Answer to Question #156090 in Statistics and Probability for Mathy

Question

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)

Answer 1

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) $(a)$ Let $C$ and $S$ be two events defined as follows,

$C=$ child survived the sinking

$S=$ someone survived the sinking

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

$P(S)=\frac{966}{2942}=\frac{483}{1471}$ and $P(C\cap S)=\frac{86}{2942}=\frac {43}{1471}$

Therefore $P(C\cup S)=P(C)+P(S)-P(C\cap S)=\frac {43}{1471}+\frac{483}{1471}-\frac {43}{1471}=\frac {483}{1471}$

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

(b) Let $W$ and $A$ be two events defined as follows,

$W=$ women did not survived the sinking

$A=$ someone did not survived the sinking

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

$P(A)=\frac{1976}{2942}=\frac{988}{1471}$ and $P(W\cap A)=\frac{418}{2942}=\frac{209}{1471}$

Therefore $P(W\cup A)=P(W)+P(A)-P(W\cap A)=\frac {209}{1471}+\frac{988}{1471}-\frac {209}{1471}=\frac {988}{1471}$

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

(c) Let $M,W$ and $A$ are three events defined as follows:

$M=$ man did not survived the sinking

$W=$ women did not survived the sinking

$A=$ someone did not survived the sinking

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

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

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

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

$=\frac{P(M\cap A)+P(W\cap A)-P(M\cap W\cap A)}{P(A)}$

Now $P(M\cap A)=\frac{1475}{2942}$ , $P(W\cap A)= \frac{418}{2942}$ and $P(M\cap W\cap A)=0$ [as $M$ and $W$ mutually independent]

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

Which is the required probability.

(d) Let $M$ and $A$ be two events such that

$M=$ man who died

$A=$ someone who died

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

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

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

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