Modeling data distributions #5 - Questions and Answers

Question 1 When establishing the differences between histogram and polygon: a) Which of the two diagrams is drawn with the help of class marks of a frequency distribution? b) Which of the two diagrams is drawn at the class boundaries of a frequency distribution? c) Which of the two diagrams contains a series of vertical rectangular bars? d) Which diagram is produced by connecting a set of consecutive points? e) Which of the two diagrams can be used to compare two or more data sets that are plotted according to the corresponding frequency distributions? Answer Step 1 "Since you have posted a question with multiple sub-parts, we will solve first three subparts for you. To get the remaining sub-parts solved, please repost the complete question and mention the sub-parts to be solved.". Histogram and polygon are two data displaying statistical techniques. Step 2 Histogram provides graphical display of data. It uses bars of different heights Frequency polygon is similar to a histogram. This is used to compare sets of data or to display a cumulative frequency distribution. A line graph is used to represent quantitative data on the plot. α) classmark = upper lim it + lowerliimit 2 Class mark is used as an internal step in making polygon. b) Both diagrams are drawn at the class boundaries of a frequency distribution. c) A series of vertical rectangular bars can be seen in both cases. Question 2 Of the following, which is untrue? A.Majority of Z scores in a right skewed distribution are negative. B.In skewed distributions the Z score of the mean might be different than 0. C.For a normal distribution, IQR is less than 2 x SD. D.Z scores are helpful for determining how unusual a data point is compared to the rest of the data in the distribution. Answer Step 1 Solution: These statement are true. B. In skewed distributions the z score of the mean might be different than 0. C. For a normal distribution, IQR is less than 2 x SD. D. Z scores are helpful for determining how unusual a data point is compared to the rest of the data in the distribution. Step 2 Z score in right skewed distribution is positive. Answer: A. Majority of Z scores in right skewed distribution are negative. Question 3 Find the equation of the Least Squares Regression Line if z = 20 sx = 2 y = 10 sy = 4 r = 0.2 Pa) y = 2 + 0.4x b) ŷ = 16 +0.4x c) ŷ = 19 – 0.1x d) ŷ = 8 -0.1x Answer Step 1 Sy y=x+n x(x - 1) ST 4 ⇒ y = 10 + 0 ×2(-20} × 2(x – 20} = 10 + 0x2x2x-0×2×2×20 =2+0x4x Question 4 The table below contains the information for the joint probability mass function of X and Y (two different measurement systems). a) Plot and calculate the marginal distributions of X and Y b) Select one of the Y values from the table and find the conditional probability mass function of X for that Y value you have selected and plot it. c) Show whether X and Y are independent or not. d) Calculate the covariance of (X,Y) i.e. Cov(Χ,Υ). Answer f(x, y) 1 2 3 4 10 0.05 0 0.1 0 20 0.1 0.1 0.05 0.05 30 0.05 0 0.15 0 40 0.1 0.15 0.05 0.05 Step 1 Given, The table below contains the information for the joint probability mass function of X and Y (two different measurement systems). f(x,y) 1 2 3 4 10 0.05 0 0.1 0 20 0.1 0.1 0.05 0.05 30 0.05 0 0.15 0 40 0.1 0.15 0.05 0.05 Step 2 a) Plot and calculate the marginal distributions of X and Y. f(x, y) 1 3 4 10 0.05 0 0.1 0 2 20 30 40 Total Total 0.15 0.1 0.1 0.05 0.05 0.3 0.05 0 0.15 0 0.2 0.1 0.15 0.05 0.05 0.35 0.3 0.25 0.35 0.1 1 The marginal distribution of X: P ( X = x) = ∑ (x = x,y = y ) y The marginal distribution of Y: P(Y = y) = ∑ P(x = x,y = y ) x X 10 20 30 40 P(X = x) 0.15 0.3 0.2 0.35 Y 1 2 3 4 P(Y = y) 0.3 0.25 0.35 0.1 Step 3 b) To determine X's conditional probability given a value of (Y = 1) P(X = x \ Y = y) = P (X = x, Y = y) P(Y = y) X 10 20 30 40 P(X= x | Y = 1) 0.167 0 0.286 0 The conditional distribution is plotted: P(X=xY=1) 0.3+ 0.25 0.27 0.15 O.lt 0.05 0 10 20 30 40 Step 4 c) Show whether X and Y are independent or not. If the variable X and Y are independent that satisfies, P(X = x \ Y = y) = P(X = x) or P(X = x \ Y = y ) = P (Y = y) Let take (X = 1) P(X = 10) = P (10, 1) + P (10, 2) + P(10, 3) + P(10, 4) = 0.05 +0 +0.1+0 = 0.15 P(X = 10 | Y = y) Hence X and Y are not independent random variables Step 5 d) Calculate the covariance of (X,Y) i.e. Cov(Χ,Υ). Cov(x,y) = E (xy) - E(x)E(y) Where, Question 5 The attendances y for two movies can be modeled by the following equations, where x is the number of days since the movies opened. Movie A: y = -x² + 35x + 100 Movie B: y = -5x + 275 Where x is number of days since the movies opened. When is the attendance for each movie the same? Answer Formula used: x -b±√b2-4ac Calculation: 2a Substituting value of equation 1 in equation 2, -x²+35x + 100 = -5x + 275 Shifting the value, -x²+35x + 100 = -5x+275 = 0 On solving, -x²+40x-175 = 0 Hence, a = -1, b = 40, c = -175, Substituting the formula x So, x x= -40 ± 402 – 4(-1)(-175) 2(-1) -40±√1600 - 700 -2 -40 ± √900 -2 -40±30 x= -2 x= -40+30 -2 x=5 Value of y, substituting in equation 2, y = -5(5) +275 y = 250 For -30, x= -40-30 x = 35 -2 Value of y, substituting in equation 2, y = -5(35) +275 y = 100 Hence, z = 5, 35 and y = 250, 100 Hence, the attendance of two movies is same on day 5 and day 35