Find the mean and standard deviation of \(X\).

The school has 1200 students and classes start at 09:00. Estimate the number of students who will be late on that day.

Maelis had not arrived by 08:30. Find the probability that she arrived late.

At 15:00 it is the end of the school day and it is assumed that the departure of the students from school can be modelled by a Poisson distribution. On average 24 students leave the school every minute.

Find the probability that at least 700 students leave school before 15:30.

At 15:00 it is the end of the school day and it is assumed that the departure of the students from school can be modelled by a Poisson distribution. On average 24 students leave the school every minute.

There are 200 days in a school year. Given that \(Y\) denotes the number of days in the year that at least 700 students leave before 15:30, find

(i)     \({\text{E}}(Y)\) ;

(ii)     \(P(Y > 150)\) .

\({\text{P}}(X < 30) = 0.4\)

\({\text{P}}(X < 55) = 0.9\)

or relevant sketch     (M1)

given \(Z = \frac{{X - \mu }}{\sigma }\)

\({\text{P}}(Z < z) = 0.4 \Rightarrow \frac{{30 - \mu }}{\sigma } =  - 0.253...\)     (A1)

\({\text{P}}(Z < z) = 0.9 \Rightarrow \frac{{55 - \mu }}{\sigma } = 1.28...\)     (A1)

\(\mu  = 30 + \left( {0.253...} \right) \times \sigma  = 55 - \left( {1.28...} \right) \times \sigma \)     M1

\(\sigma  = 16.3\) , \(\mu  = 34.1\)     A1

Note: Accept 16 and 34.

Note: Working with 830 and 855 will only gain the two M marks.

[5 marks]

\(X{\text{ ~ N}}\)(\(34.12…\), \(16.28...{^2}\) )

late to school  \( \Rightarrow X > 60\)

\({\text{P}}(X > 60) = 0.056...\)     (A1)

number of students late \({\text{ =  0}}{\text{.0560}}... \times {\text{1200}}\)     (M1)

\(= 67\) (to nearest integer) A1

Note: Accept \(62\) for use of \(34\) and \(16\).

[3 marks]

\({\text{P}}(X > 60|X > 30) = \frac{{{\text{P}}(X > 60)}}{{{\text{P}}(X > 30)}}\)     M1

\( = 0.0935\) (accept anything between \(0.093\) and \(0.094\))     A1

Note: If \(34\) and \(16\) are used \(0.0870\) is obtained. This should be accepted.

[2 marks]

let \(L\) be the random variable of the number of students who leave school in a 30 minute interval

since \(24 \times 30 = 720\)     A1

\({\text{L ~ Po(720)}}\)

\({\text{P(}}L \geqslant 700) = 1 - {\text{P}}(L \leqslant 699)\)     (M1)

\( = 0.777\)     A1

Note: Award M1A0 for \(P(L > 700) = 1 - P(L \leqslant 700)\) (this leads to \(0.765\)).

[3 marks]

(i)     \(Y{\text{ ~ B}}\)(\(200\), \(0.7767…\))     (M1)

\({\text{E}}(Y) = 200 \times 0.7767... = 155\)     A1

Note: On ft, use of \(0.765\) will lead to \(153\).

(ii)     \({\text{P}}(Y > 150) = 1 - {\text{P}}(Y \leqslant 150)\)     (M1)

\( = 0.797\)     A1

Note: Accept \(0.799\) from using rounded answer.

Note: On ft, use of \(0.765\) will lead to \(0.666\).

[4 marks]

Candidates who had been prepared to solve questions from this part of the syllabus did well on the question. As a general point, candidates did not always write down clearly which distribution was being used. There were many candidates who seemed unfamiliar with the concept of Normal distributions as well as the Poisson and Binomial distributions and did not attempt the question. Parts (a) – (c) of the question were a variation on similar problems seen on previous examinations and there were a disappointing number of candidates who seemed unable to start the question. The use of 830 and 850 rather than minutes after 8am was seen and this caused students to lose marks despite knowing the method required. In general technology was used well and this was seen in (d) when solving a problem that involved a Poisson distribution. A number of candidates were unable to identify the Binomial distribution in (e).

Candidates who had been prepared to solve questions from this part of the syllabus did well on the question. As a general point, candidates did not always write down clearly which distribution was being used. There were many candidates who seemed unfamiliar with the concept of Normal distributions as well as the Poisson and Binomial distributions and did not attempt the question. Parts (a) – (c) of the question were a variation on similar problems seen on previous examinations and there were a disappointing number of candidates who seemed unable to start the question. The use of 830 and 850 rather than minutes after 8am was seen and this caused students to lose marks despite knowing the method required. In general technology was used well and this was seen in (d) when solving a problem that involved a Poisson distribution. A number of candidates were unable to identify the Binomial distribution in (e).

Candidates who had been prepared to solve questions from this part of the syllabus did well on the question. As a general point, candidates did not always write down clearly which distribution was being used. There were many candidates who seemed unfamiliar with the concept of Normal distributions as well as the Poisson and Binomial distributions and did not attempt the question. Parts (a) – (c) of the question were a variation on similar problems seen on previous examinations and there were a disappointing number of candidates who seemed unable to start the question. The use of 830 and 850 rather than minutes after 8am was seen and this caused students to lose marks despite knowing the method required. In general technology was used well and this was seen in (d) when solving a problem that involved a Poisson distribution. A number of candidates were unable to identify the Binomial distribution in (e).

Candidates who had been prepared to solve questions from this part of the syllabus did well on the question. As a general point, candidates did not always write down clearly which distribution was being used. There were many candidates who seemed unfamiliar with the concept of Normal distributions as well as the Poisson and Binomial distributions and did not attempt the question. Parts (a) – (c) of the question were a variation on similar problems seen on previous examinations and there were a disappointing number of candidates who seemed unable to start the question. The use of 830 and 850 rather than minutes after 8am was seen and this caused students to lose marks despite knowing the method required. In general technology was used well and this was seen in (d) when solving a problem that involved a Poisson distribution. A number of candidates were unable to identify the Binomial distribution in (e).

Candidates who had been prepared to solve questions from this part of the syllabus did well on the question. As a general point, candidates did not always write down clearly which distribution was being used. There were many candidates who seemed unfamiliar with the concept of Normal distributions as well as the Poisson and Binomial distributions and did not attempt the question. Parts (a) – (c) of the question were a variation on similar problems seen on previous examinations and there were a disappointing number of candidates who seemed unable to start the question. The use of 830 and 850 rather than minutes after 8am was seen and this caused students to lose marks despite knowing the method required. In general technology was used well and this was seen in (d) when solving a problem that involved a Poisson distribution. A number of candidates were unable to identify the Binomial distribution in (e).