Question 10[Maximum mark: 6]
A manufacturer of chocolates produces them in individual packets, claiming to have an average of 85 chocolates per packet.
Talha bought 30 of these packets in order to check the manufacturer’s claim.
Given that the number of individual chocolates is x , Talha found that, from his 30 packets,
∑x = 2506 and ∑x2 = 209 738 .
a. Find an unbiased estimate for the mean number (μ) of chocolates per packet. [1]
b. Use the formula \(s_{n-1}^{2}= \frac{\sum x^{2}-\frac{(\sum x)^{2}}{n}}{n-1}\)
to determine an unbiased estimate for the variance of the number of chocolates per packet. [2]
c. Find a 95 % confidence interval for μ . You may assume that all conditions for a
confidence interval have been met. [2]
d. Suggest, with justification, a valid conclusion that Talha could make. [1]
▶️Answer/Explanation
(a) \(\bar{X}= \frac{\sum x}{n}= \frac{2506}{30}83.5 (83.5333.. )\)
(b) \(s^{2}_{n-1}= \frac{\sum x^{2}-\frac{(\sum x)^{2}}{n}}{n-1}= \frac{209738-\frac{2506^{2}}{30}}{29}= 13.9(13.9126..)\)
(c) (82.1,84.9)(82.1405…, 84.9261…)
(d) 85 is outside the confidence interval and therefore Talha would suggest that the manufacturer’s claim is incorrect
Question
The weights of potatoes in a shop are normally distributed with mean \(98\) grams and standard deviation \(16\) grams.
a.The shopkeeper places \(100\) randomly chosen potatoes on a weighing machine. Find the probability that their total weight exceeds \(10\) kilograms.[3]
b.Find the minimum number of randomly selected potatoes which are needed to ensure that their total weight exceeds \(10\) kilograms with probability greater than \(0.95\).[8]
▶️Answer/Explanation
Markscheme
let \(T\) denote the total weight, then
\(T \sim N(9800,25600)\) (M1)(A1)
\({\rm{P}}(T > 10000) = 0.106\) A1
[3 marks]
let there be \(n\) potatoes, in this case,
\(T \sim {\rm{N}}(98n,256n)\) A1
we require
\({\rm{P}}(T > 10000) > 0.95\) (M1)
or equivalently
\({\rm{P}}(T \le 10000) < 0.05\) A1
standardizing,
\({\rm{P}}\left( {Z \le \frac{{10000 – 98n}}{{16\sqrt n }}} \right) < 0.05\) A1
\(\frac{{10000 – 98n}}{{16\sqrt n }} < – 1.6449 \ldots \) (A1)
\(98n – 26.32\sqrt n – 10000 > 0\) A1
solving the corresponding equation, \(n = 104.7 \ldots \) (A1)
the required minimum value is \(105\) A1
Note: Part (b) could also be solved using SOLVER and normalcdf, or by trial and improvement.
Note: Allow the use of \( = \) instead of \( < \) and \( > \) throughout.
[8 marks]
Question
The weights of male students in a college are modelled by a normal distribution with mean 80 kg and standard deviation 7 kg.
The weights of female students in the college are modelled by a normal distribution with mean 54 kg and standard deviation 5 kg.
The college has a lift installed with a recommended maximum load of 550 kg. One morning, the lift contains 3 male students and 6 female students. You may assume that the 9 students are randomly chosen.
a.Find the probability that the weight of a randomly chosen male student is more than twice the weight of a randomly chosen female student.[6]
b.Determine the probability that their combined weight exceeds the recommended maximum.[5]
▶️Answer/Explanation
Markscheme
let \(M\), \(F\) denote the weights of the male, female
consider \(D = M – 2F\) (M1)
\({\text{E}}(D) = 80 – 2 \times 54 = – 28\) A1
\({\text{Var}}(D) = {7^2} + 4 \times {5^2}\) (M1)
\( = 149\) A1
\({\text{P}}(M > 2F) = {\text{P}}(D > 0)\) (M1)
\( = 0.0109\) A1
Note: Accept any answer that rounds correctly to 0.011.
[6 marks]
consider \({\text{S}} = \sum\limits_{i = 1}^3 {{M_i} + \sum\limits_{i = 1}^6 {{F_i}} } \) (M1)
Note: Condone the use of the incorrect notation \(3M + 6F\).
\({\text{E}}(S) = 3 \times 80 + 6 \times 54 = 564\) A1
\({\text{Var}}(S) = 3 \times {7^2} + 6 \times {5^2}\) (M1)
\( = 297\) A1
\({\text{P}}(S > 550) = 0.792\) A1
Note: Accept any answer that rounds correctly to 0.792.
