Statistics Assignment

Questions: 1. A b-ball player has the accompanying focuses for seven games: 20, 25, 32, 18, 19, 22, and 30. Register the accompanying measures: a) Compute the example mean (the normal of the purposes of each game) b) Compute the example middle c) Compute the difference and the standard deviation 2. Assume during ends of the week, 55 percent of grown-ups go to the sea shore, 45 percent go to the film, and 10 percent go to both the sea shore and the film. a) What is the likelihood that an arbitrarily picked grown-up doesn't go to the film? b) What is the likelihood that an arbitrarily picked grown-up go to the sea shore or the film or both? c) What is the likelihood that an arbitrarily picked grown-up doesn't go to the sea shore or the film? 3. A Financial Consultant has grouped his customers as indicated by their sexual orientation and the creation of their venture portfolio (basically bonds, fundamentally stocks, or a fair blend of securities and stocks). The extents of customers falling into the different classes are appeared in the accompanying table: Portfolio Composition Sexual orientation Bonds Stocks Balanced Male 0.18 0.20 0.25 Female 0.12 0.10 0.15 One customer is chosen indiscriminately, and two occasions An and B are characterized as follows: A: The customer chose is male. B: The customer chose has a reasonable portfolio. Locate the accompanying probabilities: Find the accompanying probabilities: a) P(A) b) P(B) c) P(A or B) d) P(A or B) e) P(A/B) Answers: (1). a) Let X be the variable, at that point mean of X is sum(X)/n, n being the no. of perceptions. Along these lines mean =23.71429 b) Median is that estimation of X state which with the end goal that extent of perceptions above y is 0.5. In the wake of masterminding the information in expanding request, we get Middle =22. c) The fluctuation of X is , m is the example mean. Along these lines fluctuation =30.2381 Standard deviation = =5.498918 (2). a) Let A be the occasion of going to sea shore and B be the occasion of going to film. We are required to discover P(B) =1-P(B) =0.55 b) Here we are to discover P(AUB) =P(A) +P(B) P(AB) =0.55+0.45-0.1 =0.9 c) We are to discover P(AB) =1-P(AUB) =1-0.9 =0.1 (3). A: The customer chose is male. B: The customer chose has a decent portfolio. Locate the accompanying probabilities: a) P(A) =18+0.20+ 0.25=0.63 b) P(B) =0.15+0.25 =0.4 c) P(AUB) =P(A) +P(B) P(AB) =0.63+0.4-0.25 =0.78 d) P(AUB) =P(A) +P(B) P(AB) =0.63+0.4-0.25 =0.78 e) P(A/B) =(AB)/P(B) =0.25/0.4 =0.625.