# How would you find the variance using this formula (1 Viewer)

#### Eagle Mum

##### Well-Known Member
Probability density functions weren’t in the curriculum when I was in high school, but I don’t see how it’s possible to prove this function is a valid PDF because its values aren’t stated for other values of x, outside the range of 0 to 5. Otherwise, assuming its value is zero at all other values, integrate f(x) to find the areas under the curve between 0-2 and 2-5, their sum being 1 (the area under the curve between 0-2 is (2-0)/8 = 1/4 and the area under the curve between 2-5 is (5-2)/4 = 3/4).

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• jimmysmith560

##### Active Member
Probability density functions weren’t in the curriculum when I was in high school, but I don’t see how it’s possible to prove this function is a valid PDF because it’s values aren’t stated for other values of x, outside the range of 0 to 5.
I think you just assume outside is 0.

Since when u find definite integral it's exactly 1, probably what was intended.

• Eagle Mum and jimmysmith560

#### jimmysmith560

##### Le Phénix Trilingue
Moderator
Probability density functions weren’t in the curriculum when I was in high school
I don't remember studying something like that in 2U at all either, maybe it's from 2020 and beyond.

• Eagle Mum

#### Eagle Mum

##### Well-Known Member
Thanks - I added the rest of my post at the same time as your post.

#### Eagle Mum

##### Well-Known Member
I don't remember studying something like that in 2U at all either, maybe it's from 2020 and beyond.
It’s quite a nice intersection of statistics, probability and integral calculus and shows how various branches, often taught as discrete topics, are interconnected.

#### Eagle Mum

##### Well-Known Member
µ = E(x) = ∫ xf(x)dx from 0 to 5

[(x^2)/16] for 0<x<2 plus [(x^2)/8] for 2<x<5

= 4/16 + (25-4)/8 = 23/8

E(x^2) = ∫ x^2 f(x)dx from 0 to 5

[(x^3/24] for 0<x<2 plus [(x^3/12] for 2<x<5

= 8/24 + (125-8)/12 = 121/12

V(x) = 121/12 - (23/8)^2 = 349/192

• #### Hivaclibtibcharkwa

##### Well-Known Member
µ = E(x) = ∫ xf(x)dx from 0 to 5

[(x^2)/16] for 0<x<2 plus [(x^2)/8] for 2<x<5

= 4/16 + (25-4)/8 = 23/8

E(x^2) = ∫ x^2 f(x)dx from 0 to 5

[(x^3/24] for 0<x<2 plus [(x^3/12] for 2<x<5

= 8/24 + (125-8)/12 = 121/12

V(x) = 121/12 - (23/8)^2 = 349/192
Ah I see thanks again!

#### CM_Tutor

##### Moderator
Moderator
Probability density functions weren’t in the curriculum when I was in high school, but I don’t see how it’s possible to prove this function is a valid PDF because its values aren’t stated for other values of x, outside the range of 0 to 5. Otherwise, assuming its value is zero at all other values, integrate f(x) to find the areas under the curve between 0-2 and 2-5, their sum being 1 (the area under the curve between 0-2 is (2-0)/8 = 1/4 and the area under the curve between 2-5 is (5-2)/4 = 3/4).
As a new topic, there are many questions that are poorly worded. Bill Pender posted comments on whether a PDF is defined on a closed interval or over the entire set of reals prior to the 2020 exam. He noted that the syllabus dot point concerning CDFs only considers PDFs defined on [a, b], but that question 31 of the NESA Specimen paper required calculation of a CDF from a PDF that is non-zero on [1, 4] but zero elsewhere, seemingly indicating an intention from NESA that is wider that the syllabus dot point would suggest. He also posted some examples that students using the Cambridge texts should consider. It is sad but inevitable that ambiguities arise when a new syllabus is in its early years.

I don't remember studying something like that in 2U at all either, maybe it's from 2020 and beyond.
Yes, it is new in Advanced Maths from 2020 onwards

• Eagle Mum