[5 marks]
Question
Sami is undertaking market research on packets of soap powder. He considers the brand “Gleam”. The weight of the contents of a randomly chosen packet of “Gleam” follows a normal distribution with mean 750 grams and standard deviation 20 grams.
The weight of the packaging follows a different normal distribution with mean 40 grams and standard deviation 5 grams.
a.Find:
(i) the probability that a randomly chosen packet of “Gleam” has a total weight exceeding 780 grams.
(ii) the probability that the total weight of the contents of five randomly chosen packets of “Gleam” exceeds 3800 grams.[8]
b.Sami now considers the brand “Bright”. The weight of the contents of a randomly chosen packet of “Bright” follow a normal distribution with mean 650 grams and standard deviation 16 grams. Find the probability that the contents of six randomly chosen packets of “Bright” weigh more than the contents of five randomly chosen packets of “Gleam”.[4]
▶️Answer/Explanation
Markscheme
Note: In all parts accept answers which round to the correct 2sf answer.
(i) contents: \(X \sim N(750,{\text{ }}400)\)
packaging: \(Y \sim N(40,{\text{ }}25)\)
consider \(X + Y\) (M1)
\({\text{E}}(X + Y) = 790\) A1
\({\text{Var}}(X + Y) = 425\) A1
\({\text{P}}(X + Y > 780) = 0.686\) A1
(ii) Let \({X_1} + {X_2} + {X_3} + {X_4} + {X_5} = A\) M1
\({\text{E}}(A) = 5{\text{E}}(X) = 3750\) A1
\({\text{Var}}(A) = 5{\text{Var}}(X) = 2000\) A1
\({\text{P}}(A > 3800) = 0.132\) A1
Note: Condone the notation \(A = 5X\) if the variance is correct, M0 if not
contents of Bright: \(B \sim N(650,{\text{ }}256)\)
let \(G = {B_1} + {B_2} + {B_3} + {B_4} + {B_5} + {B_6} – ({X_1} + {X_2} + {X_3} + {X_4} + {X_5})\) M1
\({\text{E}}(G) = 6 \times 650 – 5 \times 750 = 150\) A1
\({\text{Var}}(G) = 6 \times 256 + 5 \times 400 = 3536\) A1
\({\text{P}}(G > 0) = 0.994\) A1
Note: Condone the notation \(G = 6B – 5X\) if the variance is correct, M0 if not
Question
The lifetime, in years, of a randomly chosen basic vacuum cleaner is assumed to be modelled by the normal distribution \(B \sim {\text{N}}(14,{\text{ }}{3^2})\).
The lifetime, in years, of a randomly chosen robust vacuum cleaner is assumed to be modelled by the normal distribution \(R \sim {\text{N}}(20,{\text{ }}{4^2})\).
a.Find \({\text{P}}\left( {B > {\text{E}}(B) + \frac{1}{2}\sqrt {{\text{Var}}(B)} } \right)\).[2]
b.Find the probability that the total lifetime of 7 randomly chosen basic vacuum cleaners is less than 100 years.[4]
c.Find the probability that the total lifetime of 5 randomly chosen robust vacuum cleaners is greater than the total lifetime of 7 randomly chosen basic vacuum cleaners.[5]
▶️Answer/Explanation
Markscheme
\({\text{P}}(B > 15.5){\text{ }}\left( { = {\text{P}}(Z > 0.5)} \right)\) (M1)
\( = (1 – 0.69146) = 0.309\) A1
[2 marks]
consider \(V = {B_1} + {B_2} + {B_3} + {B_4} + {B_5} + {B_6} + {B_7}\) (M1)
\({\text{E}}(V) = 98\) (A1)
\({\text{Var}}(V) = 63\) or equivalent (A1)
Note: No need to state \(V\) is normal.
\({\text{P}}(V < 100) = \left( {{\text{P}}\left( {Z < \frac{2}{{\sqrt {63} }} = 0.251976 \ldots } \right)} \right) = 0.599\) A1
[4 marks]
consider \(W = {R_1} + {R_2} + {R_3} + {R_4} + {R_5} – ({B_1} + {B_2} + {B_3} + {B_4} + {B_5} + {B_6} + {B_7})\) (M1)
\({\text{E}}(W) = 2\) (A1)
\({\text{Var}}(W) = 80 + 63 = 143\) (A1)
\({\text{P}}(W > 0) = \left( {{\text{P}}\left( {Z < \frac{2}{{\sqrt {143} }}} \right)} \right)\) (M1)
\( = 0.566\) A1
[5 marks]