Suppose a class divided into two groups of students. Group 1 consists of 60% stu

ID: 2962576 • Letter: S

Question

Suppose a class divided into two groups of students. Group 1 consists of 60%

students and Group 2 the remaining 40% of the students. Each student is asked to

complete questionnaire. There are 4 questions for Group 1 students and 3 questions

for Group 2 students. Chance of correct answer for Group 1 students is 58% and for

Group 2 students is 67%. Let X be the number of questions answered incorrectly

by a randomly chosen student.

(a) (5pts) Find the probability mass function pX(x) = P(X = x).

(b) (5pts) Find the cumulative distribution function, F(x) = P(X < x)

SUCCESS IN THIS QUESTION WILL BE THE EVENT OF ANSWERING WRONG,

FOR GROUP 1 PROBABILITY OF SUCCESS= 0.58

FOR GROUP 2 PROBABILITY = 0.67

DISTRIBUTION WILL BE BINOMIAL AND WILL BE ADDED CONSIDERING STUDENT CAN BE FROM BOTH GROUPS,

FOR GROUP 1 pmf = PROBABILITY STUDENT FROM GROUP 1 AND X BE NUMBER OF WRONGLY ANSWERED QUESTIONS FROM 4 = 0.6* [ (4CX) *((0.58)^X)*(0.42)^(4-X)] ,

SIMILARLY pmf of group 2 = 0.4 * [ (3CX)*((0.67)^X)*(0.33)^(3-X)]

TOTAL PMF WILL BE PMF1 + PMF 2

B) FOR FINDING THE CUMMULATIVE DISTRIBUTION FUNCTION WE DO THE SUMMATIONS OF PMF ,

PMF 1 AT X=0 + X=1 +X=2 + X=3 + X=4

PMF 2 AT X=0+ X=1 + X=2 + X=3

AFTER SUMMING ALL THE VALUES YOU WILL GET THE CDF AS 0.996

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