Question

70% of all Americans live in cities with population greater than 100,000 people. If 35 Americans are randomly selected find the probability that a. Exactly 24 of them live in cities with population greater than 100,000 people. b. At most 23 of them live in cities with population greater than 100,000 people. e. At least 26 of them live in cities with population greater than 100,000 people. d. Between 20 and 27 (including 20 and 27) of them live in cities with population greater than 100,000 people. Hint: Hint Video on Finding Binomial Probabilities [+] Points possible: 1 This is attempt 1 of 2 Message instructor about this question Submit

Answer 1

Solution! Giren p=0.70, n=35 x ZXNB(nip) plx=2)5) pb-ph-z 5x=0,),2,--. jo.co @ p(x=24)= (25) 6.7034 610.70) 24 35- 24 = ( 35 ) (60.70,24 (1-0.70)35-24 /P(X=24) = 0.1416 0.1416 is the probability that exactly 24 of them live in cities with population greater than 100,000 people. p(x<23) = 0.3484 SPI 0.3484 is the of them live greater than probability that cet most 23 in cities with population 100,000 people.

© PCX226) = )- pex(26) = 1- P(x525) E1 - 0-6354 = 0.3646 pex>26) = 0.3646 0.3646 is the probability that 26 of them live in cities with greater than 100,000 people. at least population p(20€ X<27) Ep(x=20) + PCX=21) + P(X=22) 7 p(x=23)+ PCX= 24)+ p(x=25) + p(x226) +p (X=27) = 0.0372 +0.0619+ 0.092ot 0.1214+ 0.1416 + 0.1454+ 0.1305+ 0.1015 Sat 21 = 0.8315 p(20<x<27) = 10.8315