gsce statistics + equation of a line

[fallow wind]

unsure on the working

[fallow wind]

<@&791435371564892232>

[craggy oracle]

in order to workout an estimate of the mean from a grouped frequency table you follow these steps

multiply the midpoint of each group by the frequency. and add all these values up

then divide by the total frequency

[craggy oracle]

try to do that in terms of x and set equal to 38.25, forming an equation which can be solved

[fallow wind]

ohh ok thanks

what about the 2nd q

[sudden oar]

linear interpolation

actually is this gcse statistics or gcse maths

i can’t tell

[fallow wind]

it’s maths

But it’s just a harder question

[fallow wind]

What even is that

[ionic surge]

like

shows the position of a data value

out of however many there are

[sudden oar]

icl i think linear interpolation is the only way of doing it

and that's not in gcse maths, it's in gcse stats

[fallow wind]

can u explain it anyway

i got the q from corbett maths higher plus for practice

[sudden oar]

i would recommend learning how to find the median first by linear interpolation

just so u get the concept

yk what let's just do the question

3 steps

1. find lq by linear interpolation
2. find uq by linear interpolation
3. find iqr by subtracting both

firstly we need to find the lower quartile interval

so n/4

not n+1 cos it's grouped data

u can do that, u still get the marks. but we generally use n and not n+1 for grouped data

here's an example for finding median by linear interpolation from a histogram

firstly just make ur frequency table from the histogram

imagine the histogram wasn't there

find the median position by doing n/2, in the case of lower quartile, u do n/4

[fallow wind]

ohh so like lets say there is 60 things, then i look at the class width and do the 30th value

[sudden oar]

for median yes

and then see where that lies using the cumulative frequency

[sudden oar]

e.g here, if the median position is between 0-80 (inclusive), the median interval is 0 <= x < 20

81-180 (inclusive) would be 20 <= x < 30

etc

[sudden oar]

just have a look at this and tell me if u don't understand anything

[fallow wind]

how did they know to use the previous class width

i get the rest

to some extent but

[sudden oar]

u mean previous cumulative frequency?

previous c.f = previous cumulative frequency

that's just how the formula works

[fallow wind]

ohhh

[sudden oar]

it will be ur median position or lq position whatever - the previous cf of that interval on the numerator

[fallow wind]

ahh okay

ill do more practise on it anyway dw

[fallow wind]

how would i answer the line question tho that seems simple but idk i didnt get the right answer

[sudden oar]

find gradient of CD

and also find M by using the formula
(x1+x2)/2 to find the x value for the midpoint
and (y1+y2)/2 to find the y value for the midpoint

find gradient of AB using the fact that CD and AB are perpendicular

[fallow wind]

which x and y coordinate am i using again

[sudden oar]

points C and D, the given points which lie on the same straight line

[sudden oar]

perpendicular implies the gradient of AB is the negative reciprocal of CD

i.e a/b = -b/a

[fallow wind]

ahh okay

[sudden oar]

negative reciprocal of 3 is -1/3 (3 = 3/1)

negative reciprocal of 1/4 is -4 (-4/1)

negative reciprocal of -2 is 1/2

hope that makes sense

[sudden oar]

then sub in the coordinate for M into y=mx+c

using the gradient of AB

to find c

then u got ur equation

[fallow wind]

for AB

[sudden oar]

ye

[fallow wind]

so then how would i use that to answer the coordinate of point A

[sudden oar]

unsure for that sorry