HW 5 - 1of6ID:MST.DPD.BD.04.0010

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
-   1 of 6       ID:   MST.DPD.BD.04.0010 An analyst has been studying a particular stock and has come to the following two conclusions: the probability that the stock will rise on a particular day is 0.45 the rise or fall of the stock on any particular day is independent of the rise and fall of the stock on  any other day Calculate the probability that over the next week (7 days) the stock will rise on exactly 5 days. Give your  answer as a decimal to 2 decimal places. Probability =   [0 out of 1] -       Feedback This is not correct. Probability =   0.12 Calculation If X is the random variable that describes the possible number of days on which the stock rises then X  follows a binomial distribution. The probability that the stock will rise on 5 days can be calculated using the  following formula:   show variables p(x ) = n C x × p x × (1 - p) n-x = n C x × 0.45 5 × (1 - 0.45) 7-5 = 7! 5! 2! × 0.01845281. .. × 0.3025 = 0.11722149. ..
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
= 0.1 2 Rounded as last step [1     point] -   2 of 6       ID:   MST.DPD.LP.02.0010 A computer store offers printers, scanners and modems at discounted prices for customers that buy a  computer. A salesman notes that customers can buy none, some or all three of these extras when they  purchase a computer. He also notes the probabilities of buying 0, 1, 2 or all 3 of the extras: x 0 1 2 3 p(x) 0.61 0.2 0.16 0.03 Calculate the expected number of extras a customer will buy. Give your answer to 2 decimal places. Expected number of extras =   [1 out of 1] -       Feedback You are correct. Calculation The expected number of extras can be calculated using the following formula:   show variables E(X ) = ∑ x i p(x i ) = 0 × 0.61 + 1 × 0.2 + 2 × 0.16 + 3 × 0.03
Background image of page 2
= 0.61 [2     points] -   3 of 6       ID:   MST.DPD.PD.05.0010 The confectionery company Chocoholly make chocolate chips cookies as part of their production line.  Chocolate chips in the cookies are randomly and independently distributed according to a Poisson  distribution with an average of 12 chocolate chips per cookie. a)
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 09/29/2011 for the course STAT 225 taught by Professor Martin during the Spring '08 term at Purdue.

Page1 / 8

HW 5 - 1of6ID:MST.DPD.BD.04.0010

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online