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S1 Question

A net was used to catch swallows so that they could be ringed and examined. The weights of 55 adult birds were recorded and the results are summarised in the table below.

Weight (g) 14 - 19 20 - 21 22 - 23 24 - 25 26 - 29 30 - 35
Frequency 3 6 15 20 9 2
a) For these data calculate estimates of the 33rd percentile.
Sorry about the formatting.

I was able to do the interpolation part to get the actual value. But to start I did (55*33))/100 to find the value in the frequency but in the mark scheme it is ((55+1)*33)/100
Why do you have to add 1 to 55?
Reply 1
Avatar for Teeenbe
Teeenbe
6
My guess would be that it's to do with the data-type, i.e: discrete - as that data currently is, or continuous. That would affect your method for interpolation.
This data is continuous, and that does affect the interpolation. That is not the problem here. Ill show my working:

55*33/100 = 18.15

23.5-21.5 = 2
24-9 = 15
9-18.15 = 9.15

9.15 * 2/15 +21.5 = 22.72

The mark scheme says that the first step is wrong:
(55+1)*33/100 = 18.48
Etc
9.48 * 2/15 + 21.5 = 22.76

As these 2 values round to 22.7 and 22.8 My result is wrong
Original post by porrige51122
This data is continuous, and that does affect the interpolation. That is not the problem here. Ill show my working:

55*33/100 = 18.15

23.5-21.5 = 2
24-9 = 15
9-18.15 = 9.15

9.15 * 2/15 +21.5 = 22.72

The mark scheme says that the first step is wrong:
(55+1)*33/100 = 18.48
Etc
9.48 * 2/15 + 21.5 = 22.76

As these 2 values round to 22.7 and 22.8 My result is wrong


you should get quite a few marks. this mistake will be quite common.
I'd prefer to understand the reason why rather than accept that I will lose the mark as in statistics, mistakes build up. Plus I need as high as possible in S1 to make C3 and C4 easier
Reply 5
Avatar for Teeenbe
Teeenbe
6
Original post by porrige51122
This data is continuous, and that does affect the interpolation. That is not the problem here. Ill show my working:

55*33/100 = 18.15

23.5-21.5 = 2
24-9 = 15
9-18.15 = 9.1
9.15 * 2/15 +21.5 = 22.72

The mark scheme says that the first step is wrong:
(55+1)*33/100 = 18.48
Etc
9.48 * 2/15 + 21.5 = 22.76

As these 2 values round to 22.7 and 22.8 My result is wrong


My bad, I meant continuous. As for why you need to add 1: I'm not sure anymore tbh, sorry.
Original post by porrige51122
A net was used to catch swallows so that they could be ringed and examined. The weights of 55 adult birds were recorded and the results are summarised in the table below.

Weight (g) 14 - 19 20 - 21 22 - 23 24 - 25 26 - 29 30 - 35
Frequency 3 6 15 20 9 2
a) For these data calculate estimates of the 33rd percentile.
Sorry about the formatting.

I was able to do the interpolation part to get the actual value. But to start I did (55*33))/100 to find the value in the frequency but in the mark scheme it is ((55+1)*33)/100
Why do you have to add 1 to 55?


It's exactly the same as the reason why the median uses (n+1)/2 rather than n/2.

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