Question
A shop sells apples and pears. The weights, in grams, of the apples may be assumed to have a \({\text{N}}(200,{\text{ 1}}{{\text{5}}^2})\) distribution and the weights of the pears, in grams, may be assumed to have a \({\text{N}}(120,{\text{ 1}}{{\text{0}}^2})\) distribution.
(a) Find the probability that the weight of a randomly chosen apple is more than double the weight of a randomly chosen pear.
(b) A shopper buys 3 apples and 4 pears. Find the probability that the total weight is greater than 1000 grams.
▶️Answer/Explanation
Markscheme
(a) Let X, Y (grams) denote respectively the weights of a randomly chosen apple, pear.
Then
\(X – 2Y{\text{ is N}}(200 – 2 \times 120,{\text{ }}{15^2} + 4 \times {10^2}),\) (M1)(A1)(A1)
i.e. \({\text{N}}( – 40,{\text{ }}{25^2})\) A1
We require
\({\text{P}}(X > 2Y) = {\text{P}}(X – 2Y > 0)\) (M1)(A1)
\( = 0.0548\) A2
[8 marks]
(b) Let \(T = {X_1} + {X_2} + {X_3} + {Y_1} + {Y_2} + {Y_3} + {Y_4}\) (grams) denote the total weight.
Then
\(T{\text{ is N}}(3 \times 200 + 4 \times 120,{\text{ }}3 \times {15^2} + 4 \times {10^2}),\) (M1)(A1)(A1)
i.e. \({\text{N(1080, 1075)}}\) A1
\({\text{P}}(T > 1000) = 0.993\) A2
[6 marks]
Total [14 marks]
Question
Ahmed and Brian live in the same house. Ahmed always walks to school and Brian always cycles to school. The times taken to travel to school may be assumed to be independent and normally distributed. The mean and the standard deviation for these times are shown in the table below.
(a) Find the probability that on a particular day Ahmed takes more than 35 minutes to walk to school.
(b) Brian cycles to school on five successive mornings. Find the probability that the total time taken is less than 70 minutes.
(c) Find the probability that, on a particular day, the time taken by Ahmed to walk to school is more than twice the time taken by Brian to cycle to school.
▶️Answer/Explanation
Markscheme
(a) \(A \sim {\text{N}}(30,{\text{ }}{3^2})\)
\({\text{P}}(A > 35) = 0.0478\) (M1)A1
[2 marks]
(b) let \(X = {B_1} + {B_2} + {B_3} + {B_4} + {B_5}\)
\({\text{E}}(X) = 5{\text{E}}(B) = 60\) A1
\({\text{Var}}(X) = 5{\text{Var}}(B) = 20\) (M1)A1
\({\text{P}}(X < 70) = 0.987\) A1
[4 marks]
(c) let \(Y = A – 2B\) (M1)
\({\text{E}}(Y) = {\text{E}}(A) – 2{\text{E}}(B) = 6\) A1
\({\text{Var}}(Y) = {\text{Var}}(A) + 4{\text{Var}}(B) = 25\) (M1)A1
\({\text{P}}(Y > 0) = 0.885\) A1
[5 marks]
Total [11 marks]
Examiners report
Most candidates were able to access this question, but weaker candidates did not always realise that parts (b) and (c) were testing different things. Part (b) proved the hardest with a number of candidates not understanding how to find the variance of the sum of variables.
Question
a.Alan and Brian are athletes specializing in the long jump. When Alan jumps, the length of his jump is a normally distributed random variable with mean 5.2 metres and standard deviation 0.1 metres. When Brian jumps, the length of his jump is a normally distributed random variable with mean 5.1 metres and standard deviation 0.12 metres. For both athletes, the length of a jump is independent of the lengths of all other jumps. During a training session, Alan makes four jumps and Brian makes three jumps. Calculate the probability that the mean length of Alan’s four jumps is less than the mean length of Brian’s three jumps.[9]
b.Colin joins the squad and the coach wants to know the mean length, \(\mu \) metres, of his jumps. Colin makes six jumps resulting in the following lengths in metres.
