Multiple stage process probability

A question is this type if and only if it involves finding the probability for a process with 3+ independent stages where times/amounts are normally distributed (e.g. triathlon, multi-stage journey, game rounds).

17 questions · Standard +0.6

5.04b Linear combinations: of normal distributions
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CAIE S2 2020 Specimen Q4
10 marks Standard +0.8
4 The lifetimes, in hours, of Longlive light bulbs and Enerlow light bulbs have the independent distributions \(\mathrm { N } \left( 1020,45 ^ { 2 } \right)\) and \(\mathrm { N } \left( 2800,52 ^ { 2 } \right)\) respectively.
  1. Find the probability that the total of the lifetimes of five randomly chosen Longlive bulbs is less than 5200 hours.
  2. Find the probability that the lifetime of a randomly chosen Enerlow bulb is at least three times that of a randomly chosen Longlive bulb.
CAIE S2 2004 June Q2
6 marks Moderate -0.3
2 In athletics matches the triple jump event consists of a hop, followed by a step, followed by a jump. The lengths covered by Albert in each part are independent normal variables with means \(3.5 \mathrm {~m} , 2.9 \mathrm {~m}\), 3.1 m and standard deviations \(0.3 \mathrm {~m} , 0.25 \mathrm {~m} , 0.35 \mathrm {~m}\) respectively. The length of the triple jump is the sum of the three parts.
  1. Find the mean and standard deviation of the length of Albert's triple jumps.
  2. Find the probability that the mean of Albert's next four triple jumps is greater than 9 m .
CAIE S2 2009 June Q6
9 marks Challenging +1.2
6 When Sunil travels from his home in England to visit his relatives in India, his journey is in four stages. The times, in hours, for the stages have independent normal distributions as follows. Bus from home to the airport: \(\quad \mathrm { N } ( 3.75,1.45 )\) Waiting in the airport: \(\quad \mathrm { N } ( 3.1,0.785 )\) Flight from England to India: \(\quad \mathrm { N } ( 11,1.3 )\) Car in India to relatives: \(\quad \mathrm { N } ( 3.2,0.81 )\)
  1. Find the probability that the flight time is shorter than the total time for the other three stages.
  2. Find the probability that, for 6 journeys to India, the mean time waiting in the airport is less than 4 hours.
CAIE S2 2015 June Q3
6 marks Standard +0.3
3 The daily times, in minutes, that Yu Ming takes showering, getting dressed and having breakfast are independent and have the distributions \(\mathrm { N } \left( 9,2.2 ^ { 2 } \right) , \mathrm { N } \left( 8,1.3 ^ { 2 } \right)\) and \(\mathrm { N } \left( 17,2.6 ^ { 2 } \right)\) respectively. The total daily time that Yu Ming takes for all three activities is denoted by \(T\) minutes.
  1. Find the mean and variance of \(T\).
  2. Yu Ming notes the value of \(T\) on each day in a random sample of 70 days and calculates the sample mean. Find the probability that the sample mean is between 33 and 35 .
CAIE S2 2005 November Q7
12 marks Challenging +1.2
7 A journey in a certain car consists of two stages with a stop for filling up with fuel after the first stage. The length of time, \(T\) minutes, taken for each stage has a normal distribution with mean 74 and standard deviation 7.3. The length of time, \(F\) minutes, it takes to fill up with fuel has a normal distribution with mean 5 and standard deviation 1.7. The length of time it takes to pay for the fuel is exactly 4 minutes. The variables \(T\) and \(F\) are independent and the times for the two stages are independent of each other.
  1. Find the probability that the total time for the journey is less than 154 minutes.
  2. A second car has a fuel tank with exactly twice the capacity of the first car. Find the mean and variance of this car's fuel fill-up time.
  3. This second car's time for each stage of the journey follows a normal distribution with mean 69 minutes and standard deviation 5.2 minutes. The length of time it takes to pay for the fuel for this car is also exactly 4 minutes. Find the probability that the total time for the journey taken by the first car is more than the total time taken by the second car.
OCR S3 2006 June Q7
12 marks Challenging +1.2
7 A queue of cars has built up at a set of traffic lights which are at red. When the lights turn green, the time for the first car to start to move has a normal distribution with mean 2.2 s and standard deviation 0.75 s . This time is the reaction time for the first car. For each subsequent car the reaction time is the time taken for it to start to move after the car in front starts to move. These reaction times have identical normal distributions with mean 1.8 s and standard deviation 0.70 s . It may be assumed that all reaction times are independent.
  1. Calculate the probability that the reaction time for the second car in the queue is less than half of the reaction time for the first car.
