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31 01 13

Esame completo di Statistica per il corso di Mechanical Engineering presso Politecnico di Milano. Materiale proveniente dall’archivio storico Studwiz e classificato per la consultazione online.

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Esame completo di Statistica per il corso di Mechanical Engineering presso Politecnico di Milano. Materiale proveniente dall’archivio storico Studwiz e classificato per la consultazione online.

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Statistics - Written Examination MEC Students - BOVISA Prof.ssa A. Guglielmi 31.01.13 /EP All rights reserved. Legal action will be taken against infr ingement. Reproduction is prohibited without prior consent. Name: Student Id. Number: Properly justify all your answers. Explicitly define all the random variables you are going to use in the solution which have not already been introduced in the text. Exercise 1 Every afternoon Mary has an ice cream cone in the same ice crea m shop. She orders a small ice cream with probability 0.5, a medium one with probability 0.25 and a large one with probability 0.25, without being influenced by her pa st or future choices. The ice cream is 2 euros for a small, 3 euros for a medium and 5 euros for a large. Moreover, every Sunday afternoon Mary also buys a 10 euros ice cream tub. 1. Compute the mean and the variance of the random variable C representing the cost of one single Mary’s afternoon snack (not including the tub). 2. Compute the approximate value of the probability that in t he whole summer (= 91 days) Mary will spend more than 400 euros for her afternoon snacks, Sundays included. ( Hint : assume that the summer starts on Monday). In order to keep summer expenses under control, Mary decides to buy the ice cream tub only a few Sundays (NOT all the Sundays). 3. How many tubs (at most) can Mary buy, so that the approximat e value of the probability that during the summer she will spend more than 400 euro is at m ost 0.05? Solve an inequality. Solution 1. Let C be the cost in euros of Mary’s snacks in a weekday (from Monday to Saturday): C =      2 with probability 0 .5 3 with probability 0 .25 5 with probability 0 .25 . Therefore E(C) = 2 × 0.5 + 3 × 0.25 + 5 × 0.25 = 3 , E(C 2) = 4 × 0.5 + 9 × 0.25 + 25 × 0.25 = 11 Var(C) =…

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