| Exam Board | OCR MEI |
|---|---|
| Module | Further Statistics B AS (Further Statistics B AS) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Linear combinations of normal random variables |
| Type | Sum or total of normal variables |
| Difficulty | Standard +0.3 This is a straightforward application of normal distribution properties with linear transformations and sums. Part (i) requires a simple linear transformation and standardization. Parts (ii) and (iii) involve standard results about sums of independent normals, with (iii) requiring the insight to compare two sums by forming a single linear combination. All techniques are routine for Further Statistics students with no novel problem-solving required. |
| Spec | 2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation5.04a Linear combinations: E(aX+bY), Var(aX+bY)5.04b Linear combinations: of normal distributions |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | (i) | Y~N(0.7×345, (0.7×15)2) |
| Answer | Marks |
|---|---|
| 0.791 (0.79089...) | B1 |
| Answer | Marks |
|---|---|
| [3] | For Normal and mean |
| Answer | Marks |
|---|---|
| BC | If Normal not specified then can still |
| Answer | Marks |
|---|---|
| (ii) | Total for 5 oranges ~ N(1725, 1125) |
| 0.0127 (0.01267...) | B1 |
| Answer | Marks |
|---|---|
| [2] | For both |
| BC | Allow N(5×345, 5×225) oe |
| (iii) | Mean = 2×345 – 3×241.5 (= -34.5) |
| Answer | Marks |
|---|---|
| 0.108 (0.10847...) | M1 |
| Answer | Marks |
|---|---|
| [4] | For method for mean |
| Answer | Marks |
|---|---|
| BC | Or 27.9422 |
Question 2:
2 | (i) | Y~N(0.7×345, (0.7×15)2)
or N(241.5, 110.25)
0.791 (0.79089...) | B1
M1
A1
[3] | For Normal and mean
For variance (0.7×15)2
Can be implied by correct answer
BC | If Normal not specified then can still
score M1A1
Allow 0.79
(ii) | Total for 5 oranges ~ N(1725, 1125)
0.0127 (0.01267...) | B1
B1
[2] | For both
BC | Allow N(5×345, 5×225) oe
(iii) | Mean = 2×345 – 3×241.5 (= -34.5)
Variance = 2×152 + 3×10.52 (= 780.75)
2 unpeeled - 3 peeled ~ N(-34.5, 780.75)
0.108 (0.10847...) | M1
M1
A1
A1
[4] | For method for mean
For method for variance
For correct values
BC | Or 27.9422
Allow N(34.5, 780.75) as may be
using peeled - unpeeled2 A supermarket sells oranges. Their weights are modelled by the random variable $X$ which has a Normal distribution with mean 345 grams and standard deviation 15 grams. When the oranges have been peeled, their weights in grams, $Y$, are modelled by $Y = 0.7 X$.\\
(i) Find the probability that a randomly chosen peeled orange weighs less than 250 grams.
I randomly choose 5 oranges to buy.\\
(ii) Find the probability that the total weight of the 5 unpeeled oranges is at least 1800 grams.\\
(iii) I peel three of the oranges and leave the remaining two unpeeled. Find the probability that the total weight of the two unpeeled oranges is greater than the total weight of the three peeled ones.
\hfill \mbox{\textit{OCR MEI Further Statistics B AS 2018 Q2 [9]}}