## 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.

Determine the probability that a fish which is caught weighs less than 1.4 kg.

John catches 6 fish. Calculate the probability that at least 4 of the fish weigh more than 1.4 kg.

Determine the probability that a fish which is caught weighs less than 1 kg, given that it weighs less than 1.4 kg.

**Answer/Explanation**

## Markscheme

\({\text{P}}(x < 1.4) = 0.691\,\,\,\,\,\)(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]*

## Examiners report

Part (a) was almost universally correctly answered, albeit with an accuracy penalty in some cases. In (b) it was generally recognised that the distribution was binomial, but with some wavering about the correct value of the parameter *p*. Part (c) was sometimes answered correctly, but not with much confidence.

Part (a) was almost universally correctly answered, albeit with an accuracy penalty in some cases. In (b) it was generally recognised that the distribution was binomial, but with some wavering about the correct value of the parameter *p*. Part (c) was sometimes answered correctly, but not with much confidence.

Part (a) was almost universally correctly answered, albeit with an accuracy penalty in some cases. In (b) it was generally recognised that the distribution was binomial, but with some wavering about the correct value of the parameter *p*. Part (c) was sometimes answered correctly, but not with much confidence.

## Question

A market stall sells apples, pears and plums.

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 *.

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 \(\sigma \) .

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?

**Answer/Explanation**

## Markscheme

(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]*

## Examiners report

This was an accessible question for most students with many wholly correct answers seen. In part (b) a few candidates struggled to find the correct values from the calculator and in part (c) a small minority did not see the need to treat it as a binomial distribution.

This was an accessible question for most students with many wholly correct answers seen. In part (b) a few candidates struggled to find the correct values from the calculator and in part (c) a small minority did not see the need to treat it as a binomial distribution.

This was an accessible question for most students with many wholly correct answers seen. In part (b) a few candidates struggled to find the correct values from the calculator and in part (c) a small minority did not see the need to treat it as a binomial distribution.

## 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 \(\sigma \).

(i) Determine two simultaneous linear equations satisfied by \(\mu \) and \(\sigma \).

(ii) Find the values of \(\mu \) and \(\sigma \).

Find the probability that an office worker weighs more than \(100\) kg.

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.

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.

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.

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.

**Answer/Explanation**

## Markscheme

**Note: **In Section B, accept answers that correctly round to 2 sf.

(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]*

## Examiners report

[N/A]

[N/A]

[N/A]

[N/A]

[N/A]

[N/A]