Single tail probability P(X < a) or P(X > a)

1 It is given that \(X \sim \mathrm {~N} \left( 1.5,3.2 ^ { 2 } \right)\). Find the probability that a randomly chosen value of \(X\) is less than - 2.4 .

6. A manufacturer has a machine that fills bags with flour such that the weight of flour in a bag is normally distributed. A label states that each bag should contain 1 kg of flour.

1. The machine is set so that the weight of flour in a bag has mean 1.04 kg and standard deviation 0.17 kg . Find the proportion of bags that weigh less than the stated weight of 1 kg . The manufacturer wants to reduce the number of bags which contain less than the stated weight of 1 kg . At first she decides to adjust the mean but not the standard deviation so that only \(5 \%\) of the bags filled are below the stated weight of 1 kg .
2. Find the adjusted mean. The manufacturer finds that a lot of the bags are overflowing with flour when the mean is adjusted, so decides to adjust the standard deviation instead to make the machine more accurate. The machine is set back to a mean of 1.04 kg . The manufacturer wants \(1 \%\) of bags to be under 1 kg .
3. Find the adjusted standard deviation. Give your answer to 3 significant figures.

1. The random variables \(R , S\) and \(T\) are distributed as follows

$$R \sim \mathrm {~B} ( 15,0.3 ) , \quad S \sim \mathrm { Po } ( 7.5 ) , \quad T \sim \mathrm {~N} \left( 8,2 ^ { 2 } \right) .$$ Find

1. \(\mathrm { P } ( R = 5 )\),
2. \(\mathrm { P } ( S = 5 )\),
3. \(\mathrm { P } ( T = 5 )\).

4. A company produces jars of English Honey. The weight of the glass jars used is normally distributed with a mean of 122.3 g and a standard deviation of 2.6 g . Calculate the probability that a randomly chosen jar will weigh

1. less than 127 g ,
2. less than 121.5 g . The weight of honey put into each jar by a machine is normally distributed with a standard deviation of 1.6 g . The machine operator can adjust the mean weight of the honey put into each jar without changing the standard deviation.
3. Find, correct to 4 significant figures, the minimum that the mean weight can be set to such that at most 1 in 20 of the jars will contain less than 454 g .
   (4 marks)

8
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\end{tabular} & MATHEMATICS & 9709/52
\hline 0 & Paper 5 Probability \& Statistics 1 & October/November 2021
\hline \(\infty\) & & 1 hour 15 minutes
\hline & You must answer on the question paper. &
\hline & You will need: List of formulae (MF19) &
\hline \end{tabular} \end{center} \section*{INSTRUCTIONS}

• Answer all questions.
• Use a black or dark blue pen. You may use an HB pencil for any diagrams or graphs.
• Write your name, centre number and candidate number in the boxes at the top of the page.
• Write your answer to each question in the space provided.
• Do not use an erasable pen or correction fluid.
• Do not write on any bar codes.
• If additional space is needed, you should use the lined page at the end of this booklet; the question number or numbers must be clearly shown.
• You should use a calculator where appropriate.
• You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
• Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

\section*{INFORMATION}

• The total mark for this paper is 50.
• The number of marks for each question or part question is shown in brackets [ ].

1 Each of the 180 students at a college plays exactly one of the piano, the guitar and the drums. The numbers of male and female students who play the piano, the guitar and the drums are given in the following table. A student at the college is chosen at random.

