Answer in Statistics and Probability for Agetro #248056
October 24th, 2022
In a particular city, one in three families have a phone in their home. If 90 families are chosen at random, calculate the probability that at least 30 of them will have a phone.
Let “X=” the number of families which have a phone: “Xsim Bin(n, p).”
Given “n=90, p=dfrac{1}{3}, q=1-p=1-dfrac{1}{3}=dfrac{2}{3}.”
“P(Xgeq30)=1-P(X<30)”
“=1-displaystylesum_{x=0}^{29}dbinom{90}{x}(dfrac{1}{3})^{x}(dfrac{2}{3})^{90-x}”
“approx0.53956753215”“np=90(dfrac{1}{3})=30geq10, nq=90(dfrac{2}{3})=60geq10”
Then “X” has approximately a normal distribution with “mu=np=30” and “sigma=sqrt{npq}=sqrt{90(dfrac{1}{3})(dfrac{2}{3})}=2sqrt{5}.”
“P(Xgeq30)=P(X>29.5)=1-P(Xleq29.5)”
“=1-P(Zleqdfrac{29.5-30}{2sqrt{5}})approx1-P(Zleq-0.1118)”
“approx0.544509”
