X i A1 A2 A3 B1 B2 B3 C1 C2 C3 D F1 F2 F3 f i 4 13 7 7 6 8 17 16 18 21 0 33 12
3 Looking at this we can find the related probabilities as follows: - P(X>B1) = Number of A grades/Total grades = 24/162 = 0.15 - P(X>F1) = Number of passing grades/Total grades = 117/162 = 0.72 - P(X<D) = Number of failing grades/Total grades = 45/162 = 0.28 Now let us convert this data set into a continuous one artificially by inserting grade values instead of letters and supposing it is continuous. We have the following table (where F grades are grouped under zero): X i 4.0 3.75 3.5 3.25 3.0 2.75 2.5 2.25 2.0 1.75 0 f i 4 13 7 7 6 8 17 16 18 21 45 You will notice that the probabilities are the same when we rephrase the questions as P(X>3.25), etc. These will not be calculated again. Instead we will try to find probabilities based on arithmetic mean and standard deviation. Values for these are (supposing we have a population, which we have): g2020 = uni2211g1850 g3036 g1858 g3036 uni2211g1858 g3036 = 303.25 162 = 1.87 g1853g1866g1856 g2026 g2870 = uni2211(g1850 g3036 −g2020) g2870 g1858 g3036 uni2211g1858 g3036 = 276.9 162 = 1.71 g1853g1866g1856 g2026 = 1.31 We will now try to find the probabilities of picking a student who is one standard deviation around the arithmetic mean, who got grades more than one standard deviation from the mean, and one who got less than 1.5 standard deviations from the mean: - P(μ-σ<X< μ+σ) = P(1.87-1.31<X<1.87+1.31) = P(0.56<X<3.18) = 86/162 = 0.53 - P(X> μ+σ) = P(X>1.87+1.31) = P(X>3.18) = 31/162 = 0.19 - P(X< μ-1.5σ) = P(X<1.87-1.5*1.31) = P(X<1.87-1.965) = P(X<-0.095) = 0/162 = 0 Having seen this, it is now time to introduce once again the dividing points and ratios. These concepts were covered in previous chapters, but a quick recovery does not hurt: - Any data point (real or imaginary) is a dividing point that corresponds to a dividing ratio - We can find a dividing ratio (r/s) for a given point, or a dividing point X (r/s) for a given ratio - In empirical distributions this can be done via cumulative relative frequencies - In theoretical distributions where frequencies are given by a function f(x), it is found by: g1870 g1871 ∗g1846g1867g1872g1853g1864 g1853g1870g1857g1853 = g3505 g1858(g1876)g1856g1876 g3051 g2879g2998 - For the grouped frequency distributions, revive the related chapter - Basically we can start from the ratio to find a point or vice versa At this point we will take a closer look at the relation between dividing points and the standard deviation. Please note that this relationship is the core idea behind all confidence intervals and hypothesis testing. By any univariate distribution X (preferably given in interval or ratio scales), there exists a relationship between σ and any dividing point X (r/s) . As shown in the figure below, start from the center point g1850 g3364 and move k steps (in σ units) to the left and right to reach two different dividing points X (r1/s) and X (r2/s) . These points can be denoted as follows:
4 g1850 g3045g2869 g3046 uni2044 = g1850 g3364 − g1863 ∗g2026 g1853g1866g1856 g1850 g3045g2870 g3046 uni2044 = g1850 g3364 + g1863 ∗g2026 Taking the upper expression into account together with the graph, we can define any datum X i in reference to uniF7C3 and depending on the position of the point with respect to the center of the distribution as: X i = uniF7C3 ± k*σ The value of k will denote how many standard deviations the point is away from the center and the sign of k will simply show which side of the center the point lies on. For example, for a distribution
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- Fall '15
- Normal Distribution