1. Find the probability that the student plays the guitar.
2. Find the probability that the student is male given that the student plays the drums.
3. Determine whether the events 'the student plays the guitar' and 'the student is female' are independent, justifying your answer.
   2 A group of 6 people is to be chosen from 4 men and 11 women.
4. In how many different ways can a group of 6 be chosen if it must contain exactly 1 man?
   Two of the 11 women are sisters Jane and Kate.
5. In how many different ways can a group of 6 be chosen if Jane and Kate cannot both be in the group?
   3 A bag contains 5 yellow and 4 green marbles. Three marbles are selected at random from the bag, without replacement.
6. Show that the probability that exactly one of the marbles is yellow is \(\frac { 5 } { 14 }\).
   The random variable \(X\) is the number of yellow marbles selected.
7. Draw up the probability distribution table for \(X\).
8. Find \(\mathrm { E } ( X )\).
   4
9. In how many different ways can the 9 letters of the word TELESCOPE be arranged?
10. In how many different ways can the 9 letters of the word TELESCOPE be arranged so that there are exactly two letters between the T and the C ?
    5 In a certain region, the probability that any given day in October is wet is 0.16 , independently of other days.
11. Find the probability that, in a 10-day period in October, fewer than 3 days will be wet.
12. Find the probability that the first wet day in October is 8 October.
13. For 4 randomly chosen years, find the probability that in exactly 1 of these years the first wet day in October is 8 October.
    6 The times taken, in minutes, to complete a particular task by employees at a large company are normally distributed with mean 32.2 and standard deviation 9.6.
14. Find the probability that a randomly chosen employee takes more than 28.6 minutes to complete the task.
15. \(20 \%\) of employees take longer than \(t\) minutes to complete the task. Find the value of \(t\).
16. Find the probability that the time taken to complete the task by a randomly chosen employee differs from the mean by less than 15.0 minutes.
    7 The distances, \(x \mathrm {~m}\), travelled to school by 140 children were recorded. The results are summarised in the table below.

    Distance, \(x \mathrm {~m}\)	\(x \leqslant 200\)	\(x \leqslant 300\)	\(x \leqslant 500\)	\(x \leqslant 900\)	\(x \leqslant 1200\)	\(x \leqslant 1600\)
    Cumulative frequency	16	46	88	122	134	140

17. On the grid, draw a cumulative frequency graph to represent these results.
    \includegraphics[max width=\textwidth, alt={}, center]{93ff111b-0267-4b4b-a41c-64c3307115af-10_1593_1593_701_306}
18. Use your graph to estimate the interquartile range of the distances.
19. Calculate estimates of the mean and standard deviation of the distances.
    If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.

2 The volume, in millilitres, of lemonade in mini-cans may be assumed to be normally distributed with a standard deviation of 3.5. The volumes, in millilitres, of lemonade in a random sample of 12 mini-cans were as follows.

1. Construct a \(98 \%\) confidence interval for the mean volume of lemonade in a mini-can, giving the limits to one decimal place.
2. On each mini-can is printed " 150 ml ". Comment on this, using the given sample and your confidence interval in part (a).
3. State why, in part (a), use of the Central Limit Theorem was not necessary.

4 The weights of packets of sultanas may be assumed to be normally distributed with a standard deviation of 6 grams. The weights of a random sample of 10 packets were as follows:
\(\begin{array} { l l l l l l l l l l } 498 & 496 & 499 & 511 & 503 & 505 & 510 & 509 & 513 & 508 \end{array}\)

1. 1. Construct a \(99 \%\) confidence interval for the mean weight of packets of sultanas, giving the limits to one decimal place.
   2. State why, in calculating your confidence interval, use of the Central Limit Theorem was not necessary.
   3. On each packet it states 'Contents 500 grams'. Comment on this statement using both the given sample and your confidence interval.
2. Given that the mean weight of all packets of sultanas is 500 grams, state the probability that a 99\% confidence interval for the mean, calculated from a random sample of packets, will not contain 500 grams.

18

1. (ii)
   [0pt] [2 marks]
   \end{tabular}}
   \hline \end{tabular} \end{center}

3 Alan's journey time to work can be modelled by a normal distribution with standard deviation 6 minutes. Alan measures the journey time to work for a random sample of 5 journeys. The mean of the 5 journey times is 36 minutes. 3

1. Construct a 95\% confidence interval for Alan's mean journey time to work, giving your values to one decimal place.
   3
2. Alan claims that his mean journey time to work is 30 minutes.
   State, with a reason, whether or not the confidence interval found in part (a) supports Alan's claim.
   3
3. Suppose that the standard deviation is not known but a sample standard deviation is found from Alan's sample and calculated to be 6 Explain how the working in part (a) would change.

7. A health centre claims that the time a doctor spends with a patient can be modelled by a normal distribution with a mean of 10 minutes and a standard deviation of 4 minutes.

