In brief the word frequency is replacing by

Info icon This preview shows pages 32–35. Sign up to view the full content.

View Full Document Right Arrow Icon
variable can take. In brief, the word frequency is replacing by probability. 5.2 OBJECTIVES The main aim of this Lesson is to study the Basics of probability distribution. After going through this Lesson you should be able to Understand 1. Mathematical Expectation 2. Binomial distribution 3. Poisson distribution 4. Normal distribution 5.3. CONTENTS 1. Mathematical Expectation 2. Binomial distribution 3. Poisson distribution 4. Normal distribution 5. Illustrations Illustration - 1 A dealer of Allwyn refrigerators estimates from his past experience the probabilities of his selling refrigerators in a day. These are as follows: No. of Frigerators: 0 1 2 3 4 5 6 Probability: 0.03 0.20 0.23 0.25 0.12 0.10 0.07 Find the mean number of refrigerators sold in a day. Solution P (X)= P (X = Xi); i = 1, 2, 3, …., n P (X) = 0 × 0.03 + 1 × 02 + 2 × 0.23 + 3 × 0.225 + 4.0 × 0.12 +5 × 0.10 + 6 × 0.07
Image of page 32

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

View Full Document Right Arrow Icon
27 = 0 +0.2 + 0.46 + 0.75 + 0.48 + 0.50 + 0.42 = 2.81 Hence, mean number of refrigerators sold in a day is 2.81. Illustration - 2 A die is tossed twice. Getting an odd number is termed as a success. Find the probability distribution of number of success: Solution Getting an odd number in a throw of a die are 1, 3, 5 i.e. 3 in all So, probability of success, 3/6 = 1/2 Probability of failure = 1 1/2 = 1/2 Let us denote success by S and failure by F. In two throws of a die, X which is denoted by S becomes a random variable and takes the values of 0,1,2. P (X = 0) = P (F in both throw) = P (FF) P (F F) 4 1 2 1 2 1 P (X = 1) = P (S and F) + P (F and S) = P (S) P (F) + P (F) P (S) 2 1 4 2 2 1 2 1 2 1 2 1 P (X = 2)= P (S and S) = P (S) P (S) 4 1 2 1 2 1 Thus, the probability distribution of X is: X 0 1 2 P (X) 0.25 0.50 0.25 Illustration - 3 Four bad apples are mixed accidentally with 26 good apples. Obtain the probability distribution of number of bad applies in a draw of 3 apples at random. Solution Denote X the number of bad apples drawn. Now X is a random variable which takes of 0,1,2,3. There are (26 + 4) 30 apples in all, land the exhaustive number of cases drawing three apples is 30c 3 . We get 4060 2600 3 2 1 28 29 30 3 2 1 24 25 26 30c 26c 0) X ( P 3 3
Image of page 33
28 4060 156 3 2 1 26 25 30 1 26 1 2 3 4 30c 26c 4c 2) X ( P 3 1 2 4060 4 3 2 1 28 29 30 3 2 1 2 3 4 30c 4c 3) X ( P 3 3 Hence, the probability distribution of X is: X = 01 2 3 4060 4 4060 156 4060 1300 4060 2600 (X) P Illustration - 4 An insurance company offers a 40 year old man Rs.2300 one year term insurance policy for an annual premium of Rs.15. Assume that the number of deaths per one thousand is four persons in this group. What is the expected gain for the insurance company on a policy of this type. Solution Denote premium by X and death rate P (X). Accordingly, X = Rs.15 P(X)= = 0.004 1000 4 Probability of no death= 1-0.004 = 0.996 E(X) = 15 0.996 1185 0.004 = 14.94 4.74 = Rs.10.20 Expected gain to the insurance company for having that policy is Rs.10.20.
Image of page 34

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

View Full Document Right Arrow Icon
Image of page 35
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern