Intermediate November 2019 Unit 2 Q11
11.
(a) A biased coin is thrown 100 times.
The number of heads thrown is recorded after 20 throws, 40 throws, 60 throws, 80 throws and 100 throws.
The number of heads thrown is recorded after 20 throws, 40 throws, 60 throws, 80 throws and 100 throws.
Some of the results are recorded in the relative frequency table below.
Complete the table. [2]
| Number of throws | 20 | 40 | 60 | 80 | 100 |
|---|---|---|---|---|---|
| Number of heads | 11 | 18 | 24 | 30 | |
| Relative frequency | 0·55 | 0·45 | 0·375 | 0·37 |
(b)
5 7 8 11 14 17 17 19 26 28
The sum of the ten numbers shown above is 152.
The numbers are displayed in the grouped frequency table shown below.
| Number | 0 - 9 | 10 - 19 | 20 - 29 |
|---|---|---|---|
| Frequency | 3 | 5 | 2 |
Consider the estimated mean calculated from the table and the actual mean of the ten numbers.
Calculate the difference between these two values.
You must show all your working. [5]
| Scheme | Marks | Comments | ||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| B1 B1 |
| Scheme | Marks | Comments |
|---|---|---|
| (Mid-points are) 4·5, 14·5 and 24·5. | B1 | |
| (Estimated total =) 3 × 4·5 + 5 × 14·5 + 2 × 24·5 (= 135) | M1 | F.T. ‘their mid-points’ if within group. |
| ÷ 10 | m1 | |
| (Estimated mean =) = 13·5 | A1 | C.A.O. |
| (Difference = 15·2 – 13·5 = ) 1·7 | B1 | F.T. for difference between 15·2 and ‘their derived estimated mean (≠15·2)’. Allow −1·7. |