5.04b Linear combinations: of normal distributions

5 Still mineral water is supplied in 1.5-litre bottles. The actual volume, \(X\) millilitres, in a bottle may be modelled by a normal distribution with mean \(\mu = 1525\) and standard deviation \(\sigma = 9.6\).

1. Determine the probability that the volume of water in a randomly selected bottle is:
   1. less than 1540 ml ;
   2. more than 1535 ml ;
   3. between 1515 ml and 1540 ml ;
   4. not 1500 ml .
2. The supplier requires that only 10 per cent of bottles should contain more than 1535 ml of water. Assuming that there has been no change in the value of \(\sigma\), calculate the reduction in the value of \(\mu\) in order to satisfy this requirement. Give your answer to one decimal place.
3. Sparkling spring water is supplied in packs of six 0.5 -litre bottles. The actual volume in a bottle may be modelled by a normal distribution with mean 508.5 ml and standard deviation 3.5 ml . Stating a necessary assumption, determine the probability that:
   1. the volume of water in each of the 6 bottles from a randomly selected pack is more than 505 ml ;
   2. the mean volume of water in the 6 bottles from a randomly selected pack is more than 505 ml .
      [0pt] [7 marks]

7

1. The number of text messages, \(N\), sent by Peter each month on his mobile phone never exceeds 40. When \(0 \leqslant N \leqslant 10\), he is charged for 5 messages.
   When \(10 < N \leqslant 20\), he is charged for 15 messages.
   When \(20 < N \leqslant 30\), he is charged for 25 messages.
   When \(30 < N \leqslant 40\), he is charged for 35 messages.
   The number of text messages, \(Y\), that Peter is charged for each month has the following probability distribution:

   \(\boldsymbol { y }\)	5	15	25	35
   \(\mathbf { P } ( \boldsymbol { Y } = \boldsymbol { y } )\)	0.1	0.2	0.3	0.4

   1. Calculate the mean and the standard deviation of \(Y\).
   2. The Goodtime phone company makes a total charge for text messages, \(C\) pence, each month given by $$C = 10 Y + 5$$ Calculate \(\mathrm { E } ( C )\).
2. The number of text messages, \(X\), sent by Joanne each month on her mobile phone is such that $$\mathrm { E } ( X ) = 8.35 \quad \text { and } \quad \mathrm { E } \left( X ^ { 2 } \right) = 75.25$$ The Newtime phone company makes a total charge for text messages, \(T\) pence, each month given by $$T = 0.4 X + 250$$ Calculate \(\operatorname { Var } ( T )\).

7

1. The random variable \(X\) has a Poisson distribution with \(\mathrm { E } ( X ) = \lambda\).
   1. Prove, from first principles, that \(\mathrm { E } ( X ( X - 1 ) ) = \lambda ^ { 2 }\).
   2. Hence deduce that \(\operatorname { Var } ( X ) = \lambda\).
2. The independent Poisson random variables \(X _ { 1 }\) and \(X _ { 2 }\) are such that \(\mathrm { E } \left( X _ { 1 } \right) = 5\) and \(\mathrm { E } \left( X _ { 2 } \right) = 2\). The random variables \(D\) and \(F\) are defined by $$D = X _ { 1 } - X _ { 2 } \quad \text { and } \quad F = 2 X _ { 1 } + 10$$
   1. Determine the mean and the variance of \(D\).
   2. Determine the mean and the variance of \(F\).
   3. For each of the variables \(D\) and \(F\), give a reason why the distribution is not Poisson.
3. The daily number of black printer cartridges sold by a shop may be modelled by a Poisson distribution with a mean of 5 . Independently, the daily number of colour printer cartridges sold by the same shop may be modelled by a Poisson distribution with a mean of 2. Use a distributional approximation to estimate the probability that the total number of black and colour printer cartridges sold by the shop during a 4 -week period ( 24 days) exceeds 175.