1. Using this model, find the probability that the time spent with a randomly selected patient is more than 15 minutes. Some patients complain that the mean time the doctor spends with a patient is more than 10 minutes. The receptionist takes a random sample of 20 patients and finds that the mean time the doctor spends with a patient is 11.5 minutes.
2. Stating your hypotheses clearly and using a \(5 \%\) significance level, test whether or not there is evidence to support the patients' complaint. The health centre also claims that the time a dentist spends with a patient during a routine appointment, \(T\) minutes, can be modelled by the normal distribution where \(T \sim \mathrm {~N} \left( 5,3.5 ^ { 2 } \right)\)
3. Using this model,
   1. find the probability that a routine appointment with the dentist takes less than 2 minutes
   2. find \(\mathrm { P } ( T < 2 \mid T > 0 )\)
   3. hence explain why this normal distribution may not be a good model for \(T\). The dentist believes that she cannot complete a routine appointment in less than 2 minutes.
      She suggests that the health centre should use a refined model only including values of \(T > 2\)
4. Find the median time for a routine appointment using this new model, giving your answer correct to one decimal place. Name: □ \section*{U8th A LEVEL Single Mathematics Assessment Mechanics \(7 ^ { \text {th } }\) January 2021 } Instructions
   • Answer all the questions
   • Write your answer to each question in the space provided under each question. The question number(s) must be clearly shown.
   • Use black or blue ink. Pencil may be used for graphs and diagrams only.
   • You should clearly write your name at the top of this page and on any additional sheets that you use.
   • You are permitted to use a scientific or graphical calculator in this paper.
   • Final answers should be given to a degree of accuracy appropriate to the context.
   Information
   • The total mark for this paper is 55 marks.
   • The marks for each question are shown in brackets ( ).
   • You are reminded of the need for clear presentation in your answers.
   • You should allow approximately 60 minutes for this section of the test
   \section*{Formulae} \section*{A Level Mathematics A (H240)} \section*{Arithmetic series} \(S _ { n } = \frac { 1 } { 2 } n ( a + l ) = \frac { 1 } { 2 } n \{ 2 a + ( n - 1 ) d \}\) \section*{Geometric series} \(S _ { n } = \frac { a \left( 1 - r ^ { n } \right) } { 1 - r }\)
   \(S _ { \infty } = \frac { a } { 1 - r }\) for \(| r | < 1\) \section*{Binomial series} \(( a + b ) ^ { n } = a ^ { n } + { } ^ { n } \mathrm { C } _ { 1 } a ^ { n - 1 } b + { } ^ { n } \mathrm { C } _ { 2 } a ^ { n - 2 } b ^ { 2 } + \ldots + { } ^ { n } \mathrm { C } _ { r } a ^ { n - r } b ^ { r } + \ldots + b ^ { n } \quad ( n \in \mathbb { N } )\), where \({ } ^ { n } \mathrm { C } _ { r } = { } _ { n } \mathrm { C } _ { r } = \binom { n } { r } = \frac { n ! } { r ! ( n - r ) ! }\)
   \(( 1 + x ) ^ { n } = 1 + n x + \frac { n ( n - 1 ) } { 2 ! } x ^ { 2 } + \ldots + \frac { n ( n - 1 ) \ldots ( n - r + 1 ) } { r ! } x ^ { r } + \ldots \quad ( | x | < 1 , n \in \mathbb { R } )\) \section*{Differentiation}

   \(\mathrm { f } ( x )\)	\(\mathrm { f } ^ { \prime } ( x )\)
   \(\tan k x\)	\(k \sec ^ { 2 } k x\)
   \(\sec x\)	\(\sec x \tan x\)
   \(\cot x\)	\(- \operatorname { cosec } ^ { 2 } x\)
   \(\operatorname { cosec } x\)	\(- \operatorname { cosec } x \cot x\)

