How did STATISTICS 1 go?

[Kamihus]

i am getting 40+ easily...what will be expected gt....i think 35-37

38 or 39.

 

[IGCSE13]

You have to multiply the square of mid-points by frequencies and then subtract the square of mean.

I was talking about the mean

 

[thementor]

and your variance should be 0.600 your midpoints are not right.10.25,10.75,11.5 so on.

 

[saadgujjar]

38 or 39.

37 was of may 13 which was easier than this paper

 

[randomcod]

BTW, due to accuracy issues, it should have been 11.7 for the mean, not 11.6 (11.66 rounds up). You used 10.2 and 10.7, should have been 10.25 and 12.75. CIE clearly states you should work to 4 s.f.'s and give answers to 3 s.f.'s

 

[randomcod]

And variance was 0.6! Once again ,due to accuracy!

 

[thementor]

Exactly. Thank you for stating that

BTW, due to accuracy issues, it should have been 11.7 for the mean, not 11.6 (11.66 rounds up). You used 10.2 and 10.7, should have been 10.25 and 12.75. CIE clearly states you should work to 4 s.f.'s and give answers to 3 s.f.'s

 

[Kamihus]

I was talking about the mean

Yes mean will be calculated that way.

and your variance should be 0.600 your midpoints are not right.10.25,10.75,11.5 so on.

Check the Q3 (iii) for this paper. Less than was used here too.

37 was of may 13 which was easier than this paper

I found this one easier. Although its better to wait for the actual thresholds but it surely will be 35-40 range.

 

[IGCSE13]

The

Yes mean will be calculated that way.

Check the Q3 (iii) for this paper. Less than was used here too.

I found this one easier.
Although its better to wait for the actual thresholds but it surely will be 35-40 range
.

The method you stated was for the variance

 

[Kamihus]

BTW, due to accuracy issues, it should have been 11.7 for the mean, not 11.6 (11.66 rounds up). You used 10.2 and 10.7, should have been 10.25 and 12.75. CIE clearly states you should work to 4 s.f.'s and give answers to 3 s.f.'s

You should also look at this paper, less than was used.

 

[Kamihus]

The

The method you stated was for the variance

The method you mentioned was correct for the mean. My mean and variance are correct for MY mid-points.

 

[IGCSE13]

The method you mentioned was correct for the mean. My mean and variance are correct for MY mid-points.

I got the same midpoints and frequencies as you but I seem to get a different answer , that's what's confusing me I thought that my midpoints are wrong that's I am not getting the same mean and variance as everybody else

 

[randomcod]

BTW For the histograms, did you ensure the line did not touch exactly the line for the boundaries for the right? As in <12 so being at like 11.95

 

[Kamihus]

I got the same midpoints and frequencies as you but I seem to get a different answer , that's what's confusing me I thought that my midpoints are wrong that's I am not getting the same mean and variance as everybody else

My mean is 11.6122807. Use this exact value in finding the variance.

 

[thementor]

.

Yes mean will be calculated that way.

Check the Q3 (iii) for this paper. Less than was used here too.

I found this one easier. Although its better to wait for the actual thresholds but it surely will be 35-40 range.

me.genius that questionhas discrete data. In that boundaries are different. -_-

 

[Kamihus]

BTW For the histograms, did you ensure the line did not touch exactly the line for the boundaries for the right? As in <12 so being at like 11.95

I wanted to ask for the histogram myself. I used 9.95, 10.45 and so on. Q4 in this paper is a similar one and exact 20, 30 are used according to the mark scheme. Both are correct perhaps.

 

[Kamihus]

.

me.genius that questionhas discrete data. In that boundaries are different. -_-

The boundaries are similar. The only difference is the decimals so guess the same method will be applied.

 

[thementor]

The boundaries are similar. The only difference is the decimals so guess the same method will be applied.

oh god. Dude. In discrete data, there is a different way. In continous data there is a different way. Very small difference.

 

[thementor]

The boundaries are similar. The only difference is the decimals so guess the same method will be applied.

oh god. Dude. In discrete data, there is a different way. In continous data there is a different way. Very small difference.

 

[Shadow]

the question stated repetition is allowed and less than and the numbers shkuld be less than 1000 and multiples of 5. so the last digit has onr possibility. and there were 7 no.s in total. since repetitions are allowed and it was a three digit no. 7x7 =49

u r forgetting that there could be 2 digits or single digit numbers also
so the answer is 57!

^Yupp thats exactly what was i going to tell u btw, the number cud have been 555 or even only 5 as well so 2 more possibilities..