Question
The fish in a lake have weights that are normally distributed with a mean of 1.3 kg and a standard deviation of 0.2 kg.
a.Determine the probability that a fish which is caught weighs less than 1.4 kg.[1]
b.John catches 6 fish. Calculate the probability that at least 4 of the fish weigh more than 1.4 kg.[3]
c.Determine the probability that a fish which is caught weighs less than 1 kg, given that it weighs less than 1.4 kg. [2]
▶️Answer/Explanation
Markscheme
a.\({\text{P}}(x < 1.4) = 0.691\,\,\,\,\,\)(accept 0.692) A1
[1 mark]
METHOD 1
\(y \sim {\text{B(6, 0.3085…)}}\) (M1)
\({\text{P}}(Y \geqslant 4) = 1 – {\text{P}}(Y \leqslant 3)\) (M1)
\( = 0.0775\,\,\,\,\,\)(accept 0.0778 if 3sf approximation from (a) used) A1
METHOD 2
\(X \sim {\text{B(6, 0.6914…)}}\) (M1)
\({\text{P}}(X \leqslant 2)\) (M1)
\( = 0.0775\,\,\,\,\,\)(accept 0.0778 if 3sf approximation from (a) used) A1
[3 marks]
\({\text{P}}(x < 1|x < 1.4) = \frac{{{\text{P}}(x < 1)}}{{{\text{P}}(x < 1.4)}}\) M1
\( = \frac{{0.06680…}}{{0.6914…}}\)
\( = 0.0966\,\,\,\,\,\)(accept 0.0967) A1
[2 marks]
Question
A market stall sells apples, pears and plums.
a.The weights of the apples are normally distributed with a mean of 200 grams and a standard deviation of 25 grams.
(i) Given that there are 450 apples on the stall, what is the expected number of apples with a weight of more than 225 grams?
(ii) Given that 70 % of the apples weigh less than m grams, find the value of m .[5]
b.The weights of the pears are normally distributed with a mean of ∝ grams and a standard deviation of \(\sigma \) grams. Given that 8 % of these pears have a weight of more than 270 grams and 15 % have a weight less than 250 grams, find ∝ and \(\sigma \) .[6]
c.The weights of the plums are normally distributed with a mean of 80 grams and a standard deviation of 4 grams. 5 plums are chosen at random. What is the probability that exactly 3 of them weigh more than 82 grams? [3]
▶️Answer/Explanation
Markscheme
a.(i) \({\text{P}}(X > 225) = 0.158…\) (M1)(A1)
expected number \( = 450 \times 0.158… = 71.4\) A1
(ii) \({\text{P}}(X < m) = 0.7\) (M1)
\( \Rightarrow m = 213{\text{ (grams)}}\) A1
[5 marks]
\(\frac{{270 – \mu }}{\sigma } = 1.40…\) (M1)A1
\(\frac{{250 – \mu }}{\sigma } = – 1.03…\) A1
Note: These could be seen in graphical form.
solving simultaneously (M1)
\(\mu = 258,{\text{ }}\sigma = 8.19\) A1A1
[6 marks]
\(X \sim {\text{N}}({80,4^2})\)
\({\text{P}}(X > 82) = 0.3085…\) A1
recognition of the use of binomial distribution. (M1)
\(X \sim {\text{B}}(5,\,0.3085…)\)
\({\text{P}}(X = 3) = 0.140\) A1
[3 marks]
Question
a.Farmer Suzie grows turnips and the weights of her turnips are normally distributed with a mean of \(122g\) and standard deviation of \(14.7g\).
(i) Calculate the percentage of Suzie’s turnips that weigh between \(110g\) and \(130g\).
(ii) Suzie has \(100\) turnips to take to market. Find the expected number weighing more than \(130g\).
(iii) Find the probability that at least \(30\) of the \(100g\) turnips weigh more than \(130g\). [6]
Farmer Ray also grows turnips and the weights of his turnips are normally distributed with a mean of \(144g\). Ray only takes to market turnips that weigh more than \(130g\). Over a period of time, Ray finds he has to reject \(1\) in \(15\) turnips due to their being underweight.
(i) Find the standard deviation of the weights of Ray’s turnips.
(ii) Ray has \(200\) turnips to take to market. Find the expected number weighing more than \(150g\). [6]
▶️Answer/Explanation
Markscheme
a.(i) \(P(110 < X < 130) = 0.49969 \ldots = 0.500 = 50.0\% \) (M1)A1
Note: Accept \(50\)
Note: Award M1A0 for \(0.50\) (\(0.500\))
(ii) \(P(X > 130) = (1 – 0.707 \ldots ) = 0.293 \ldots \) M1
expected number of turnips \( = 29.3\) A1
Note: Accept \(29\).
