Welcome, Guest: Register On Nairaland / LOGIN! / Trending / Recent / NewStats: 3,194,066 members, 7,953,251 topics. Date: Thursday, 19 September 2024 at 01:02 PM |
Nairaland Forum / Nairaland / General / Education / Great Statisticians In The House (318 Views)
Statisticians In The House Please Help Me / Mathematicians/Statisticians in the house, Kindly solve / Statisticians In The House (2) (3) (4)
Great Statisticians In The House by adebayo18015(m): 8:10pm On Nov 24, 2020 |
1. A bank received 3 bad cheques per day what is the probability that in a particular day the bank will receive (I)no bad cheque (II)at least 2 bad cheques (iii) at most 1 bad cheque. 2. In a manufacturing company,for every 15 bulbs produced 3 were found defective.if 5 bulbs were produced,find the probability that (I)none is defective (II)at most 4 are defective (iii) at least 4 are defective 3. Two coins were tossed once,what is the probability that (I) exactly two tails appear? (II) at least two tails appear? (III) at most three tails appear? 4. The marks scored by 1000 candidates in a test are normally distributed with a mean mark of 50 and a variance of 16,find the number of candidates whose marks are (I) below 25 (II) above 45 (III) between 25 and 45 5.In a sta 203 exam 60% of the student passed,if a random num of 3 students were selected for investigation.what is the probability that 1.no student passed 2. Atleast one student passed 3. At most one student passed 4. No student failed 5. Atleast one student failed 6. Atmost no student failed |
Re: Great Statisticians In The House by adebayo18015(m): 5:04am On Nov 25, 2020 |
This questions are base on probability in statistics |
Re: Great Statisticians In The House by adebayo18015(m): 11:22am On Nov 25, 2020 |
please I need help on this question |
Re: Great Statisticians In The House by Spartancosta(m): 1:23pm On Nov 25, 2020 |
adebayo18015:omoh I no sure say I go get time to solve all, (2) 15 produced,3 defective probability of defective bulbs=0.2 probability of good bulb=1-(0.2)=0.8 p=0.2 q=0.8 the probability of there being a defective bulb is 0,1,2...5 is given by successive terms of (q +p)^5 In a sample of 5, the probability of no defective bulb= q^5=(0.^5=0.32768 (ii) at most 4= probability of 0,1,2,3,4 probability of zero defective=q^5=0.32768 probability of one defective =5q⁴p =5(0.8⁴)(0.2) =0.4096 two defective=10q³p² =10(0 .8³)(0 2²) =0.2048 three=10q²p³ =10(0.8²)(0.2³) =0.0512 four=5qp⁴ =5(0.(0.2⁴) =0 0064 at most 4 =0+1+2+3+4(defective bulbs) =0.32768+0.4096+0.2048+0.0512+0.0064 =0.99968 (iii)at least 4 =sum of probability of having 1,2,3,4,5 defective bulbs =5q^5p + 10q³p² + 10q²p³+5qp⁴+p^5 already solved all except p^5 which is equal to (0.2)^5 =0.00032 therefore, probability of at least 4 defective bulbs equals the sum of having 1...5 defective bulbs =0.4096 + 0.2048 + 0.0512 + 0.0064 + 0.00032 =0.67242 |
Re: Great Statisticians In The House by Spartancosta(m): 1:32pm On Nov 25, 2020 |
adebayo18015:(3) When we toss two coins simultaneously then the possible of outcomes are: (two heads) or (one head and one tail) or (two tails) i.e., in short (H, H) or (H, T) or (T, T) respectively; where H is denoted for head and T is denoted for tail. (i) the outcomes are (H,H),(H,T),(T,T),(T,H)(total number of outcome is 4) probability of getting exactly two tails =1/4 (ii) atleast 2 tails at most 2 tails I think there's a mistake in the question it should be 3 coins |
Re: Great Statisticians In The House by Spartancosta(m): 2:15pm On Nov 25, 2020 |
adebayo18015:(5) 60% passed p(passed)=0.6 p(failed)=0.4 q=passed p=failed probability is given by successive terms of (q+p)³ =q³ +3q²p +3qp² +p³ none failed(all passed)=q³=0.6³=0.216 one failed(two passed)=3q²p=3(0.6²)(0.4)=0.432 two failed (one passed)=3qp²=3(0.6)(0.4²)=0.288 three failed (none passed)=p³=0.4³=0.064 in probability, at least = greater than or equal to at most= less than or equal to (i)none passed (all failed) =p³=0.4³=0.064 (ii) at least one passed equals probability of 1,2,3 passed= 0.288 + 0.432 + 0.216=0.936 (iii)at most one passed =p(0 passed) + p(one passed) =0.064+0.288=0.352 (iv) at least one failed =p(1 failed) + p(2 failed) + p( 3 failed) =0.432 + 0.288 + 0.064 =0.784 (v)at most none failed =q³=0.216 will drop the remaining solution if I'm chanced tonight |
Re: Great Statisticians In The House by adebayo18015(m): 8:09pm On Nov 25, 2020 |
Spartancosta: Thank you sir for interest in helping solve this problem. I really appreciate. More knowledge and grace to you sir. |
Re: Great Statisticians In The House by Spartancosta(m): 10:58pm On Nov 25, 2020 |
adebayo18015:Amen boss |
Re: Great Statisticians In The House by Spartancosta(m): 11:53pm On Nov 25, 2020 |
adebayo18015:(1) the question is a question on poisson distribution p(X=x)= (¥^x * e^-¥)/x! ¥= lambda (i) no bad cheques p(3,0)=(3^0 * e^-3)/0! p(no bad cheques=0.0498 (ii) at least 2 bad cheques = probability 2 and 3 bad cheques p(3,2)=3² * e^-³/2! =0.2240 p(3,3)= 3³ * e^-³/3! =0.2240 p(at least 2 bad cheques)=0.2240 + 0.2240 =0.448 (iii) at most one bad cheque =p(0) +p(1) p(3,0)=0.0498 p(3,1)= 3¹ * e^-³/1! =0.1494 p(at most one bad cheque)= 0.0498 + 0.1494 =0.1992 |
(1) (Reply)
Work In Poland / What Is Special About Your >Waec Result< That You Can't Forget? / Dr. Jen Jacob Bags Global Awards
(Go Up)
Sections: politics (1) business autos (1) jobs (1) career education (1) romance computers phones travel sports fashion health religion celebs tv-movies music-radio literature webmasters programming techmarket Links: (1) (2) (3) (4) (5) (6) (7) (8) (9) (10) Nairaland - Copyright © 2005 - 2024 Oluwaseun Osewa. All rights reserved. See How To Advertise. 37 |