in desperate need of help

hi all,

could you please explain to me part c and d in question 10 and question 11.
much appreciated

You have the frequency density, can you find the frequency using the class width? (hint: frequency density = frequency / class width so frequency = class width * frequency). Then you can find the midpoint of each histogram bar, for example the first one is (70+80)/2 = 75. Then the mean is sum of (frequency * midpoint) / total frequency.

So it would start as, for the first bar:

class width = 10
midpoint = 75
frequency = 6 * 10 = 60

So frequency * midpoint = (60 * 75).

Then do the same for the second bar, and the third and add all the (frequency * midpoints) up and then divide by the total frequency.

see i did that and i get 100.78125 but the stupid mark scheme says the answer is 75 and i dont know why :/

Could you post up a picture of your working? Make sure all your frequency numbers are correct, see if they line up with the number on the axes properly, etc...

these are my workings out

I would agree with your answer to part (c) unless I'm missing something blindingly obvious.

For part (d), you know that 288 people passed the test, yes. This means the test score is somewhere in the 130-150 region as the cumulative frequency in that region is 270-320. So now you need to be able to somehow find the test score corresponding to a frequency of 288.

i got this but i dont think it's right :/

Again (it's been a long time since I've done any of this) but I agree with your answer.

omg thank you

so would i do the same thing with the second question?

Well, not exactly the same thing. You'll want to look up 7.5 on the x-axis and see what the cumulative frequency up to 7.5 is and then do that cumulative frequency / total frequency to find the probability.

i get what you're saying it's just that how can i find the frequency of each bar, im just given one piece of info, so the f.d. would be 9 for the last bar but other than that idk what to do :/