Mathematical Methods Units 3/4 Student Learning Plan Continuous Probability Distributions & Normal Distribution 7 lessons Notes: Students need practice in recognising whether a problem involves a discrete or continuous variable, binomial or normal, and in extended response type questions in which the type of variable keeps changing in each related part.
Learning Goals Students should understand and be able to do: -what a continuous random variable is and how to recognise it -why the probability of a continuous variable cannot be evaluated for a single value of the variable and must always be given for an interval -determine whether a given function represents a probability density function or not 1. f(x) 0 for all x. f( x) dx = 1 -find probabilities of the variable between certain values by finding the appropriate area under the probability density function Assessment for Learning Select from Ex 15A -use integrals to evaluate areas -use of Define to simplify working if needing to find multiple integrals involving the same function -*calculate the mean, mode median, variance and standard deviation for continuous random distributions Select from Ex 15B and 15C Mode Occurs at the maximum value of the probability density function, f( x ), which can be found by solving f ʹ ( x) = 0or by using the graph of f( x) to find the maximum. There can be more than one mode. Variance and standard deviation
-interpret the significance of different values for the mean and variance of a distribution and how they define the position and shape of a graph Eg: A probability density function is given by: x (100 x ) 0 x 10 f( x) = 500 0 elsewhere Find: a) the mean of the distribution b) the median and mode of the distribution c) the variance of the distribution
- recognise and apply the normal distribution as a special case of a continuous random distribution where the probability density function is given by: *,)- / +. f x = 1 σ π e) (not needed to be known for exam) *the symmetry properties of this probability density function and how it behaves as the mean and standard deviation changes *In general, approximately 68% of the distribution lies within one standard deviation of the mean, approximately 95% lies within two standard deviations and approx 99.7% lies within three standard deviations Ie Properties of the normal distribution - A bell-shaped curve which is symmetrical about the mean. 1 - It has a maximum value of σ π when x = µ - The normal distribution with mean, µ and variance, σ is written as X ~ N ( µ, σ ) - The mean, median and mode ( µ ) of the normal distribution are the same. The confidence intervals associated with the normal distribution are as follows: 1. Approximately 68% of the values lie within 1 standard deviation of the mean. ie. Pr( µ σ X µ + σ) 0.68. Approximately 95% of the values lie within standard deviations of the mean (very probable) ie. Pr( µ σ X µ + σ) 0.95 3. Approximately 99.7% of the values lie within 3 standard deviations of the mean (almost certain) ie. Pr( µ 3σ X µ + 3 σ) 0.997 Select from Ex 16A Khan Academy video on the Normal Distribution http://www.khanacademy.org/math/statistics/v/introdu ction-to-the-normal-distribution Interactive showing what happens as std deviation change: http://www.shodor.org/interactivate/activities/normal Distribution/ The parameters µ and σ have an effect on the normal distribution curve as follows: 1. the mean, µ, controls the position of the curve.. The standard deviation, σ, controls the width of the curve. -The standard normal distribution is a way in which all normal distributions can be compared to each other on the same scale -The standard normal distribution is written as Z ~ N(0, 1 ) since µ = 0 and σ = 1. x µ -Also z = σ - Standardised values can be positive or negative: - A positive z-value indicates that the data value it represents lies above the mean. - A negative z-value indicates that the data value it represents lies below the mean. Select from Ex 16B, 16C, 16D Symmetry properties of the standard normal distribution:
Ø Pr( Z > z) = 1 Pr( Z < z) Ø Pr(Z < z) = 1 Pr(Z < z) Ø Pr(Z > z) = Pr(Z < z) Ø Pr(a < Z < b) = Pr(Z < b) Pr(Z < a) Eg: Suppose X is normally distributed with a mean µ =100 and a standard deviation σ = 6. Find: a) Pr (X < 105) b) k such that Pr (X < k) = 0.955 Revision Tips Examination Probability Tips worksheet on STL Link