   Quotient rule \(y = \frac { u } { v } , \frac { \mathrm {~d} y } { \mathrm {~d} x } = \frac { v \frac { \mathrm {~d} u } { \mathrm {~d} x } - u \frac { \mathrm {~d} v } { \mathrm {~d} x } } { v ^ { 2 } }\) \section*{Differentiation from first principles} \(\mathrm { f } ^ { \prime } ( x ) = \lim _ { h \rightarrow 0 } \frac { \mathrm { f } ( x + h ) - \mathrm { f } ( x ) } { h }\) \section*{Integration} \(\int \frac { \mathrm { f } ^ { \prime } ( x ) } { \mathrm { f } ( x ) } \mathrm { d } x = \ln | \mathrm { f } ( x ) | + c\)
   \(\int \mathrm { f } ^ { \prime } ( x ) ( \mathrm { f } ( x ) ) ^ { n } \mathrm {~d} x = \frac { 1 } { n + 1 } ( \mathrm { f } ( x ) ) ^ { n + 1 } + c\)
   Integration by parts \(\int u \frac { \mathrm {~d} v } { \mathrm {~d} x } \mathrm {~d} x = u v - \int v \frac { \mathrm {~d} u } { \mathrm {~d} x } \mathrm {~d} x\) Small angle approximations
   \(\sin \theta \approx \theta , \cos \theta \approx 1 - \frac { 1 } { 2 } \theta ^ { 2 } , \tan \theta \approx \theta\) where \(\theta\) is measured in radians \section*{Trigonometric identities} \(\sin ( A \pm B ) = \sin A \cos B \pm \cos A \sin B\)
   \(\cos ( A \pm B ) = \cos A \cos B \mp \sin A \sin B\)
   \(\tan ( A \pm B ) = \frac { \tan A \pm \tan B } { 1 \mp \tan A \tan B } \quad \left( A \pm B \neq \left( k + \frac { 1 } { 2 } \right) \pi \right)\) \section*{Numerical methods} Trapezium rule: \(\int _ { a } ^ { b } y \mathrm {~d} x \approx \frac { 1 } { 2 } h \left\{ \left( y _ { 0 } + y _ { n } \right) + 2 \left( y _ { 1 } + y _ { 2 } + \ldots + y _ { n - 1 } \right) \right\}\), where \(h = \frac { b - a } { n }\)
   The Newton-Raphson iteration for solving \(\mathrm { f } ( x ) = 0 : x _ { n + 1 } = x _ { n } - \frac { \mathrm { f } \left( x _ { n } \right) } { \mathrm { f } ^ { \prime } \left( x _ { n } \right) }\) \section*{Probability} \(\mathrm { P } ( A \cup B ) = \mathrm { P } ( A ) + \mathrm { P } ( B ) - \mathrm { P } ( A \cap B )\)
   \(\mathrm { P } ( A \cap B ) = \mathrm { P } ( A ) \mathrm { P } ( B \mid A ) = \mathrm { P } ( B ) \mathrm { P } ( A \mid B )\) or \(\mathrm { P } ( A \mid B ) = \frac { \mathrm { P } ( A \cap B ) } { \mathrm { P } ( B ) }\) \section*{Standard deviation} \(\sqrt { \frac { \sum ( x - \bar { x } ) ^ { 2 } } { n } } = \sqrt { \frac { \sum x ^ { 2 } } { n } - \bar { x } ^ { 2 } }\) or \(\sqrt { \frac { \sum f ( x - \bar { x } ) ^ { 2 } } { \sum f } } = \sqrt { \frac { \sum f x ^ { 2 } } { \sum f } - \bar { x } ^ { 2 } }\) \section*{The binomial distribution} If \(X \sim \mathbf { B } ( n , p )\) then \(\mathbf { P } ( X = x ) = \binom { n } { x } p ^ { x } ( 1 - p ) ^ { n - x }\), Mean of \(X\) is \(n p\), Variance of \(X\) is \(n p ( 1 - p )\) \section*{Hypothesis test for the mean of a normal distribution} If \(X \sim \mathrm {~N} \left( \mu , \sigma ^ { 2 } \right)\) then \(\bar { X } \sim \mathrm {~N} \left( \mu , \frac { \sigma ^ { 2 } } { n } \right)\) and \(\frac { \bar { X } - \mu } { \sigma / \sqrt { n } } \sim \mathrm {~N} ( 0,1 )\) \section*{Percentage points of the normal distribution} If \(Z\) has a normal distribution with mean 0 and variance 1 then, for each value of \(p\), the table gives the value of \(z\) such that \(\mathrm { P } ( Z \leqslant z ) = p\).

