Question #156090

1.What is the level of measurement for

The height of people is a sample of airplane passengers.(nominal,ratio,interval,ordinal) why


2.Summary of passengers in ferry.

Man Woman Boy Girl

Survive 447 433 44 42

Dead 1475 418 50 33


i.the probability of getting a child or someone who survived the sinking(with steps)


ii.the probability of getting a woman or someone who did not survived the sinking. (with steps)


iii.What is the probability of getting a man or woman, given that the randomly selected person is someone who died? (with steps)


iv.If we randomly select someone who died, what is the probability of getting a man? (with steps)


1
Expert's answer
2021-01-18T16:52:47-0500

1) The height of people is a sample of airplane passengers is a ratio level data.Because the interval between the numbers are comparable and there is an absolute zero for height.It is likely interval scale data but it has a 0 point.you will not have negative value in ratio level data. It takes care of ratio problem and gives most information.


2) (a)(a) Let CC and SS be two events defined as follows,

C=C= child survived the sinking

S=S= someone survived the sinking

Then P(C)=862942=431471P(C)=\frac{86}{2942}=\frac{43}{1471}

P(S)=9662942=4831471P(S)=\frac{966}{2942}=\frac{483}{1471} and P(CS)=862942=431471P(C\cap S)=\frac{86}{2942}=\frac {43}{1471}

Therefore P(CS)=P(C)+P(S)P(CS)=431471+4831471431471=4831471P(C\cup S)=P(C)+P(S)-P(C\cap S)=\frac {43}{1471}+\frac{483}{1471}-\frac {43}{1471}=\frac {483}{1471}

\therefore The probability of getting a child or someone who survived the sinking is =4831471=\frac{483}{1471} .


(b) Let WW and AA be two events defined as follows,

W=W= women did not survived the sinking

A=A= someone did not survived the sinking

Then P(W)=4182942=2091471P(W)=\frac{418}{2942}=\frac{209}{1471}

P(A)=19762942=9881471P(A)=\frac{1976}{2942}=\frac{988}{1471} and P(WA)=4182942=2091471P(W\cap A)=\frac{418}{2942}=\frac{209}{1471}

Therefore P(WA)=P(W)+P(A)P(WA)=2091471+98814712091471=9881471P(W\cup A)=P(W)+P(A)-P(W\cap A)=\frac {209}{1471}+\frac{988}{1471}-\frac {209}{1471}=\frac {988}{1471}

\therefore The probability of getting a women or someone who did not survived the sinking is =9881472=\frac{988}{1472}


(c) Let M,WM,W and AA are three events defined as follows:

M=M= man did not survived the sinking

W=W= women did not survived the sinking

A=A= someone did not survived the sinking

Then, P(A)=19762942=9881471P(A)=\frac {1976}{2942}=\frac {988}{1471}

Now we have to find P[(MW)/A]P[({M\cup W})/{A}]

\therefore P[(MW)/A]=P[(MW)A]P(A)P[({M\cup W})/{A}]=\frac{P[({M\cup W})\cap{A}]}{P(A)}

=P[(MA)(WA)]P(A)=\frac{P[(M\cap A)\cup (W\cap A)]}{P(A)}

=P(MA)+P(WA)P(MWA)P(A)=\frac{P(M\cap A)+P(W\cap A)-P(M\cap W\cap A)}{P(A)}

Now P(MA)=14752942P(M\cap A)=\frac{1475}{2942} , P(WA)=4182942P(W\cap A)= \frac{418}{2942} and P(MWA)=0P(M\cap W\cap A)=0 [as MM and WW mutually independent]

P[(MW)/A]=14752942+418294219762942\therefore P[({M\cup W})/{A}]=\frac{\frac{1475}{2942}+\frac{418}{2942}}{\frac{1976}{2942}} =18931976=\frac{1893}{1976}

Which is the required probability.


(d) Let MM and AA be two events such that

M=M= man who died

A=A= someone who died

Then we have to find P(M/A)P(M/A)

Now P(M/A)=P(MA)P(A)P(M/A)=\frac{P(M\cap A)}{P(A)}

Here P(MA)=14752942P(M\cap A)=\frac{1475}{2942} , P(A)=19762942P(A)=\frac{1976}{2942}

Therefore the required probability P(M/A)=1475294219762942=14751976P(M/A)=\frac{\frac{1475}{2942}}{\frac{1976}{2942}}=\frac{1475}{1976}


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