Question

please help me solve this! show me by hand or calculator commands, please no excel.

(5 points) Suppose the number of cell phones in a household has a binomial distribution with parameters n = 20 and p = 15%. Find the probability of a household having (a) 14 or 16 cell phones 4.29*10^-8 (b) 14 or fewer cell phones (c) 17 or more cell phones (d) fewer than 16 cell phones (e) more than 14 cell phones Note: You can eam partial credit on this problem

Answer 1

Answer:-

PCX=*) : "px quis 1850,1,2... Hete n=20, P: 0.15 = 4 : 0.85 a) pr 14 or 16 cell phones) - pex = 147 + prx=16). = 2047460.159'46o.ps)6 + 2016 (0.18) 16 (0.2374 = 4.2841x100 = 0.00000004284 PC14 OY 16 cell phones) O 0 14 or fewer cell phones p (14 or fewer cell phones) = P(x214) p(x $14) : p(x=0) + P(x-1) + ... + p (X=14) = 0.0387 + 0.1367 + ... + 0.00000 PCX < 14): 0. 9 9 9 9 ♡ P(x2,10):1- pcxc107) :1- plx = 0) + ... + p (X=16) 4 - |- Ý 0:03 87 + + + 06 0 0% > = 0.000001 PCX 7,17) u o o pix <16) = P(x=0) + ... + p (x = 15) P(x216) : 0.9999 . e) PCX714) : 1- prx514) = 1- px : 0) + ... + Plx1413 1- {0.0387+ ... + 0.000001} pex>14) 0000001 P(X>14) no IP (X>14 hot

Remember , if the value upto 4th decimal is zero ( i.e , if p(x) = 0.000001 ) then usually we consider probability zero.