Question

E. n = 3, p = 0.4 Question 5 According to a survey, 30% of employed women have never been married. If 10 employed women are selected at random, the probability (correct to four decimal places) that at least 7 have not been married is A. 0.0016 B. 0.0090 C. 0.0106 D. 0.9894 E. 0.9984

Answer 1

Answes- Given: and calculation : According to a suguel, 307. of employed women have never been maddied, hence, 30 = 0,3 =P 100 P be the probability of women never been morded we know that, P+9=1 Cie Total poobability sum is one) q=1 - P= 1-0.3 -0.7 q=0.7 q be the probability of women have been mosoſed Here, consideo X be the sandom variable X: No.of women not been modied. Here, lo employed select at gondom . N=10 From the above discussion we con conclude that Jhoe will be foial I no of employed goeteo than Cien)) and also theøe will be two poobabilities ie p = empolyed women neues been married eitheo q . employed maadled Hence, it follow s, Binomial Distojbution. je xa Bcn, P) xa B (10,0.3)

XO B (nip) Coup be the pasometer of dista) P (X=2) = n px qnoz ; X=011,2....n. o.w In=10 P=0.3 q=0.7 / Now, we hove poobability that of least have not been masdied. ie P (X77) • P(X21) = P(x-1) + PC X=8) + P(x= 9) + P(x =lo) 10-7 1- P1X2A) =(80,) 0:3)*(017) + (10.) (03) 8 60:7)-8 + (1) 60-35°C07)009+ (10.) 0:2) 0 (02) *** [ Ang consect to be fous decimal 1. PCX77) = 0.0090 + 0.0014 +0.0001 +(5.9048 x 10-6) 1. P(X7,72 = 0.01099 = 0.0106 Hence, Probability that at least 7 have not been maodied = P(X77) - 0.0016 ©