6 The table shows the probability distribution for the number of weekday (Monday to Friday) morning newspapers, \(X\), purchased by the Reed household per week.

1. Find values for \(\mathrm { E } ( X )\) and \(\operatorname { Var } ( X )\).
2. The number of weekday (Monday to Friday) evening newspapers, \(Y\), purchased by the same household per week is such that $$\mathrm { E } ( Y ) = 2.0 , \quad \operatorname { Var } ( Y ) = 1.5 \quad \text { and } \quad \operatorname { Cov } ( X , Y ) = - 0.43$$ Find values for the mean and variance of:
   1. \(S = X + Y\);
   2. \(\quad D = X - Y\).
3. The total cost per week, \(L\), of the Reed household's weekday morning and evening newspapers may be assumed to be normally distributed with a mean of \(\pounds 2.31\) and a standard deviation of \(\pounds 0.89\). The total cost per week, \(M\), of the household's weekend (Saturday and Sunday) newspapers may be assumed to be independent of \(L\) and normally distributed with a mean of \(\pounds 2.04\) and a standard deviation of \(\pounds 0.43\). Determine the probability that the total cost per week of the Reed household's newspapers is more than \(\pounds 5\).

5 In the manufacture of desk drawer fronts, a machine cuts sheets of veneered chipboard into rectangular pieces of width \(W\) millimetres and height \(H\) millimetres. The 4 edges of each of these pieces are then covered with matching veneered tape. The distributions of \(W\) and \(H\) are such that $$\mathrm { E } ( W ) = 350 \quad \operatorname { Var } ( W ) = 5 \quad \mathrm { E } ( H ) = 210 \quad \operatorname { Var } ( H ) = 4 \quad \rho _ { W H } = 0.75$$

1. Calculate the mean and the variance of the length of tape, \(T = 2 W + 2 H\), needed for the edges of a drawer front.
2. A desk has 4 such drawers whose sizes may be assumed to be independent. Given that \(T\) may be assumed to be normally distributed, determine the probability that the total length of tape needed for the edges of the desk's 4 drawer fronts does not exceed 4.5 metres.
   \includegraphics[max width=\textwidth, alt={}]{b855b5b3-097e-4894-aaec-d77f515949b0-13_2484_1709_223_153}

6 The weight, \(X\) grams, of a dressed pheasant may be modelled by a normal random variable with a mean of 1000 and a standard deviation of 120 . Pairs of dressed pheasants are selected for packing into boxes. The total weight of a pair, \(Y = X _ { 1 } + X _ { 2 }\) grams, may be modelled by a normal distribution with a mean of 2000 and a standard deviation of 140 .

1. 1. Show that \(\operatorname { Cov } \left( X _ { 1 } , X _ { 2 } \right) = - 4600\).
   2. Given that \(X _ { 1 } - X _ { 2 }\) may be assumed to be normally distributed, determine the probability that the difference between the weights of a selected pair of dressed pheasants exceeds 250 grams.
2. The weight of a box is independent of the total weight of a pair of dressed pheasants, and is normally distributed with a mean of 500 grams and a standard deviation of 40 grams. Determine the probability that a box containing a pair of dressed pheasants weighs less than 2750 grams.
   \includegraphics[max width=\textwidth, alt={}]{fa3bf9d6-f064-4214-acff-d8b88c33a81e-16_2486_1714_221_153}

6 Alyssa lives in the country but works in a city centre.
Her journey to work each morning involves a car journey, a walk and wait, a train journey, and a walk. Her car journey time, \(U\) minutes, from home to the village car park has a mean of 13 and a standard deviation of 3 . Her time, \(V\) minutes, to walk from the village car park to the village railway station and wait for a train to depart has a mean of 15 and a standard deviation of 6 . Her train journey time, \(W\) minutes, from the village railway station to the city centre railway station has a mean of 24 and a standard deviation of 4 . Her time, \(X\) minutes, to walk from the city centre railway station to her office has a mean of 9 and a standard deviation of 2 . The values of the product moment correlation coefficient for the above 4 variables are $$\rho _ { U V } = - 0.6 \quad \text { and } \quad \rho _ { U W } = \rho _ { U X } = \rho _ { V W } = \rho _ { V X } = \rho _ { W X } = 0$$

