Determine the probability of ##P(x>4)##

[chwala]

Homework Statement

see attached

Relevant Equations

Knowledge of distributions

Find the solution here;

Ok my interest is on part (b) and (c) only. Let's start with (b),
My take is,
$$\int_4^5 \dfrac{2}{75}x\,dx=\left.[\frac{x^2}{75}]\right|_4^5$$
$$=(0.33333333-0.21333333)+\frac{2}{15}×5$$
$$=0.12+0.6666666666=0.78666666$$
note that at $f(x)$=$\dfrac{2}{15}$, the probability value will be the area subtended by values of $x$ from $x=5$ to $10$ hence $10-5=5$...in short, probability $P(x>4)$ at this point will be the area subtended by the straight line within the given domain $5<x≤10$.

 

[nuuskur]

You are confusing the reader with all those decimals. The general rule is
<br /> \mathbb P(X\leqslant a) = \int _{-\infty}^a f(t)\mathrm{d}t,<br />
if $X$ has a probability density $f$. For b) apply what you know about the probabilities of an event and its opposite event.

If you are looking for feedback, explain your thought process that leads you to a certain computation. The probability you are looking for in c) is that at least one battery fails within the next $40$ hours, because both batteries are required for the device to work.

 

[pbuk]

You are confusing the reader with all those decimals.

Yes, you should not use decimals in a problem like this; express your answer as a rational number. And look up the meaning of the word "subtend", it has nothing to do with area.

As to method: you have correctly worked out $P(x>4)$ by evaluating an integral at 2 points and adding on an area of a rectangle. Can you think of a way to arrive at the same answer simply by calculating the area of a triangle? (Perhaps it would have been better not to ignore part (a).)

 

[chwala]

Yes, you should not use decimals in a problem like this; express your answer as a rational number. And look up the meaning of the word "subtend", it has nothing to do with area.

As to method: you have correctly worked out $P(x>4)$ by evaluating an integral at 2 points and adding on an area of a rectangle. Can you think of a way to arrive at the same answer simply by calculating the area of a triangle? (Perhaps it would have been better not to ignore part (a).)

Ok I will look at that...I had already seen how the graph looks like in part (a)...we have a right angle triangle and a rectangle...noted on 'subtended'...

 

[pbuk]

Ok I will look at that...I had already seen how the graph looks like in part (a)...we have a right angle triangle and a rectangle...noted on 'subtended'...

OK, so if you divide the area under the graph into two pieces by drawing a vertical line at $x = 4$, you have a triangle with a known base and a height which can be calculated with a trivial substitution. You also have an irregular pentagon. For which piece is it easier to calculate the area?

 

[chwala]

Thanks @pbuk ...kindly let me look at this later...when I am through with other tasks...

 

[chwala]

Ok we shall have,
$\left[(\dfrac{1}{2}×5×\dfrac{2}{15})-(\dfrac{1}{2}×4×\dfrac{8}{75})\right]+5×\dfrac{2}{15}=\dfrac{59}{75}$

 

[pbuk]

Ok we shall have,
$\left[(\dfrac{1}{2}×5×\dfrac{2}{15})-(\dfrac{1}{2}×4×\dfrac{8}{75})\right]+5×\dfrac{2}{15}=\dfrac{59}{75}$

You are still over-complicating this, you should be able to see on inspection that
$P(x \gt 4) = 1 - P(x \le 4) = 1 - \dfrac{1}{2}×4×\dfrac{8}{75} = \dfrac{59}{75}$.

 

[chwala]

You are still over-complicating this, you should be able to see on inspection that
$P(x \gt 4) = 1 - P(x \le 4) = 1 - \dfrac{1}{2}×4×\dfrac{8}{75} = \dfrac{59}{75}$.

Thanks mate.

 

[chwala]

Find the solution below from the Markscheme.

Now for part (c), i was not understanding how they came up with using part (b) to realize the solution i.e

...then i realized from the question itself indicate...in tens of hours... implying in our case, $4$ × tens of hours = $40$ hours. The steps to final solution is very much clear to me. Cheers guys.