Apostolos, believes there are twice as many OU undergraduate students aged between 25 and 35 years of age, as there are

Business, Finance, Economics, Accounting, Operations Management, Computer Science, Electrical Engineering, Mechanical Engineering, Civil Engineering, Chemical Engineering, Algebra, Precalculus, Statistics and Probabilty, Advanced Math, Physics, Chemistry, Biology, Nursing, Psychology, Certifications, Tests, Prep, and more.
Post Reply
answerhappygod
Site Admin
Posts: 899559
Joined: Mon Aug 02, 2021 8:13 am

Apostolos, believes there are twice as many OU undergraduate students aged between 25 and 35 years of age, as there are

Post by answerhappygod »

Apostolos Believes There Are Twice As Many Ou Undergraduate Students Aged Between 25 And 35 Years Of Age As There Are 1
Apostolos Believes There Are Twice As Many Ou Undergraduate Students Aged Between 25 And 35 Years Of Age As There Are 1 (629.52 KiB) Viewed 86 times
Apostolos, believes there are twice as many OU undergraduate students aged between 25 and 35 years of age, as there are below 25 years of age. He defines a random variable Y, which assigns a numerical value to each of three age categories; under 25 years, 25–35 years and over 35 years of age. He also defines 6 to be the probability that a student is less than 25 years old. Using these definitions he created the following table which represents the p.m.f. for the random variable Y. Age 35 Y 1 2 3 Probability 0 20 (1-30) (a) Explain why @ must be less than :
(b) Apostolos uses the ages of 544 students who started to study M248 in October 2020 to produce the following table. Outcome 1 2 3 Frequency 83 197 264 Show that the likelihood of A based on this data is L(O) = C0280 (1 - 30)264, where C is a positive constant not dependent on 0.
Produce a sketch, by hand, of the likelihood function given in part (b), carefully labelling your axes and the maximum likelihood estimate.
Join a community of subject matter experts. Register for FREE to view solutions, replies, and use search function. Request answer by replying!
Post Reply