Essay About 0.6P And Number Of The Residents

Business Stat
GROUP 2CHAN KA LOK50966632MCBADCDBDCDABAABDDBLQ1)There are 25 people conducted this survey and 11 people rated 50 or below.11/25 x 100%=44.0000%The percentage of the respondents rated overall television quality with a rating of 50 or below is 44.0000%.2) Let A be the number of the residents own 2 cars.Let A’ be the number of the residents not own 2 cars.Let B be the number of the residents’ annual household income is over $25,000.Let B’ be the number of the residents’ annual household income is not over $25,000.P (A l B) = P (A and B)/ P (B)0.8 = P (A and B)/ 0.6P (A and B) = 0.4800EventAA’TotalB0.48000.12000.6000B’0.22000.18000.4000Total0.70000.30001.0000The probability that the residents of a household do not own 2 cars and have an income over $25,000 a year is 0.1200.3a)We have to prove P (A and B) = P (A) P (B)L.S = P (A and B)   = 0.4800R.S = P (A) P (B)   = 0.7000 x 0.6000   = 0.4200L.S =/ R.SThe events of the residents of a household own 2 cars and income over $25,000 a year are not statistically dependent.3b)We have to prove P (A and B) = 0L.S = P (A and B)   = 0.4800   =/ R.S = 0The events of the residents of a household own 2 cars and income over $25,000 a year are not mutually exclusive.3c)We have to prove P (A or B) = 1L.S = P (A or B)   = P (A) + P (B) – P (A and B)   = 0.7000 + 0.6000 – 0.4800   = 0.8200 =/ R.S =1The events of the residents of a household own 2 cars and income over $25,000 a year are not collectively exhaustive.4i)Let A be the number of student who did well on the Midterm.Let A’ be the number of student who did poorly on the Midterm.