Question

Question 2 (2 points) Suppose Tori has an unfair coin which lands on Tails with probability 0.28 when flipped. If she flips the coin 10 times, find each of the following: The standard deviation of the number of Tails PExactly 1 Tail) 1. 0.1798 2. 0.7021 PExactly 4 Tails) 3. 2.8 PiNo Tails) 4 0.1181 P[No more than 3 Tails) 5. 0.1456 PlMore than 3 Tails) 6. 0.0374 7. 0.9626 PlAt least 1 Tail) 8. 0.2979 PAt least 5 Tails) 9. 0.4378 The mean number of Tails 10. 1.42

Answer 1

guestion-1: n=10 X: No. of tails Pro28 9-02 1.e xaBcn, P) XGB (10,0.28) P (x=x) = pr qu-x ; o<Pel C=0,,2 - o 0. P (X=X) = 10 c (0.28) * (0.72) out 1) PCExactly 4 Toil) = POX= = 100, (0.28)' (0.72) = 0.1455 2) PC Exactly 4 Toils) = P(x=4) = 100, (0.28) 4 (0:72) 10-4 = 0.1798 37 PINO Tail) = P(x=0) - 10cm (0.28)° (0.72) = 0.0374 4) PC No more than 310ils) = P(X<3) = P(X=0) + P(x-1) + P(X-2) +P(x-3) = 10cm (028 ) °60.72) + 10 (0.28) 10:42)|-| + 10c2 (0.28)2(0:72) 8 + 10% (0.28) C0-42) 7 0.0374 + 0.1455 t 0.2567 +0.2642 10.2018

5) P (Mode thon 3 Toils) - P(X73) -1-PIX83) - 1-0.7018 = 0.2982 from @ 6) PC At least 1 Toil) = P(x>1) -P-PCX<1 - 1-PCX=o) - 1- 0.087 4 = 0.9626 7) PC At leost 5 Tails) = P(X75) = 1-PCX (5) = 1-[(x-03 +P[X-1] +P [X-21 +P [X-3] + P(x=417 - 1- [0.03747 0.1455 +0.2547 +0.26427 0.1798) -1- 0.8816 = 0.1184 8) The mean numbeo of Tails. XSB (IP) E(x)=nP = 10x0.28 = 2.8 9) The stondood deviation of the no of tails & Xe Brnipl 9.D(x) = Jual е у ҮРq c':'ycx) = Pe - Vloxo-28 X0,72 114198

Question-2- X; -Goals foequency (fi) Xifi 16 3 99 8 0 4 0 164 5 25 l Totals 2 Celior 30 Joo J - 72 604 25 1581 182 EX) = {xifi = 58 = 2,6363 {fi 22 Cov =X Population Standood deviation + NVO VX) = 5.0(X) - Exifi --X2 = 182 - (2,6363) 2 <fi 01 - 1:1500 / 22 S. Dex) = 1.1500 S .