(iii) no of turnips weighing more than \(130\) is \(Y \sim B(100,{\text{ }}0.293)\) M1
\(P(Y \ge 30) = 0.478\) A1 [6 marks]
(i) \(X \sim N(144,{\text{ }}{\sigma ^2})\)
\(P(X \le 130) = \frac{1}{{15}} = 0.0667\) (M1)
\(P\left( {Z \le \frac{{130 – 144}}{\sigma }} \right) = 0.0667\)
\(\frac{{14}}{\sigma } = 1.501\) (A1)
\(\sigma = 9.33{\text{ g}}\) A1
(ii) \(P(X > 150|X > 130) = \frac{{P(X > 150)}}{{P(X > 130)}}\) M1
\( = \frac{{0.26008 \ldots }}{{1 – 0.06667}} = 0.279\) A1
expected number of turnips \( = 55.7\) A1 [6 marks] Total [12 marks]
Question
A survey is conducted in a large office building. It is found that \(30\% \) of the office workers weigh less than \(62\) kg and that \(25\% \) of the office workers weigh more than \(98\) kg.
The weights of the office workers may be modelled by a normal distribution with mean \(\mu \) and standard deviation \(\sigma \).
a.(i) Determine two simultaneous linear equations satisfied by \(\mu \) and \(\sigma \).
(ii) Find the values of \(\mu \) and \(\sigma \). [6]
Find the probability that an office worker weighs more than \(100\) kg. [1]
There are elevators in the office building that take the office workers to their offices.
Given that there are \(10\) workers in a particular elevator,
find the probability that at least four of the workers weigh more than \(100\) kg. [2]
Given that there are \(10\) workers in an elevator and at least one weighs more than \(100\) kg,
find the probability that there are fewer than four workers exceeding \(100\) kg. [3]
The arrival of the elevators at the ground floor between \(08:00\) and \(09:00\) can be modelled by a Poisson distribution. Elevators arrive on average every \(36\) seconds.
Find the probability that in any half hour period between \(08:00\) and \(09:00\) more than \(60\) elevators arrive at the ground floor. [3]
An elevator can take a maximum of \(10\) workers. Given that \(400\) workers arrive in a half hour period independently of each other,
find the probability that there are sufficient elevators to take them to their offices. [3]
▶️Answer/Explanation
Markscheme
Note: In Section B, accept answers that correctly round to 2 sf.
a.(i) let \(W\) be the weight of a worker and \(W \sim {\text{N}}(\mu ,{\text{ }}{\sigma ^2})\)
\({\text{P}}\left( {Z < \frac{{62 – \mu }}{\alpha }} \right) = 0.3\) and \({\text{P}}\left( {Z < \frac{{98 – \mu }}{\sigma }} \right) = 0.75\) (M1)
Note: Award M1 for a correctly shaded and labelled diagram.
\(\frac{{62 – \mu }}{\sigma } = {\Phi ^{ – 1}}(0.3)\;\;\;( = – 0.524 \ldots )\;\;\;\)and
\(\frac{{98 – \mu }}{\sigma } = {\Phi ^{ – 1}}(0.75)\;\;\;( = 0.674 \ldots )\)
or linear equivalents A1A1
Note: Condone equations containing the GDC inverse normal command.
(ii) attempting to solve simultaneously (M1)
\(\mu = 77.7,{\text{ }}\sigma = 30.0\) A1A1
[6 marks]
Note: In Section B, accept answers that correctly round to 2 sf.
\({\text{P}}(W > 100) = 0.229\) A1
[1 mark]
Note: In Section B, accept answers that correctly round to 2 sf.
let \(X\) represent the number of workers over \(100\) kg in a lift of ten passengers
\(X \sim {\text{B}}(10,{\text{ }}0.229 \ldots )\) (M1)
\({\text{P}}(X \ge 4) = 0.178\) A1
[2 marks]
Note: In Section B, accept answers that correctly round to 2 sf.
\({\text{P}}(X < 4|X \ge 1) = \frac{{{\text{P}}(1 \le X \le 3)}}{{{\text{P}}(X \ge 1)}}\) M1(A1)
Note: Award the M1 for a clear indication of a conditional probability.
\( = 0.808\) A1
[3 marks]
Note: In Section B, accept answers that correctly round to 2 sf.
\(L \sim {\text{Po}}(50)\) (M1)
\({\text{P}}(L > 60) = 1 – {\text{P}}(L \le 60)\) (M1)
\( = 0.0722\) A1
[3 marks]
Note: In Section B, accept answers that correctly round to 2 sf.
\(400\) workers require at least \(40\) elevators (A1)
\({\text{P}}(L \ge 40) = 1 – {\text{P}}(L \le 39)\) (M1)
\( = 0.935\) A1
[3 marks]
Total [18 marks]