In chapter 8, Question 8.5, part (vi), I am unable to understand how to get the lower limit of the range as 0.5.?
It's because of the continuity correction. Here we need to get the value of \(5<X\le{10}\), we will segregated it as \(X>5\) which transforms as by using continuity correction \(X>5.5\) and \(X\le{10}\) which transforms as \(X<10.5\). Therefore the whole transform occurs as \(5.5<X<10.5\). We used to use the continuity correction because when dealing with the normal approximation to the Binomial and Poisson Distribution, which are both discrete distribution is being approximated by a continuous one. When using such an approximation the change from discrete to continuous must be allowed for. To compensate for the ‘gaps’ between the bars of the Discrete Distributions, we suppose that they are actually rounded to the nearest integer. For example, the \(X=6\) bar is assumed to represent values between \(X=5.5\) and \(X=6.5\). From here the \(0.5\) concept comes.
Mr. Gupta is absolutely correct. It was my faux pas considering the (vi) as (iv). They have approximated the \(4.8\) as \(4.5\).
Can you please explain me how to do the continuity correction for this part? For all other parts its either whole numbers or .5 fractions.
We got a problem as \(4\le10X\le48\), so \(10X\) takes values as, if \(X=1\) then \(10\), \(X=2\) then \(20\), \(X=3\) then \(30\), if \(X=4\) then \(40\). It must me within the limit from \(4-48\), so we could not consider beyond \(40\) or below \(10\). Therefore the actual possible figures for \(10X\) would be \(10,20,30,40\). Now we are arranging again the problem as \(10\le10X\le40\). Now by using the continuity correction we get \(10\le10X\) as \(5<10X\) which is \(0.5<X\) and \(10X\le40\) which become by using continuity correction as \(10X<45\) and \(X<4.5\). Therefore the entire data became as \(0.5\le{X}\le4.5\).
