Question #108806
A sample of 200 persons is asked about their handedness. A two way table of observed counts follow
1
Expert's answer
2020-04-10T17:21:08-0400
LefthandedRighthandedTotalMen117990Women9101110Total20180200\def\arraystretch{1.5} \begin{array}{c:c:c:c} & Left-handed & Right-handed& Total\\ \hline Men & 11 &79 & 90 \\ \hdashline Women & 9 & 101 & 110 \\ \hdashline Total & 20 & 180 & 200 \end{array}

Let M=M= selected person is a men, W=W= selected person is a women, L=L= selected person is left-handed, R=R= selected person is right-handed.

If one person is randomly selected

a. P(W)=110200=1120=0.55P(W)=\dfrac{110}{200}=\dfrac{11}{20}=0.55


b. P(R)=180200=910=0.9P(R)=\dfrac{180}{200}=\dfrac{9}{10}=0.9


c. P(MR)=79200=0.395P(M\cap R)=\dfrac{79}{200}=0.395


d. P(WL)=20+1109200=121200=0.605P(W\cup L)=\dfrac{20+110-9}{200}=\dfrac{121}{200}=0.605


e. P(ML)=1120=0.55P(M|L)=\dfrac{11}{20}=0.55


f. P(RW)=1011100.9182P(R|W)=\dfrac{101}{110}\approx0.9182


If one person is randomly selected with replacement


P(first ML, second MR)=1120079200=P(first\ M\cap L, \ second\ M\cap R )=\dfrac{11}{200}\cdot\dfrac{79}{200}==86940000=0.021725=\dfrac{869}{40000}=0.021725

P(first WL, second WL)=92009200=P(first\ W\cap L, \ second\ W\cap L )=\dfrac{9}{200}\cdot\dfrac{9}{200}==8140000=0.002025=\dfrac{81}{40000}=0.002025

If one person is randomly selected without replacement


P(first ML, second MR)=1120079199=P(first\ M\cap L, \ second\ M\cap R )=\dfrac{11}{200}\cdot\dfrac{79}{199}==869398000.021834=\dfrac{869}{39800}\approx0.021834


P(first WL, second WL)=92009199=P(first\ W\cap L, \ second\ W\cap L )=\dfrac{9}{200}\cdot\dfrac{9}{199}==81398000.002035=\dfrac{81}{39800}\approx0.002035



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