I am having problems with understanding the solution to the following exercise, which is as follows:
The weights of a large number of miniature poo- dles are approximately normally distributed with a mean of 8 kilograms and a standard deviation of 0.9 kilogram. If measurements are recorded to the nearest tenth of a kilogram, find the fraction of these poodles with weights (a) over 9.5 kilograms; (b) of at most 8.6 kilograms; (c) between 7.3 and 9.1 kilograms inclusive
now, the solution is:
(a) $ = P(X > 9.55) = P(Z > 1.72) = 0.0427. $
(b) $ = P(X < 8.65) = P(Z < 0.72) = 0.7642. $
(c) $= P(7.25 < X < 9.15) = P(−0.83 < Z < 1.28) = 0.8997 − 0.2033 = 0.6964. $
Now, I have successfully completed the exercise a). The exercise said that the measurements are recorded to the nearest tenth of a kilogram. What I did in a) was add half a kilogram to 9.5 which gives us 9.55 because it's the first higher value (>) which is recorded
However, in b) my solution was
b) $P(X≤8.6)$,
and in
c) $P(7.25 ≤ X ≤ 9.00).$
My method was to add half a kilogram to the x value whenever the sign was "greater or equal to", and to subtract half a kilogram from the x value whenever the sign was "less or equal to".
Obviously my "trick" did not work in those two exercises and I have no clue why. Can anyone help?
One way to conceptualize the model here is to think of two random variables, say $Y$, which is the weighed value to the nearest $0.1$ kg, and the true value $X$, which is not observable due to the limited instrument precision. So if $Y = 8.2$ kg observed, this means that $X$ is some value in the interval $[8.15, 8.25)$.
Then what we are given is that $$X \sim \operatorname{Normal}(\mu = 8, \sigma = 0.9),$$ not $Y$, because the latter is a rounded value, thus it is not possible to observe $Y = 8.2353$, for example. You'd observe $Y = 8.2$ instead.
We are asked in Part (a) to compute $\Pr[Y > 9.5]$ the probability of poodles whose observed weights exceed $9.5$ kg. This of course must correspond to $\Pr[X \ge 9.55]$, and since the outcome $X = 9.55$ has measure zero, this is equivalent to computing $\Pr[X > 9.55]$. The other parts of the question are handled similarly.
Your reasoning in Part (a) doesn't make sense. $9.55$ is not the next "half-kilogram" value exceeding $9.5$. Half a kilogram would be $0.5$, thus $9.5 + 0.5 = 10$. If you instead meant the next $0.05$ kilogram--i.e., a twentieth of a kilogram, then you have to justify why this works; e.g., why add instead of subtract. This part of your reasoning is what is flawed.
To understand the relationship between the desired probability for $Y$ and the corresponding set of outcomes for $X$, all one needs to do is think about what $X$-values will round to a $Y$-value that satisfies the inequality. For example, if we want $$\Pr[8.3 \le Y < 9.2],$$ then the smallest $X$-value that rounds up to $Y = 8.3$ is $X = 8.25$. The largest $X$-value that rounds down to $Y = 9.1$ is $X = 9.15$ (technically there is no supremum but any number less than $9.15$ results in $Y \le 9.1$). The key here is that if the inequality for $Y$ is strict--for instance, $Y < 9.2$, then because $Y$ is discrete, we can't observe $Y = 9.199999$; thus $$Y < 9.2 \iff Y \le 9.1.$$ Then we find the biggest $X$ that rounds down to this value.