Question

WINST Calle Timelimit hours 41-08 remaining 10% of all Americans live in poverty. If 43 Americans are randomly selected, find the probability that a. Exactly 6 of them live in poverty. b. At most 5 of them live in poverty c. At least 4 of them live in poverty d. Between 3 and 10 (including 3 and 10) of them live in poverty Hint Hint Textbook Pages Video on Expected Value [+] Get Help Points possible: 1 This is attempt 1 of 2 Subm 2020 XYZ Homework I Privacy Policy I ADA Compliance PDF 1 Contact Us xyzHomework is powered by Malhas 2006-2020 by David Lippman

Answer 1

Answer: Given that, probability that an americances live in poverty P=0.10 probability that an american doesot live in poverty q=1- pol-0.10=0.90 n=43 Q) ? (X 6) p[x -x]= [1] Br (2) 0-2 3.P{ x=6] = 0] (0.105 (0.90) 43-6 3 ./P(x=6] = 0.1236 b) PCX35) - PCX = 0) + PCX=1) + PCX=2) +PCX 23) + P(x = 4) +PCX25) = 0.0108 + 0.0515 + 0.1201 + 0.1824 +0.20 277 0.1757 PCXS 5) -0.7432 P (X²4) - - $ 1- P[x< 3] 1-{P[x =o) + P(x=13 + PC x= 2 ) +P[x = 3)] 31-(0.0108 +0.0515 + 0.1201+0.1824] - 1-0.3648 (P(x24] = 0.6352 d) PC*6 35x310) - P(x=374p(x-43+9x -9]+ P(x26] + P(x = 7}+ ...n+ P(x=10] 0.1824 +0. 2027 +0.1757 +0.1236 +0.0726 + 0.0363+ (0.0157)+0.0060 (PC35x510) 0-815 - Thank you -