| Exam Board | OCR |
|---|---|
| Module | Further Statistics (Further Statistics) |
| Year | 2022 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Single period profit/loss calculation |
| Difficulty | Standard +0.3 This is a straightforward application of linear transformations of normal distributions and sums of independent normals. Part (a) requires converting a cost threshold to a distance and using normal tables. Part (b) involves forming a linear combination of independent normals and standardizing. While it requires careful bookkeeping of coefficients, the techniques are standard for Further Statistics with no novel insight required. |
| Spec | 5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| 5 | (a) | 80 + 2X |
| Answer | Marks |
|---|---|
| P(< £235) = 0.386 | M1 |
| Answer | Marks |
|---|---|
| [4] | 3.1b |
| Answer | Marks |
|---|---|
| 3.4 | Consider distribution of 80 + 2X |
| Answer | Marks |
|---|---|
| Awrt 0.386 | OR: make X subject M1 |
| Answer | Marks |
|---|---|
| (b) | Using costs A , A , A , B , B : |
| Answer | Marks |
|---|---|
| P(diff 300) = 0.254 | B1 |
| Answer | Marks |
|---|---|
| A1 [6] | 1.1 |
| Answer | Marks |
|---|---|
| 3.4 | Means 246, 244.5, can be implied by later work |
| Answer | Marks |
|---|---|
| Awrt 0.254, allow 0.02 if cc used | If in effect using 3X, 2Y: |
| Answer | Marks |
|---|---|
| OR | Using distances X , X , X , X , X : |
| 1 2 3 4 5 | If in effect using 3X, 2Y: |
| Answer | Marks | Guidance |
|---|---|---|
| 1 2 3 | B1 | Means 249, 166, can be implied by later work |
| Answer | Marks | Guidance |
|---|---|---|
| 4 5 | B1 | Variances 1080, 720, ditto |
| A’s charge ~ N(738, 4320) | B1 | Means 738 & 249, allow 498 & 249 |
| B’s charge ~ N(489, 1620) | B1 | Variances 4320 and 1620 |
| Difference ~ N(249, 5940) | M1 | Use normal for difference, add variances |
| P(diff 300) = 0.254 | A1 | Awrt 0.254, allow 0.02 if cc used |
| OR | Diff in costs is 2(X +X +X ) – 1.5(X +X ) | |
| 1 2 3 4 5 | M1A1 | M1 needs attempt to include constants |
| ~ N(249, 5940) | M1A2 | M1 N(249, ...), A2 variance correct, A1 for 16200 |
| P(diff 300) = 0.254 | A1 | Awrt 0.254, allow 0.02 if cc used |
| SC1: insufficient working shown | 0.254 www gets 6/6; 0.746 www gets 5/6 [A0] | 0.344 www gets 4/6 [A0A0]; all else 0/6 |
| SC2: assume all/both variables equal: | M1A1 | 3X > 300 M1 N(249, 3240) or N(51, 3240) A1 |
Question 5:
5 | (a) | 80 + 2X
~ N(246, …
… 1440)
P(< £235) = 0.386 | M1
M1
A1
A1
[4] | 3.1b
1.1
1.1
3.4 | Consider distribution of 80 + 2X
Normal with mean 246
Correct variance
Awrt 0.386 | OR: make X subject M1
X < 77.5 (or 77) A1
P(X < 77.5) from N(83, 360) M1, allow 77
P(< £235) = 0.386 A1 (0.376 A0)
Or P(2X < 155) where 2X ~ N(166, 1440)
(b) | Using costs A , A , A , B , B :
1 2 3 1 2
A = 80 + 2X ~ N(246, 1440)
1 1
B = 120 + 1.5X ~ N(244.5, 810)
1 4
A + A + A ~ N(738, 4320)
1 2 3
B + B ~ N(489, 1620)
1 2
Difference ~ N(249, 5940)
P(diff 300) = 0.254 | B1
B1
B1
B1
M1
A1 [6] | 1.1
2.2a
1.1
1.1
3.1b
3.4 | Means 246, 244.5, can be implied by later work
Variances 1440, 810, ditto
Means 738 & 489, allow 498 & 249
Variances 4320 and 1620
Use normal for difference, add variances
Awrt 0.254, allow 0.02 if cc used | If in effect using 3X, 2Y:
Means 246, 244.5 B1
Variances 1440, 810 B1
Means 738, 489 B1
Variances 12960, 3240 B0
N(249, 16200) M1
0.344 A0
OR | Using distances X , X , X , X , X :
1 2 3 4 5 | If in effect using 3X, 2Y:
X + X + X ~ N(249, 1080)
1 2 3 | B1 | Means 249, 166, can be implied by later work | Means 249, 166 B1
X + X ~ N(166, 720)
4 5 | B1 | Variances 1080, 720, ditto | Variances 3240, 1440 B0
A’s charge ~ N(738, 4320) | B1 | Means 738 & 249, allow 498 & 249 | Means 738, 489 B1
B’s charge ~ N(489, 1620) | B1 | Variances 4320 and 1620 | Variances 12960, 3240 B1 NB
Difference ~ N(249, 5940) | M1 | Use normal for difference, add variances | N(249, 16200) M1
P(diff 300) = 0.254 | A1 | Awrt 0.254, allow 0.02 if cc used | 0.344 A0
OR | Diff in costs is 2(X +X +X ) – 1.5(X +X )
1 2 3 4 5 | M1A1 | M1 needs attempt to include constants
~ N(249, 5940) | M1A2 | M1 N(249, ...), A2 variance correct, A1 for 16200 | SC: variance formula seen correct but
P(diff 300) = 0.254 | A1 | Awrt 0.254, allow 0.02 if cc used | wrongly calculated: (M1)M1 SCM1
SC1: insufficient working shown | 0.254 www gets 6/6; 0.746 www gets 5/6 [A0] | 0.344 www gets 4/6 [A0A0]; all else 0/6
SC2: assume all/both variables equal: | M1A1 | 3X > 300 M1 N(249, 3240) or N(51, 3240) A1 | Or N(83, 360) and compare 100: A1 (0.185)5 A company uses two drivers for deliveries.\\
Driver $A$ charges a fixed rate of $\pounds 80$ per day plus $\pounds 2$ per mile travelled on that day.
Driver $B$ charges a fixed rate of $\pounds 120$ per day plus $\pounds 1.50$ per mile travelled on that day.\\
On each working day the total distance, in miles, travelled by each driver is a random variable with the distribution $\mathrm { N } ( 83,360 )$.
\begin{enumerate}[label=(\alph*)]
\item Find the probability that driver $A$ charges the company less than $\pounds 235.00$ for a randomly chosen day's deliveries.
\item Find the probability that the total charge to the company of three randomly chosen days' deliveries by driver $A$ is at least $\pounds 300$ more than the total charge of two randomly chosen days' deliveries by driver $B$.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Statistics 2022 Q5 [10]}}