1. Determine values for the mean and the variance of:
   1. \(M = U + V\);
   2. \(D = W - 2 U\);
   3. \(T = M + W + X\), given that \(\rho _ { M W } = \rho _ { M X } = 0\).
2. Assuming that the variables \(M , D\) and \(T\) are normally distributed, determine the probability that, on a particular morning:
   1. Alyssa's journey time from leaving home to leaving the village railway station is exactly 30 minutes;
   2. Alyssa's train journey time is more than twice her car journey time;
   3. Alyssa's total journey time is between 50 minutes and 70 minutes.

5 The schedule for an organisation's afternoon meeting is as follows.
Session A (Speaker 1) 2.00 pm to 3.15 pm
Session B (Discussion) 3.15 pm to 3.45 pm
Session C (Speaker 2) \(\quad 3.45 \mathrm { pm }\) to 5.00 pm
Records show that:
the duration, \(X\), of Session A has mean 68 minutes and standard deviation 10 minutes;
the duration, \(Y\), of Session B has mean 25 minutes and standard deviation 5 minutes;
the duration, \(Z\), of Session C has mean 73 minutes and standard deviation 15 minutes;
and that: $$\rho _ { X Z } = 0 \quad \rho _ { X Y } = - 0.8 \quad \rho _ { Y Z } = 0$$

1. Determine the means and the variances of:
   1. \(L = X + Z\);
   2. \(M = X + Y\).
2. Assuming that \(L\) and \(M\) are each normally distributed, determine the probability that:
   1. the total time for the two speaker sessions is less than \(2 \frac { 1 } { 2 }\) hours;
   2. Session C is late in starting.

6 The demand for a WWSatNav at a superstore may be modelled by a Poisson distribution with a mean of 2.5 per day. The superstore is open 6 days each week, from Monday morning to Saturday evening.

1. 1. Determine the probability that the demand for WWSatNavs during a particular week is at most 18 .
   2. The superstore receives a delivery of WWSatNavs on each Sunday evening. The manager, Meena, requires that the probability of WWSatNavs being out of stock during a week should be at most \(5 \%\). Determine the minimum number of WWSatNavs that Meena requires to be in stock after a delivery.
2. 1. Use a distributional approximation to estimate the probability that the demand for WWSatNavs during a period of \(\mathbf { 2 }\) weeks is more than 35.
   2. Changes to the superstore's delivery schedule result in it receiving a delivery of WWSatNavs on alternate Sunday evenings. Meena now requires that the probability of WWSatNavs being out of stock during the 2 weeks following a delivery should be at most \(5 \%\). Use a distributional approximation to determine the minimum number of WWSatNavs that Meena now requires to be in stock after a delivery.
      (3 marks)

3. The time that a school pupil spends on French homework each week is normally distributed with a mean of 55 minutes and a standard deviation of 10 minutes. The time that this pupil spends on English homework each week is normally distributed with a mean of 1 hour 30 minutes and a standard deviation of 18 minutes. Find the probability that in a randomly chosen week

1. the pupil spends more than 2 hours in total doing French and English homework,
2. the pupil spends more than twice as long doing English homework as he spends doing French homework.
   (6 marks)

5. A child is playing with a set of red and blue wooden cubes. The side length of the red cubes is normally distributed with a mean of 14.5 cm and a variance of \(16.0 \mathrm {~cm} ^ { 2 }\). The side length of the blue cubes is normally distributed with a mean of 12.2 cm and a variance of \(9.0 \mathrm {~cm} ^ { 2 }\).

