S1 June 2015 Q6
6. The random variable \(Z \sim \mathrm{N}(0, 1)\)
\(A\) is the event \(Z \gt 1.1\)
\(B\) is the event \(Z \gt -1.9\)
\(C\) is the event \(-1.5 \lt Z \lt 1.5\)
The random variable \(X\) has a normal distribution with mean 21 and standard deviation 5
| Scheme | Marks |
|---|---|
| (i) \(\mathrm{P}(A) = \mathrm{P}(Z \gt 1.1) = 1 - 0.8643 = \)0.1357 (accept awrt 0.136) | B1 |
| (ii) \(\mathrm{P}(B) = \mathrm{P}(Z \gt -1.9) = \)0.9713 (accept awrt 0.971) | B1 |
| (iii) \(\mathrm{P}(C) = [\mathrm{P}(-1.5 \lt Z \lt 1.5)] = 0.9332 - (1 - 0.9332)\) or \((0.9332 - 0.5) \times 2\) \(= \)0.8664 (accept awrt 0.866) | M1 A1 |
| (iv) \(\mathrm{P}(A \cup C) = \mathrm{P}(Z \gt -1.5)\) or \(\mathrm{P}(Z \lt 1.5)\) or \(= \mathrm{P}(A) + \mathrm{P}(C) - \mathrm{P}(A \cap C) = \text{"}0.1357\text{"} + \text{"}0.8664\text{"} - (0.9332 - 0.8643)\) \(= \)0.9332 (accept awrt 0.933) | M1 A1 |
| (6) |
Notes
Mark final answer here so in (ii) 0.9713 followed by 1 – 0.9713 is B0 but for rounding errors e.g. 29.245 followed by 29.3 apply ISW and award for 29.245
(iii) M1 for correct expression with probability values . Correct ans implies M1A1
(iv) M1 for a correct addition formula with some correct substitution (or correct ft) or \(\mathrm{P}(Z \gt -1.5)\) (o.e) or for a fully correct expression with correct probabilities
A1 for 0.9332 (accept 0.933) Correct answer only is M1A1
| Scheme | Marks |
|---|---|
| \(\left[\mathrm{P}(X \gt w \mid X \gt 28) =\right] \dfrac{\mathrm{P}(X \gt w)}{\mathrm{P}(X \gt 28)} = [0.625]\) | M1 |
| \(\mathrm{P}(X \gt 28) = \mathrm{P}\left(Z \gt \dfrac{28 - 21}{5}\right) = \mathrm{P}(Z \gt 1.4) = [0.0808\ \text{ calc: } 0.080756..]\) | M1 |
| \(\mathrm{P}(X \gt w) = 0.0808 \times 0.625\ (= 0.0505)\) or ( \(\mathrm{P}(X \lt w) = 0.9495\)) | A1 |
| \(\dfrac{w - 21}{5} = 1.64\) | M1 B1 |
| \(w = \) awrt 29.2 | A1 |
| (6) | |
| (12 marks) |
Notes
M1 for correct expression for conditional probability- must have \(\mathrm{P}(X \gt w)\) as num’
May be implied by \(\mathrm{P}(X \gt w) = 0.625 \times\) (any probability)
M1 for standardising 28 with 21 and 5 Allow \(\pm\)
(May be implied by 0.0808 [or awrt 0.081] seen in correct position)
A1 for \(\mathrm{P}(X \gt w) = 0.0808 \times 0.625\) or \(\mathrm{P}(X \gt w) = 0.0505\) or \(\mathrm{P}(X \lt w) = 0.9495\))
This A1 depends on both Ms but seeing \(\mathrm{P}(X \gt w) = 0.0808 \times 0.625\) scores M1M1A1
1st 3 marks Allow \(\mathrm{P}\left(Z \gt \dfrac{w - 21}{5}\right)\) instead of \(\mathrm{P}(X \gt w)\) for these first 3 marks
M1 for standardising \(w\) with 21 and 5 (allow \(\pm\)) and setting equal to a \(z\)-value \(|z| \gt 1\)
Allow any letter instead of \(w\)
B1 for 1.64 (or better) used correctly. [Calculator gives: 1.6402851...]
A1 allow awrt 29.2
(calc: 0.080756.. corrected from the printed mark scheme: 0.80756..)