Question

someone please help me out i will really apperaciate you

(1 point) Suppose the number of children in a household has a binomial distribution with parameters n = 23, and p = 60 %. Find the probability of a household having: (a) 19 or 21 children (b) 19 or fewer children (c) 19 or more children (d) fewer than 21 children (e) more than 19 children Remark: If necessary, round off your result to 5 decimal places.

Answer 1

(1) X: Number of children in a house hold Xw Binomial (n=23, p=0.6) or 21 children) (a) p (Household having 19 or P(x - 3) + (x =))) =Ggp 19 (0.4)23-19 23 C20, (0.6)241 (0.4) n. 201 19 n-1g 281 + 아 291 q 23 Gg (0.6) + 0.01381 + 00414 0.00089 = 0.01795 0.01470 Probability that the household having 19 or 21 children - Fororogo [0, 01470 (b) PC House bold having 19 or fewer children) - P ( X < 9) ) - ( x > ) ] - [ P(X=20) + ( X = 2\) + ( X = 22) + ( X = 23) Co (0.6)20 (0.4923-20 3 Cg (6.62 (0.4) 23-21 (0.6) 22 (0.4223-22 (0.6) 2 (0.4) 23-23 23 23 20 + 23 C qr + + 23 C 23 = 0.00414 + 0.00089 + 0.00012 + 0.0000@1] 0.99484 Probability that the household having 19 or fewer children 0.99484 -

= - 23 23(23 (0.6) (c) PC Household having 19 or more children) =P ( X 219) PCX=19) + P(X=20) + P ( X =21) + P(X=22) + P (X=23) 23 Cg (0.6) 19 (0.4) 23-19 23 C20 (0.6) 20 (0.4323-20 0.01381 to 004 14 t 8 9 t 0012+ 0.00001 0.01795 + 0.00102 = 0.018 9 7 Probability that the household having 19 or more children 0.018971 + + 23 3-23 OOO (d) P(Household having Fewer than 21 children.) =P ( X <21) =) - P ( x 721) = 1 - [PCX = 21) + PL X =22) + P(X=237 = 1 -[0.00089 +0.00012 + 0.00001 = 0.99898 that the household having Fewer than 21 children = [0.99898 Probability = (e) PC Household baving Presser more than 1g children) =P ( X>19) = P ( X = 20) + P(x-21) + PIX=22) + P(X=23) = 0.00414 + 000089 +0.00012 + 0.00001 = 0.00516 Probability that the household having more than 19 children (0.00516)

ane (2) When a pair of dice is rolled, the total number of outcomes 36. Sum of 7 can be obtained 6 times as the possible outcomes ane, (1,6), (2,5), (3, 4), (43), (5,2), (6, 1) Since the experiment is Binomial experiment le either you get above possible' 6 outcomes or not. Probability of success = p = Probability of failure 6 36 6 1 9 36 36 Number of trials = n = 7

Probability of throwing a sum of 7 at least 4 times is 7-4 3 ? C4 P 9 Cu 30 36 36 = 0.0563 The probability of throwing a least 4 times 0.01563 sum of 7 at

(3) P Probability of best rate = 0.8 n=number of flights = 20 X = Number of flights arriving on time, flights arrive on time (a) P( at least 19 = P(X29) =P ( X =19) + P1 X=20) 20 (68) 20 (0.2/20-20 20 20 6g (0.8) 19 (0.2) 20-19+ = 0.05765 t 0.01 153 = 0.06918 Probability that at least 19 flights arrive on time 10.06918 (b) P (Two flights will be late) =/- P ( Two Flights arrive 1-P (X=2) = ) - 2002 008)? (0.2) 18 on time) Yes it would be unusal for southwest to have 2 flights arrive late.