1. Find the probability that a randomly chosen red cube will have a side length of more than 3 cm greater than a randomly chosen blue cube. The child makes two towers, one from 4 red cubes and one from 5 blue cubes. Assuming that the cubes for each colour of tower were chosen at random,
2. find the probability that the red tower is taller than the blue tower.
3. Explain why the assumption that the cubes for each tower were chosen at random is unlikely to be realistic.

6. Four swimmers, \(A , B , C\) and \(D\), are to be used in a \(4 \times 100\) metres freestyle relay. The time for each swimmer to complete a leg follows a normal distribution. The mean and standard deviation, in seconds, of the time for each swimmer to complete a leg and the order in which they are to swim are shown in the table below.

1. Find the probability that the total time for first two legs is less than the total time for the last two.
   (6 marks)
   The total time for another team to complete this relay is normally distributed with a mean of 259.0 seconds and a standard deviation of 3.4 seconds. The two teams are to compete over four races.
2. Find the probability that the first team wins all four races, assuming that the team's performances are not affected by previous results.
   (8 marks)

2 A supermarket sells oranges. Their weights are modelled by the random variable \(X\) which has a Normal distribution with mean 345 grams and standard deviation 15 grams. When the oranges have been peeled, their weights in grams, \(Y\), are modelled by \(Y = 0.7 X\).

1. Find the probability that a randomly chosen peeled orange weighs less than 250 grams. I randomly choose 5 oranges to buy.
2. Find the probability that the total weight of the 5 unpeeled oranges is at least 1800 grams.
3. I peel three of the oranges and leave the remaining two unpeeled. Find the probability that the total weight of the two unpeeled oranges is greater than the total weight of the three peeled ones.

3 A bus runs from point A on the outskirts of a city, stops at point B outside the rail station, and continues to point C in the city centre.
The journey times for the sections A to B and B to C vary according to traffic conditions, and are modelled by independent Normal distributions with means and standard deviations as shown in the table.

1. Find the probability that a randomly chosen journey from A to B takes less than the scheduled time of 23 minutes. For every journey, the bus stops for 1 minute when it reaches B to drop off and pick up passengers.
2. Find the probability that a randomly chosen journey from A to C takes less than the scheduled time of 50 minutes. Mary travels on the bus from the station at B to her workplace at C every working day. You should assume that times for her bus journeys on different days are independent.
3. Find the probability that the total time taken for her five journeys on the bus in a randomly chosen week is at least \(2 \frac { 1 } { 2 }\) hours.
4. Comment on the assumption that times on different days are independent.

6 The length \(L\) of a particular type of fence panel is Normally distributed with mean 179.2 cm and standard deviation 0.8 cm . You should assume that the lengths of individual fence panels are independent of each other.

1. Find the probability that the length of a randomly chosen fence panel is at least 180 cm .
2. Find the probability that the total length of 5 randomly chosen fence panels is less than 895 cm . The width \(W\) of a fence post is Normally distributed with mean 9.8 cm and standard deviation 0.3 cm . A straight fence is constructed using 6 posts and 5 panels with no gaps between them. Fig. 6 shows a view from above of the first two posts, the first panel and the start of the second panel. You should assume that the lengths of fence panels and widths of fence posts are independent. \begin{figure}[h]
   \includegraphics[alt={},max width=\textwidth]{4caa7409-cb32-41da-ad64-012a45753296-6_213_1522_934_244} \captionsetup{labelformat=empty} \caption{Fig. 6}
   \end{figure}
3. Determine the probability that the total length of the fence, including the posts, is less than 9.5 m .
4. State another assumption that is necessary for the calculation of the probability in part (c) to be valid.

5 A food company makes mini apple pies. The weight of pastry in a pie is Normally distributed with mean 75 g and standard deviation 4 g . The weight of filling in a pie is Normally distributed with mean 130 g and standard deviation 8 g . You should assume that the weights of pastry and filling in a pie are independent.

