June 2025 Paper 3 Q5
5. The heights of men in a tennis club are normally distributed with a mean of 183 cm and a standard deviation of 4.9 cm.
A man from the tennis club is selected at random.
The heights of women in the tennis club are normally distributed with a mean of \(\mu\) cm and a standard deviation of \(\sigma\) cm.
Given that 40% of the women are shorter than 170 cm and 15% are taller than 175 cm,
Show your working clearly. (5)
A man from the tennis club and a woman from the tennis club are selected at random.
| Scheme | Marks | AO |
|---|---|---|
| (i) [Let \(M\) be the height of a man from the club, \(\mathrm{P}(M > 186) =\) ] \(0.270187\ldots = \) awrt 0.270 | B1 | 1.1b |
| (ii) \([\mathrm{P}(175 < M < 185) =]\ 0.6071520\ldots =\) awrt 0.607 | B1 | 3.4 |
| (2) |
Notes
(i) B1: for awrt 0.270 (accept just 0.27)
(ii) B1: for awrt 0.607
| Scheme | Marks | AO |
|---|---|---|
| [Let \(F\) be the height of a women from the club] \(\mathrm{P}(F < 170) = 0.40 \Rightarrow \dfrac{170 - \mu}{\sigma} = -0.2533\) [ calc gives \(-0.253347\ldots\)] | M1 | 3.1b |
| \(\mathrm{P}(F > 175) = 0.15 \Rightarrow \dfrac{175 - \mu}{\sigma} = 1.0364\) [ calc gives \(1.036433\ldots\)] | M1 | 3.4 |
| Solving: \(5 = \sigma(1.0364 + 0.2533)\) | M1 | 1.1b |
| \(\sigma = 3.87687\ldots\) awrt 3.88 | A1 | 1.1b |
| \(\mu = 170.9820\ldots\) awrt 171 | A1 | 3.2b |
| (5) |
Notes
M1: for standardising 170 with \(\mu\) and \(\sigma\) and set equal to a \(z\) value \(0 < |z| < 0.3\)
M1: for standardising 175 with \(\mu\) and \(\sigma\) and set equal to a \(z\) value \(1 < |z| < 2\)
M1: for solving their 2 linear equations, reducing to a linear equation in one variable
This can be implied by two answers that are correct to 2sf.
To award the A marks they must have scored 1st M1 and 2nd M1
A1: for = \(\sigma\) awrt 3.88
A1: for = \(\mu\) awrt 171 (Must not be from incorrect working e.g. \(\sigma < 0\))
Ans only: We need to see the linear equations in \(\mu\) and \(\sigma\) to start awarding marks.
If 1st M1 and 2nd M1 are scored and correct answers are seen we can award 5 marks.
MR: (15% as 0.015) 2nd M1 for \(z = 2.17\ldots\), 3rd M1 implied by \(\sigma = 2.06\ldots\) and \(\mu = 170.5\ldots\)
i.e. Maximum M1M1M1A0A0
| Scheme | Marks | AO |
|---|---|---|
| \([\mathrm{P}(170 < F < 175) =]\ 1 - (0.40 + 0.15) = 0.45\) (allow awrt 0.45) \([\mathrm{P}(170 < M < 175) =]\ 0.047282\ldots =\) (allow awrt 0.047) | M1 | 3.4 |
| \([\mathrm{P}(170 < F < 175) \times \mathrm{P}(170 < M < 175) =]\ \text{``}0.45\text{''} \times \text{``}0.047\text{''}\) or \(0.021\ldots\) | A1ft | 1.1b |
| So \(\mathrm{P}(\text{both}) = 0.021277\ldots\) awrt 0.021 | A1 | 1.1b |
| (3) | ||
| (10 marks) |
Notes
M1: for correct probability for women or correct probability for men
or clear intention to multiply \(\mathrm{P}(170 < F < 175)\) by \(\mathrm{P}(170 < M < 175)\)
A1ft: for using \(\mathrm{P}(170 < F < 175) \times \mathrm{P}(170 < M < 175)\) with one probability correct to 2sf
A1: for awrt 0.021. Correct answer only is 3 marks.
MR: (if used in (b)) \(\mathrm{P}(170 < F < 175) = 0.585\) (or awrt 0.58 or 0.59) answer awrt 0.027 or 0.028
All 3 marks can be scored here.