MATHSTAT 231

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MATH/STAT 231

• Chapter 6
• Normal Distribution

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Introduction

• Histograms of many naturally occurring
  characteristics are bell-shaped.
• Example
• 1. Heights of Female Students
• 2. Weights of Nickels

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Heights of Female Students The histogram below
give the heights of 123 women in a statistics
class at Penn State University in the 1970s.
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Weights of Nickels The histogram below give the
weights, to the nearest hundredth of a gram, of a
sample of 100 new nickels.
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Dissolution Times For a chemistry experiment,
students measured the time for a solute to
dissolve. The experiment was repeated 50
times. The results are shown in the following
chart and histogram.
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• A theoretical model --- a normal distribution
  happens to describe this bell-shaped pattern
  very well. Variables such as human height, IQ
  score, repeated measurements of a same object,
  body temperature are approximately normally
  distributed.

The normal distribution is also called the
Gaussian distribution, in honor of Carl Friedrich
Gauss, who was among the first to use the
distribution.
Similar to histogram, a normal curve has mean,
median, mode and standard deviation.
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Normal Distribution
N(µ,s2) A normal distribution with mean µ and
standard deviation s.
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Properties of normal distribution

• The mean, median and mode are equal.
• A normal distribution is bell-shaped, unimodal
  and symmetric.
• The mean µ and the standard deviation s
  completely determine the normal distribution. The
  mean µ controls the center, and the standard
  deviation s controls the spread.
• A normal curve is always above 0.
• The total area under a normal curve and above the
  horizontal line is 1.00 or 100 (the probability
  of all possible values).

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• Area underneath a normal curve probability
  (Percentage)
• The test score approximately follows a normal
  distribution. It is a normal random variable.
  What is the chance to have a score above 6?

When you want to describe probability for a
normal variable, you do so by describing a
certain area (underneath the curve). A large area
implies a large probability and a small area
implies a small probability.
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Reviewing z-score

• If x is an observation (data value) from a
  distribution that has mean µ and standard
  deviation s, the z-score (standardized value) of
  x is
• The z score for an data value, indicates how far
  and in what direction, that item deviates from
  its distribution's mean, expressed in units of
  its distribution's standard deviation.

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• Example
• 1. Find the z-score corresponding to a raw
  score of 132 from a normal distribution with mean
  100 and standard deviation 15.
• 2.A z-score of 1.7 was found from an
  observation coming from a normal distribution
  with mean 14 and standard deviation 3. Find the
  value of the observation .

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• Five students have taken different forms of the
  spelling test. The scores of different forms are
  normally distributed with different mean and
  standard deviation. How to compare their
  performance?

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The standard normal distribution

• The standard normal distribution is a normal
  distribution that has a mean of 0 and a standard
  deviation of 1. A standard normal variable is
  denoted by Z as will be discussed later.
  ZN(0,1)
• A variable X with the normal distribution N(µ,s2)
  can be transformed to a variable Z with the
  standard normal distribution by the formula

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Probability (area) calculation --- Standard
normal distribution

• The Precision Scientific Instrument Company
  manufactures thermometers that are supposed to
  give readings of 0C at the freezing point of
  water. Tests on a large sample of these
  thermometers reveal that at the freezing point of
  water, some give readings below 0C (denoted by
  negative numbers) and some give readings above
  0C (denoted by positive numbers). Assume that
  the mean reading is 0C and the standard
  deviation of the readings is 1.00C. Also assume
  that the frequency distribution closely resembles
  the normal distribution. Approximately, find the
  proportion of thermometers that, at the freezing
  point of water, the readings are less than 1.2
  C.

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• The chance (proportion) of thermometers that, at
  the freezing point of water, the readings are
  less than 1.2 C
• P( Reading lt 1.2)
• The area of zlt1.2 underneath the standard
  normal curve
• (use the z-table)
• Z-table is available at the class website

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• Find the proportion of observations from the
  standard normal are
• 1. less than 1.12?
• P(Zlt1.12)
• 2. greater than 1.12?
• P(Zgt1.12)
• 3. less than -1.12
• P(Zlt -1.12)
• 4. less than 0?

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• Find the probability (proportion) of observations
  from the standard normal are
• 1. less than -1.2 or greater than 1.0
• 2. greater -1.2 and less than 1.0
• 3. greater 0.76 and less than 2.5

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Formula

• Z is a standard normal variable , a is a number
  and bgta.
• P(Z lt a) is available in the table.
• P(Z gt a) 1- P(Z lta) (or)
• P(Zlt -a) (symmetric
  about the mean 0)
• P(altZltb) P(Zltb) P(Zlta)
• P(Zlta or Zgtb) P(Zlta) P(Zgtb)
• P(Zlta) 1 P(Zltb)
  (or)
• P(Zlta) P(Zlt-b)

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Symmetric
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Probability (area) calculation --- Normal
distribution

• Example Let X be the level of cholesterol (in
  units of mg/dl) of 14 year old boys. X
  approximately follows a normal distribution with
  mean 170 and standard deviation 30 ( X N(170,
  302) ). Find the probability that a randomly
  selected 14 year old boy has a level of blood
  cholesterol between 170 and 240 mg/dl? (Find the
  proportion that 14 year old boys have a level of
  blood cholesterol between 170 and 240 mg/dl )
• Note X does not follow a standard normal
  --- use the table directly?

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Step 1 Calculate the z-scores (Standardization).
z1 z-score of 170 z2 z-score of 240
Step 2 Proportion (area) calculation for a
standard normal distribution P( blood
cholesterol between 170 and 240) P(170 lt blood
cholesterol lt 240 ) P(Zlt z2) P(Z lt z1)
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• X is a normal variable with mean µ and standard
  deviation s, a is a number and bgta.
• Let Z(X- µ)/ s is a standard normal
  variable
• za (a- µ)/ s z score of the
  number a
• zb (b- µ)/ s z score of the
  number b
• The rule is simple replace X by Z (standard
  normal variable), and replace numbers a, b by
  their z scores respectively.
• For example
• P(X lt a) P(Zlt za), which is available in
  the table.
• P(X gt a) P(Z gt zb)
• P(altXltb) P(za ltZlt zb)
• P(Xlta or Xgtb) P(Zltza or Zgtzb)

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• Inverse normal calculations
• --- Finding the data value given a specific
  probability.
• Example
• 1. Find the value z such that 60 of
  observations from the standard normal are less
  than the value z. (Use the z-table)

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• Example Find the value c such that 60 of
  observations from the standard normal are between
  c and c . (Use the z-table)
• P(-cltZltc) 0.6 gt c ?
• P(-cltZltc)
• P(Zltc) P(Z lt -c) (Make a plot)
• 1 - P(Zgt c) P(Zlt-c)
• 1 2 P(Z lt -c)
• gt P(Z lt -c) 0.2 gt -c (use Z-table) gt
  c

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• Example SAT verbal test scores roughly follow
  a normal distribution with mean 505 and standard
  deviation 110. distribution. How high must a
  student score to place in the top 10 of all
  students taking the exam?

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• Example For a medical study, a researcher wishes
  to select people in the middle 60 of the
  population based on blood pressure. If the blood
  pressures are normal distributed with mean 120
  and the standard deviation 8, find the upper and
  lower readings that would qualify people to
  participate the study.
• Step 1 Make a plot and mark the percentage
  and cutoff points.
• Step 2 Find the z-scores corresponding to
  the middle 60.
• Step 3 Find the data values corresponding
  to the z-scores.