   \(p\)	0.75	0.90	0.95	0.975	0.99	0.995	0.9975	.0 .999	0.9995
   \(z\)	0.674	1.282	1.645	1.960	2.326	2.576	2.807	3.090	3.291

   \section*{Kinematics} Motion in a straight line
   \(v = u + a t\)
   \(s = u t + \frac { 1 } { 2 } a t ^ { 2 }\)
   \(s = \frac { 1 } { 2 } ( u + v ) t\)
   \(v ^ { 2 } = u ^ { 2 } + 2 a s\)
   \(s = v t - \frac { 1 } { 2 } a t ^ { 2 }\) Motion in two dimensions
   \(\mathbf { v } = \mathbf { u } + \mathbf { a } t\)
   \(\mathbf { s } = \mathbf { u } t + \frac { 1 } { 2 } \mathbf { a } t ^ { 2 }\)
   \(\mathbf { s } = \frac { 1 } { 2 } ( \mathbf { u } + \mathbf { v } ) \boldsymbol { t }\)
   \(\mathbf { s } = \mathbf { v } t - \frac { 1 } { 2 } \mathbf { a } t ^ { 2 }\)
   [0pt] [BLANK PAGE]
   1. A vehicle is driven at a constant speed of \(12 \mathrm {~ms} ^ { - 1 }\) along a straight horizontal road. Only one of the statements below is correct. Identify the correct statement.
   Tick ( \(\checkmark\) ) one box. The vehicle is accelerating □ The vehicle's driving force exceeds the total force resisting its motion □ The resultant force acting on the vehicle is zero □ The resultant force acting on the vehicle is dependent on its mass □
   2. A number of forces act on a particle such that the resultant force is \(\binom { 6 } { - 3 } \mathrm {~N}\)
   One of the forces acting on the particle is \(\binom { 8 } { - 5 } \mathrm {~N}\)
   Calculate the total of the other forces acting on the particle.
   Circle your answer.
   [0pt] [1 mark] $$\binom { 2 } { - 2 } \mathrm {~N} \quad \binom { 14 } { - 8 } \mathrm {~N} \quad \binom { - 2 } { 2 } \mathrm {~N} \quad \binom { - 14 } { 8 } \mathrm {~N}$$ 3. A rough plane is inclined to the horizontal at an angle \(\alpha\), where \(\tan \alpha = \frac { 3 } { 4 }\)
   A brick \(P\) of mass \(m\) is placed on the plane.
   The coefficient of friction between \(P\) and the plane is \(\mu\)
   Brick \(P\) is in equilibrium and on the point of sliding down the plane.
   Brick \(P\) is modelled as a particle.
   Using the model,
5. find, in terms of \(m\) and \(g\), the magnitude of the normal reaction of the plane on brick \(P\)
6. show that \(\mu = \frac { 3 } { 4 }\) For parts (c) and (d), you are not required to do any further calculations.
   Brick \(P\) is now removed from the plane and a much heavier brick \(Q\) is placed on the plane. The coefficient of friction between \(Q\) and the plane is also \(\frac { 3 } { 4 }\)
7. Explain briefly why brick \(Q\) will remain at rest on the plane. Brick \(Q\) is now projected with speed \(0.5 \mathrm {~ms} ^ { - 1 }\) down a line of greatest slope of the plane.
   Brick \(Q\) is modelled as a particle.
   Using the model,
8. describe the motion of brick \(Q\), giving a reason for your answer.
   4. A particle \(P\) moves with acceleration \(( 4 \mathbf { i } - 5 \mathbf { j } ) \mathrm { ms } ^ { - 2 }\)
   At time \(t = 0 , P\) is moving with velocity \(( - 2 \mathbf { i } + 2 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }\)
9. Find the velocity of \(P\) at time \(t = 2\) seconds. At time \(t = 0 , P\) passes through the origin \(O\).
   At time \(t = T\) seconds, where \(T > 0\), the particle \(P\) passes through the point \(A\).
   The position vector of \(A\) is \(( \lambda \mathbf { i } - 4.5 \mathbf { j } ) \mathrm { m }\) relative to \(O\), where \(\lambda\) is a constant.
10. Find the value of \(T\).
11. Hence find the value of \(\lambda\)
    5.
    1. At time \(t\) seconds, where \(t \geqslant 0\), a particle \(P\) moves so that its acceleration \(\mathbf { a } \mathrm { ms } ^ { - 2 }\) is given by $$\mathbf { a } = ( 1 - 4 t ) \mathbf { i } + \left( 3 - t ^ { 2 } \right) \mathbf { j }$$ At the instant when \(t = 0\), the velocity of \(P\) is \(36 \mathbf { i } \mathrm {~m} \mathrm {~s} ^ { - 1 }\)
12. Find the velocity of \(P\) when \(t = 4\)
13. Find the value of \(t\) at the instant when \(P\) is moving in a direction perpendicular to \(\mathbf { i }\)