  2. Calculate the probability that the fifth car in the queue starts to move less than 10 seconds after the lights turn green.
  3. State where, in part (i), independence is required.
OCR MEI S3 2007 January Q3
18 marks Standard +0.3
3 Bill and Ben run their own gardening company. At regular intervals throughout the summer they come to work on my garden, mowing the lawns, hoeing the flower beds and pruning the bushes. From past experience it is known that the times, in minutes, spent on these tasks can be modelled by independent Normally distributed random variables as follows.
MeanStandard deviation
Mowing444.8
Hoeing322.6
Pruning213.7
  1. Find the probability that, on a randomly chosen visit, it takes less than 50 minutes to mow the lawns.
  2. Find the probability that, on a randomly chosen visit, the total time for hoeing and pruning is less than 50 minutes.
  3. If Bill mows the lawns while Ben does the hoeing and pruning, find the probability that, on a randomly chosen visit, Ben finishes first. Bill and Ben do my gardening twice a month and send me an invoice at the end of the month.
  4. Write down the mean and variance of the total time (in minutes) they spend on mowing, hoeing and pruning per month.
  5. The company charges for the total time spent at 15 pence per minute. There is also a fixed charge of \(\pounds 10\) per month. Find the probability that the total charge for a month does not exceed \(\pounds 40\).
OCR MEI S3 2012 June Q3
18 marks Standard +0.3
3 The triathlon is a sports event in which competitors take part in three stages, swimming, cycling and running, one straight after the other. The winner is the competitor with the shortest overall time. In this question the times for the separate stages are assumed to be Normally distributed and independent of each other. For a particular triathlon event in which there was a very large number of competitors, the mean and standard deviation of the times, measured in minutes, for each stage were as follows.
Mean
Standard
deviation
Swimming11.072.36
Cycling57.338.76
Running24.233.75
  1. For a randomly chosen competitor, find the probability that the swimming time is between 10 and 13 minutes.
  2. For a randomly chosen competitor, find the probability that the running time exceeds the swimming time by more than 10 minutes.
  3. For a randomly chosen competitor, find the probability that the swimming and running times combined exceed \(\frac { 2 } { 3 }\) of the cycling time.
  4. In a different triathlon event the total times, in minutes, for a random sample of 12 competitors were as follows. $$\begin{array} { l l l l l l l l l l l l } 103.59 & 99.04 & 85.03 & 81.34 & 106.79 & 89.14 & 98.55 & 98.22 & 108.87 & 116.29 & 102.51 & 92.44 \end{array}$$ Find a 95\% confidence interval for the mean time of all competitors in this event.
  5. Discuss briefly whether the assumptions of Normality and independence for the stages of triathlon events are reasonable.
OCR MEI S3 2016 June Q1
18 marks Standard +0.8
1 A game consists of 20 rounds. Each round is denoted as either a starter, middle or final round. The times taken for each round are independently and Normally distributed with the following parameters (given in seconds).
Type of roundMeanStandard deviation
Starter20015
Middle22025
Final25020
The game consists of 4 starter, 12 middle and 4 final rounds. Find the probability that
  1. the mean time per round for the 4 final rounds will exceed 260 seconds,
  2. all 20 rounds will be completed in a total time of 75 minutes or less,
  3. the 12 middle rounds will take at least 3.5 times as long in total as the 4 starter rounds,
  4. the mean time per round for the 12 middle rounds will be at least 25 seconds less than the mean time per round for the 4 final rounds.
Edexcel S3 2022 January Q5
17 marks Standard +0.3
  1. Charlie is training for three events: a 1500 m swim, a 40 km bike ride and a 10 km run.
From past experience his times, in minutes, for each of the three events independently have the following distributions. $$\begin{aligned} & S \sim \mathrm {~N} \left( 41,5.2 ^ { 2 } \right) \text { represents the time for the swim } \\ & B \sim \mathrm {~N} \left( 81,4.2 ^ { 2 } \right) \text { represents the time for the bike ride } \\ & R \sim \mathrm {~N} \left( 57,6.6 ^ { 2 } \right) \text { represents the time for the run } \end{aligned}$$
  1. Find the probability that Charlie's total time for a randomly selected swim, bike ride and run exceeds 3 hours.
  2. Find the probability that the time for a randomly selected swim will be at least 20 minutes quicker than the time for a randomly selected run. Given that \(\mathrm { P } ( S + B + R > t ) = 0.95\)
  3. find the value of \(t\) A triathlon consists of a 1500 m swim, immediately followed by a 40 km bike ride, immediately followed by a 10 km run. Charlie uses the answer to part (a) to find the probability that, in 6 successive independent triathlons, his time will exceed 3 hours on at least one occasion.