1. Find the probability that the weight of pastry in a randomly chosen pie is between 70 g and 80 g .
2. Find the probability that the mean weight of filling in 10 randomly chosen pies is at least 125 g. The pies are sold in packs of 4 . The weight of the packaging is Normally distributed with mean 165 g and standard deviation 6 g .
3. In order to find the probability that the total weight of a pack of 4 pies is less than 1 kg , you must assume that the weight of the packaging is independent of the weight of the pies.
   1. State another necessary assumption.
   2. Given that the assumptions are valid, calculate this probability.

7 Two flatmates work at the same location. One of them takes the bus to work and the other one cycles. Journey times, measured in minutes, are distributed as follows.

• By bus: Normally distributed with mean 23 and standard deviation 6
• By bicycle: Normally distributed with mean 21 and standard deviation 2

You should assume that all journey times are independent.

1. One morning the two flatmates set out at the same time. Find the probability that the person who takes the bus arrives before the cyclist.
2. Find the probability that the total time taken for 5 bus journeys is less than 2 hours.
3. Comment on the assumption that all journey times are independent. \section*{END OF QUESTION PAPER} }{www.ocr.org.uk}) after the live examination series. If OCR has unwittingly failed to correctly acknowledge or clear any third-party content in this assessment material, OCR will be happy to correct its mistake at the earliest possible opportunity.
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6 The discrete random variable \(X\) has a uniform distribution over \(\{ n , n + 1 , \ldots , 2 n \}\).

1. Given that \(n\) is odd, find \(\mathrm { P } \left( X < \frac { 3 } { 2 } n \right)\).
2. Given instead that \(n\) is even, find \(\mathrm { P } \left( X < \frac { 3 } { 2 } n \right)\), giving your answer as a single algebraic fraction.
3. The sum of 6 independent values of \(X\) is denoted by \(Y\). Find \(\operatorname { Var } ( Y )\).

3 The weights of bananas sold by a supermarket are modelled by a Normal distribution with mean 205 g and standard deviation 11 g .

1. Find the probability that the total weight of 5 randomly selected bananas is at least 1 kg . When a banana is peeled the change in its weight is modelled as being a reduction of \(35 \%\).
2. Find the probability that the weight of a randomly selected peeled banana is at most 150 g Andy makes smoothies. Each smoothie is made using 2 peeled bananas and 20 strawberries from the supermarket, all the items being randomly chosen. The weight of a strawberry is modelled by a Normal distribution with mean 22.5 g and standard deviation 2.7 g .
3. Find the probability that the total weight of a smoothie is less than 700 g .

2 A manufacturer is testing how long coloured LED lights will last before the battery runs out, using two different battery types. The times in hours before the battery runs out are modelled by independent Normal distributions with means and standard deviations as shown in the table.

1. In a particular test, a battery of type A is used and the time taken for it to run out is recorded. This process is repeated until a total of 5 randomly selected batteries have been used. Determine the probability that the total time the 5 batteries last is at least 120 hours.
2. In a similar test, 3 randomly selected batteries of type A are used, one after the other. Then 2 randomly selected batteries of type B are used, one after the other. Determine the probability that the 3 type A batteries last longer in total than the 2 type B batteries.
3. Explain why it is necessary that the Normal distributions are independent in order to be able to find the probability in part (b).

3 The table shows the probability distribution of the random variable \(X\), where \(a\) and \(b\) are constants.

1. Given that \(\mathrm { E } ( X ) = 1.8\), determine the values of \(a\) and \(b\). The random variable \(Y\) is given by \(Y = 10 - 3 X\).
2. Using the values of \(a\) and \(b\) which you found in part (a), find each of the following.

