| Date | November 2010 | Marks available | 1 | Reference code | 10N.1.sl.TZ0.7 |
| Level | SL only | Paper | 1 | Time zone | TZ0 |
| Command term | Find | Question number | 7 | Adapted from | N/A |
Question
The weights of 90 students in a school were recorded. The information is displayed in the following table.
Write down the mid interval value for the interval \(50 \leqslant w \leqslant 60\).
Use your graphic display calculator to find an estimate for the mean weight.
Use your graphic display calculator to find an estimate for the standard deviation.
Find the weight that is 3 standard deviations below the mean.
Markscheme
55 (A1) (C1)
[1 mark]
\(62.\bar 5{\text{ }}\left( {62.6} \right)\) (A2)(ft) (C2)
[2 marks]
8.86 (A1) (C1)
Note: Follow through from their answer to part (a).
[1 mark]
62.6 – 3 × 8.86 = 36.0 (M1)(A1)(ft) (C2)
Note: Accept 36.
Follow through from their values in part (b) only if working is seen.
[2 marks]
Examiners report
This question was not well answered. Many candidates could not find the mid interval value, and used their graphic display calculator incorrectly to find the mean and standard deviation.
This question was not well answered. Many candidates could not find the mid interval value, and used their graphic display calculator incorrectly to find the mean and standard deviation.
This question was not well answered. Many candidates could not find the mid interval value, and used their graphic display calculator incorrectly to find the mean and standard deviation.
This question was not well answered. Many candidates could not find the mid interval value, and used their graphic display calculator incorrectly to find the mean and standard deviation. The candidates who showed working and correct method in part c) were awarded the final two marks. If working was not shown then these marks could not be awarded.
