below the mean ? so
  this gets subtracted from the first value ?
  0.8413 0.0288 0.8185 as we saw before with
  the graph and graphing software

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(D) Examples

• Use the table to evaluate p(zlt1.5). Interpret the
  value.
• The table gives us the value 0.9332, which means
  that 93.32 of our data lies 1.5 SD units above
  the mean and below ? or the probability of
  getting a random data point that is at most 1.5
  SD units above the mean is 0.9332
• We can see this illustrated on the graph

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(D) Example Using Standard Normal Tables

• For the standard normal variable, find
• (i) p(z lt 1)
• (ii) p(z lt 0.96)
• (iii) p(z lt 0.03)
• Some slightly more challenging examples 
• (i) p(z gt 1.7)
• (ii) p(z lt -0.88)
• (iii) p(z gt -1.53)
• And now some in-between values 
• (i) p(1.7 lt z lt 2.5)
• (ii) p(-1.12 lt z lt 0.67)
• (iii) p(-2.45 lt z lt -0.08)
• WE can also do some Ainverse_at_ problems
• (i) p(z lt a) 0.5478
• (ii) p(z gt a) 0.6
• (iii) p(z lt a) 0.05

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(E) Homework

• SL Math text, Chap 29H.1, p736, Q1-4
• HL Math text, Chap30K.1, p757, Q2-5

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(F) Standardizing Normal Distributions

• When we have applications wherein we apply a
  normal distribution (i.e. with any continuous R/V
  like height, weight of people), each unique
  application has its own unique mean and standard
  deviation along with its unique distribution
  graph
• What we wish to accomplish now ? can we somehow
  standardize a normal distribution so that one
  single standardized normal distribution applies
  for every single possible normal distribution
• We can accomplish this by a combination of
  transformations of our unique data with its
  unique normal distribution

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(F) Standardizing Normal Distributions

• So from every data point in our distribution, we
  will subtract the populations mean and then
  divide this difference by the populations
  standard deviation ? we will call this result a
  z-score
• So our formula for this data transformation is
  z (x - ?)/?
• So we then graph the newly transformed data
  points and we get a standardized normal
  distribution curve
• The two key features on the standardized normal
  distribution curve are (i) the mean is 0 and (ii)
  the standard deviation is 1

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(G) Graph of Standardized Normal Distribution
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(H) Working with a Standardized Normal
Distribution

• Ex 1 ? The heights of all rugby players from
  India is normally distributed with a mean of 179
  cm with a standard deviation of 5 cm. Find the
  probability that a randomly selected player
• (i) was less than 181 cm tall
• (ii) was at least 177.5 cm tall
• (iii) was between 175 and 190 cm

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(H) Working with a Standardized Normal
Distribution

• Solution 1(i) is to use the z-score tables
• z (181-179)/5 0.40
• So find 0.40 on the tables, which is 0.6554
• So given that the table gives us the cumulative
  area under the curve until the specified z-score
  (0.40), then we can conclude that 65.5 of the
  players would be less than 181 cm

• Alternatively, we can use a GDC
• We simply select the normalcdf( command and enter
  the specifics as follows
• Normalcdf(-EE99,181,179,5) which tells the GDC
  that you want the heights less than 181
  (basically from 181 down to -?) and that the
  population mean is 179 and the SD is 5
• Our result is 0.6554 .. similar to the result
  from the table

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(H) Working with a Standardized Normal
Distribution

• Solution 1(ii) ? use the z-score tables ?
  however we must realize that the table gives us a
  cumulative area under the curve up to the given
  z-score ? now however we are looking for a value
  GREATER than the given area
• So, using the table, simply find the area under
  the curve BELOW the given z-score
• Then, using the complement idea, simply
  subtract the area from 1
• z-score (177.5-179)/5 -0.30
• Table value is 0.4404 (so 44.04 of the area
  under the curve is to the left of 0.30 on the
  z-axis)
• Therefore, the area representing the probability
  of our players being GREATER than 177.5 cm would
  be 1 0.4404 0.5596 ? (so this would be the
  area under the curve, to the right of z -0.30)
• In using the GDC, we again simply enter the
  command normalcdf(177.5, EE99, 179, 5) and get
  0.5596 as our answer

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(H) Working with a Standardized Normal
Distribution

• Solution 1(iii) ? use the z-score tables ?
  however we must realize that the table gives us a
  cumulative area under the curve up to the given
  z-score ? now however we are looking for a value
  BETWEEN 2 given values
• So our two z-scores for 175 and 190 are z 0.80
  and z 2.1, which we can illustrate below

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(H) Working with a Standardized Normal
Distribution

• So, again our tables require several steps in the
  calculation
• (i) find the area under the curve that is LESS
  THAN 0.80 ? 0.2119
• (ii) Now find the area under the curve that is
  less than 2.1 ? 0.9821
• So clearly, the 0.9821 total cumulative area
  includes the 0.2119 that we DO NOT have within
  our specified range of z-scores (player heights
  less than 175 cm)
• Which suggests that we need to subtract the
  0.2119 from 0.9821 0.7702
• Alternatively, using the GDC, we enter
  normalcdf(175,190,179,5) and get the same
  0.7702..

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(I) Homework

• HL Math text
• Chap30K.2, p759, Q1-3
• Chap 30K.3, p760, Q1-4
• Chap 30L, p761, Q1-7

• SL Math text
• Chap 29H.2, p738, Q1-3
• Chap 29H.3, p739, Q1-3
• Chap 29I, p740, Q1-8