Answer to Question #105000 in Statistics and Probability for Riya

Question

State which of the following statements are True and which are False. In case of the false
statement, give the correct statement: (10)
i) The probability of any event is always a proper fraction.
ii) with usual notations, show that
Cov aX( + bY cX, + dY) = acVar(X) + bdVar(Y) + ad( + bc)Cov( ,X Y)
iii) In the chi-square test of goodness of fit, if the calculated value of 2
χ is zero then the
fit is a bad fit.
iv) For binomial distribution, Mean = Mode = Median.
v) Mean lies between median and mode.

Answer 1

i) The probability of any event is always a proper fraction.  

False. For any event $0\leq P(A)\leq 1.$ A probability of means the event will never occur. A probability of $1$ means that the event will occur 100% of the time.

ii) with usual notations, show that

Cov (aX+bY, cX+dY) = acVar(X) + bdVar(Y) + (ad+bc)Cov(X,Y) 

For constants $a,b,c,d$ and random variables $X, Y$

$$Cov(aX+bY, cX+dY)=$$ $$=acCov(X,X)+adCov(X,Y)+bcCov(Y,X)+bdCov(Y,Y)=$$ $$=acVar(X)+adCov(X,Y)+bcCov(Y,X)+bdVar(Y)=$$ $$=acVar(X)+bdVar(Y)+(ad+bc)Cov(X,Y)$$

iii) False.

$\chi^2=0$ corresponds to a perfect fit.

iv) For binomial distribution, Mean = Mode = Median.  

False.

The mean, median, and mode coincide for binomial distribution and equal $np$ only if  $np$ is an integer.

v) False.

In a moderately skewed distribution median lies between mean and mode.