2.Find the turning point of the function given in question 1. 3.Solve 3x 2 8x + 1 = 0 giving your answers to 2 decimal places. Suppose the heights of a very large number of 17 year olds was surveyed The total area under the graph is 1 and particular areas represent probabilities Suppose the class widths were refined to 0.1m Assuming that enough data was collected the class widths could be further reduced until the relative frequency polygon approximates to a curve. When the function f(x) is found for the curve, it is called the probability density function, or pdf To be a probability density function (pdf), f(x) must satisfy these basic properties: f(x) 0 for all x, so that no probabilities are negative You can find the probability that a random variable lies between x = a and x = b from the area under the curve represented by f(x) between these two points. f(x) 0 for all x, so that no probabilities are negative f(x) is positive for all values of x If f(x) is a pdf, then 1 f(x) x 102 P(0.5 < x < 1.3) (a) (b) For any CONTINUOUS distribution, P(X = a) = 0 P(X = 2) = 0 In S1, the cumulative distribution function (cdf) F(x 0 ) was defined as P(X x 0 ) for discrete random variables. For continuous random variables, you can find the cdf by integrating the pdf. Similarly, you can find the pdf from the cdf by differentiating. Probability Density Cumulative Distribution Function (pdf) Function (cdf) f(x) F(x) The mean or expected value of a discrete probability distribution is defined as: = E(X) = px For a continuous random variable: where in practice, the limits will be the interval over which f(x) is defined. Var(X) = E(X 2 ) 2 * The mode of a continuous random variable is the value of x for which f(x) is a maximum over the interval in which f(x) exists. This is either a stationary point, at which f (x) = 0, or the end value of the interval over which f(x) is defined. There may not be a mode if no single value occurs more often that any other. (Bimodal distributions do occur commonly in real life). You can use the cumulative distribution function to find the median and quartiles for a continuous random variable: If m is the median, then F(m) = 0.5 If Q 1 is the lower quartile then F(Q 1 ) = 0.25 If Q 3 is the upper quartile then F(Q 3 ) = 0.75 You can set up an appropriate algebraic equation and solve it to find the required quartile, decile or percentile. (a) Median is when F(X)=0.5 (b) Q 1 is when F(X) = 0.25 Q 3 is when F(X) = 0.75 Q 2 Q 1 = 0.41 Q 3 Q 2 = 0.32 Q 2 Q 1 > Q 3 Q 2 Negative Skew If you were asked for the SHAPE of the distribution or the SKEW