    (ii) At time \(t\) seconds, where \(t \geqslant 0\), a particle \(Q\) moves so that its position vector \(\mathbf { r }\) metres, relative to a fixed origin \(O\), is given by $$\mathbf { r } = \left( t ^ { 2 } - t \right) \mathbf { i } + 3 t \mathbf { j }$$ Find the value of \(t\) at the instant when the speed of \(Q\) is \(5 \mathrm {~m} \mathrm {~s} ^ { - 1 }\)
    6. \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{d1809ec7-dccf-446b-8a1d-f04a4252ebec-28_529_993_374_529} \captionsetup{labelformat=empty} \caption{Figure 2}
    \end{figure} A small ball is projected with speed \(U \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from a point \(O\) at the top of a vertical cliff.
    The point \(O\) is 25 m vertically above the point \(N\) which is on horizontal ground.
    The ball is projected at an angle of \(45 ^ { \circ }\) above the horizontal.
    The ball hits the ground at a point \(A\), where \(A N = 100 \mathrm {~m}\), as shown in Figure 2 .
    The motion of the ball is modelled as that of a particle moving freely under gravity.
    Using this initial model,
14. show that \(U = 28\)
15. find the greatest height of the ball above the horizontal ground \(N A\). In a refinement to the model of the motion of the ball from \(O\) to \(A\), the effect of air resistance is included. This refined model is used to find a new value of \(U\).
16. How would this new value of \(U\) compare with 28 , the value given in part (a)?
17. State one further refinement to the model that would make the model more realistic.
    7. Block \(A\), of mass 0.2 kg , lies at rest on a rough plane.
    The plane is inclined at an angle \(\theta\) to the horizontal, such that \(\tan \theta = \frac { 7 } { 24 }\)
    A light inextensible string is attached to \(A\) and runs parallel to the line of greatest slope until it passes over a smooth fixed pulley at the top of the slope. The other end of this string is attached to particle \(B\), of mass 2 kg , which is held at rest so that the string is taut, as shown in the diagram below.
    \includegraphics[max width=\textwidth, alt={}, center]{d1809ec7-dccf-446b-8a1d-f04a4252ebec-32_424_1070_815_486}
18. \(\quad B\) is released from rest so that it begins to move vertically downwards with an acceleration of \(\frac { 543 } { 625 } \mathrm {~g} \mathrm {~m} \mathrm {~s} ^ { - 2 }\) Show that the coefficient of friction between \(A\) and the surface of the inclined plane is 0.17
19. In this question use \(g = 9.81 \mathrm {~m} \mathrm {~s} ^ { - 2 }\) When \(A\) reaches a speed of \(0.5 \mathrm {~ms} ^ { - 1 }\) the string breaks.
20. 1. Find the distance travelled by \(A\) after the string breaks until first coming to rest.
21. (ii) State an assumption that could affect the validity of your answer to part (b)(i).

7. A health centre claims that the time a doctor spends with a patient can be modelled by a normal distribution with a mean of 10 minutes and a standard deviation of 4 minutes.

1. Using this model, find the probability that the time spent with a randomly selected patient is more than 15 minutes. Some patients complain that the mean time the doctor spends with a patient is more than 10 minutes. The receptionist takes a random sample of 20 patients and finds that the mean time the doctor spends with a patient is 11.5 minutes.
2. Stating your hypotheses clearly and using a \(5 \%\) significance level, test whether or not there is evidence to support the patients' complaint. The health centre also claims that the time a dentist spends with a patient during a routine appointment, \(T\) minutes, can be modelled by the normal distribution where \(T \sim \mathrm {~N} \left( 5,3.5 ^ { 2 } \right)\)
3. Using this model,
   1. find the probability that a routine appointment with the dentist takes less than 2 minutes
   2. find \(\mathrm { P } ( T < 2 \mid T > 0 )\)
   3. hence explain why this normal distribution may not be a good model for \(T\). The dentist believes that she cannot complete a routine appointment in less than 2 minutes.
      She suggests that the health centre should use a refined model only including values of \(T > 2\)
4. Find the median time for a routine appointment using this new model, giving your answer correct to one decimal place.