4 A machine manufactures batches of 100 titanium sheets. The thickness of every sheet in a batch is Normally distributed with mean \(\mu \mathrm { mm }\) and standard deviation 0.03 mm . You should assume that each sheet is of uniform thickness and that the thicknesses of different sheets are independent of each other. The values of \(\mu\) for three different batches, A, B and C, are 3.125, 3.117 and 3.109 respectively.

1. Determine the probability that the total thickness of 10 sheets from Batch A is less than 31.0 mm .
2. Determine the probability that, if a single sheet from Batch A is cut into pieces and 10 of the pieces are stacked together, the total thickness of the stack is less than 31.0 mm .
3. Determine the probability that, if one sheet from each of Batches A, B and C are stacked together, the total thickness of the stack is at least 9.4 mm .
4. Determine the probability that the total thickness of 10 sheets from Batch A is less than the total thickness of 10 sheets from Batch B.

3 At a launderette the process of cleaning a load of clothes consists of three stages: washing, drying and folding. The times in minutes for each process are modelled by independent Normal distributions with means and standard deviations as shown in the table.

1. Find the probability that drying a randomly chosen load of clothes takes more than 50 minutes.
2. It is given that for \(99 \%\) of loads of clothes the washing time is less than \(k\) minutes. Find the value of \(k\).
3. Determine the probability that the drying time for a randomly chosen load of clothes is less than the total of the washing and folding times.
4. Determine the probability that the mean time for cleaning 5 randomly chosen loads of clothes is less than 90 minutes. You should assume that the time for cleaning any load is independent of the time for cleaning any other load.

10 Ben takes an underground train to work and back home each day. The waiting time is defined as the time from when he reaches the station platform until he boards the train. On his way to work the waiting time is \(X\) minutes, where \(X\) is modelled by a continuous uniform distribution on \([ 0,6 ]\). On his way back from work, the waiting time is \(Y\) minutes, where \(Y\) is modelled by a continuous uniform distribution on [0,4]. Ben's total waiting time for both journeys is \(Z\) minutes, where \(Z = X + Y\). You should assume that \(X\) and \(Y\) are independent.

1. Find \(\mathrm { E } ( \mathrm { Z } )\).
2. Ben thinks that \(Z\) will be well modelled by a continuous uniform distribution on \([ 0,10 ]\). By considering variances, show that he is not correct.
3. Ben's friend Jamila constructs the spreadsheet below, which shows a simulation of 20 values of \(X , Y\) and \(Z\). All of the values have been rounded to 2 decimal places.

   \multirow[b]{3}{*}{ 1 2 }	A	B	C
   	X	Y	Z
   	1.17	3.83	5.01
   3	2.01	0.81	2.82
   4	1.27	1.52	2.78
   5	1.41	3.94	5.35
   6	4.11	2.94	7.05
   7	1.76	0.96	2.72
   8	3.29	0.98	4.27
   9	0.77	0.22	0.99
   10	0.99	1.44	2.43
   11	4.79	2.43	7.22
   12	3.82	3.93	7.75
   13	5.25	2.74	7.99
   14	2.64	0.48	3.12
   15	1.54	2.18	3.72
   16	2.71	1.66	4.36
   17	0.04	3.24	3.28
   18	5.95	3.12	9.07
   19	5.22	1.21	6.42
   20	4.16	0.11	4.27
   21	1.02	0.99	2.01
   22

   Write down an estimate of \(\mathrm { P } ( Z > 6 )\).
4. Use a Normal approximation to determine the probability that Ben's total waiting time when travelling to and from work on 40 days is more than 210 minutes.

3 A supermarket sells cashew nuts in three different sizes of bag: small, medium and large. The weights in grams of the nuts in each type of bag are modelled by independent Normal distributions as shown in Table 3. \begin{table}[h] \end{table}

1. Find the probability that the mean weight of two randomly selected large bags is at least 200 g .
2. Find the probability that the total weight of eight randomly selected small bags is greater than the total weight of two randomly selected medium bags and one randomly selected large bag.