5.21, 5.30, 5.22, 5.19, 5.28, 5.18
(i) Calculate an unbiased estimate of both the mean \(\mu \) and the variance of the lengths of his jumps.
(ii) Assuming that the lengths of these jumps are independent and normally distributed, calculate a 90 % confidence interval for \(\mu \) .[10]
▶️Answer/Explanation
Markscheme
let \(\bar A,{\text{ }}\bar B\) denote the means of Alan’s and Brian’s jumps
attempting to find the distributions of \(\bar A,{\text{ }}\bar B\) (M1)
\(\bar A{\text{ is N}}\left( {5.2,\frac{{{{0.1}^2}}}{4}} \right)\) A1
\(\bar B{\text{ is N}}\left( {5.1,\frac{{{{0.12}^2}}}{3}} \right)\) A1
attempting to find the distribution of \(\bar A – \bar B\) (M1)
\(\bar A – \bar B{\text{ is N}}\left( {5.2 – 5.1,\frac{{{{0.1}^2}}}{4} + \frac{{{{0.12}^2}}}{3}} \right)\) (A1)(A1)
i.e. \({\text{N}}(0.1,{\text{ }}0.0073)\) A1
\({\text{P}}(\bar A < \bar B) = {\text{P}}(\bar A – \bar B < 0)\) M1
\( = 0.121\) A1
[9 marks]
(i) \(\sum {x = 31.38,{\text{ }}\sum {{x^2} = 164.1294} } \)
\(\bar x = \frac{{31.38}}{6} = 5.23\) (M1)A1
EITHER
\(s_{n – 1}^2 = \frac{{164.1294}}{5} – \frac{{{{31.38}^2}}}{{5 \times 6}} = 0.00240\) (M1)(A1)A1
OR
\({s_{n – 1}} = 0.04899 \Rightarrow s_{n – 1}^2 = 0.00240\) (M1)(A1)A1
Note: Accept the exact answer 0.0024 without an arithmetic penalty.
(ii) using the t-distribution with DF = 5 (A1)
critical value of t = 2.015 A1
90 % confidence limits are \(5.23 \pm 2.015\sqrt {\frac{{0.0024}}{6}} \) M1A1
giving [5.19, 5.27] A1 N5
[10 marks]
Examiners report
In (a), it was disappointing to note that many candidates failed to realise that the question was concerned with the mean lengths of the jumps and worked instead with the sums of the lengths.
Most candidates obtained correct estimates in (b)(i), usually directly from the GDC. In (b)(ii), however, some candidates found a z-interval instead of a t-interval.
Question
A shop sells apples, pears and peaches. The weights, in grams, of these three types of fruit may be assumed to be normally distributed with means and standard deviations as given in the following table.
Alan buys 1 apple and 1 pear while Brian buys 1 peach. Calculate the probability that the combined weight of Alan’s apple and pear is greater than twice the weight of Brian’s peach.
▶️Answer/Explanation
Markscheme
let X, Y, Z denote respectively the weights, in grams, of a randomly chosen apple, pear, peach
then \(U = X + Y – 2Z{\text{ is N}}(115 + 110 – 2 \times 105,{\text{ }}{5^2} + {4^2} + {2^2} \times {3^2})\) (M1)(A1)(A1)
Note: Award M1 for attempted use of U.
i.e. N(15, 77) A1
we require
\({\text{P}}(X + Y > 2Z) = {\text{P}}(U > 0)\) M1A1
\( = 0.956\) A2
Note: Award M0A0A2 for 0.956 only.
[8 marks]
Examiners report
Solutions to this question again illustrated the fact that many candidates are unable to distinguish between nX and \(\sum\limits_{i = 1}^n {{X_i}} \) so that many candidates obtained an incorrect variance to evaluate the final probability.