  4. Find the answer Charlie should obtain. Jane says that Charlie should not have used the answer to part (a) for the calculation in part (d).
  5. Explain whether or not Jane is correct.
Edexcel S3 2018 June Q7
15 marks Standard +0.8
7.(i)As part of a recruitment exercise candidates are required to complete three separate tasks.The times taken,\(A , B\) and \(C\) ,in minutes,for candidates to complete the three tasks are such that $$A \sim \mathrm {~N} \left( 21,2 ^ { 2 } \right) , B \sim \mathrm {~N} \left( 32,7 ^ { 2 } \right) \text { and } C \sim \mathrm {~N} \left( 45,9 ^ { 2 } \right)$$ The time taken by an individual candidate to complete each task is assumed to be independent of the time taken to complete each of the other tasks. A candidate is selected at random.
  1. Find the probability that the candidate takes a total time of more than 90 minutes to complete all three tasks.
  2. Find \(\mathrm { P } ( A > B )\) (ii)A simple random sample,\(X _ { 1 } , X _ { 2 } , X _ { 3 } , X _ { 4 }\) ,is taken from a normal population with mean \(\mu\) and standard deviation \(\sigma\) Given that $$\bar { X } = \frac { X _ { 1 } + X _ { 2 } + X _ { 3 } + X _ { 4 } } { 4 }$$ and that $$\mathrm { P } \left( X _ { 1 } > \bar { X } + k \sigma \right) = 0.1$$ where \(k\) is a constant,
    find the value of \(k\) ,giving your answer correct to 3 significant figures.
    END
AQA S3 2012 June Q6
17 marks Standard +0.8
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.
AQA S3 2013 June Q5
10 marks Standard +0.3
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.
Edexcel S3 Q6
14 marks Challenging +1.2
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.
meanstandard deviation
\(1 ^ { \text {st } }\) leg \(- A\)63.11.2
\(2 ^ { \text {nd } }\) leg \(- B\)65.71.5
\(3 ^ { \text {rd } } \operatorname { leg } - C\)65.41.8
\(4 ^ { \text {th } }\) leg - \(D\)62.50.9
  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)
OCR MEI Further Statistics Major 2024 June Q3
8 marks Standard +0.3
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.
\cline { 2 - 3 } \multicolumn{1}{c|}{}MeanStandard deviation
Washing352.4
Drying463.1
Folding122.2
  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.
OCR Further Statistics 2018 December Q1
5 marks Standard +0.8
1 The performance of a piece of music is being recorded. The piece consists of three sections, \(A , B\) and \(C\). The times, in seconds, taken to perform the three sections are normally distributed random variables with the following means and standard deviations.
SectionMeanStandard deviation
\(A\)26413
\(B\)1739
\(C\)26413
  1. Assume first that the times for the three sections are independent. Find the probability that the total length of the performance is greater than 720.0 seconds.
  2. In fact sections \(A\) and \(C\) are musically identical, and the recording is made by using a single performance of section \(A\) twice, together with a performance of section \(B\). In this case find the probability that the total length of the performance is greater than 720.0 seconds.
OCR Further Statistics 2021 June Q1
5 marks Standard +0.3
1 The performance of a piece of music is being recorded. The piece consists of three sections, \(A , B\) and \(C\). The times, in seconds, taken to perform the three sections are normally distributed random variables with the following means and standard deviations. \end{table}
QuestionAnswerMarkAOGuidance
\multirow[t]{3}{*}{1}\multirow[t]{3}{*}{(a)}
\(A + B + C \sim \mathrm {~N} ( 701 , \ldots\)
.. 419)
M11.1aNormal, mean \(\mu _ { A } + \mu _ { B } + \mu _ { C }\)\multirow{3}{*}{}
A11.1Variance 419
\(\mathrm { P } ( > 720 ) = 0.176649\)A11.1Answer, 0.177 or better, www
\multirow[t]{2}{*}{1}\multirow[t]{2}{*}{(b)}\(2 A + B \sim \mathrm {~N} ( 701,757 )\)M11.1aNormal, same mean, \(4 \sigma _ { A } { } ^ { 2 } + \sigma _ { B } { } ^ { 2 }\)\multirow{2}{*}{}
\(\mathrm { P } ( > 720 ) = 0.244919\)A1 [2]1.1Answer, art 0.245
\multirow{2}{*}{2}\multirow{2}{*}{(a)}\(\frac { { } ^ { 8 } C _ { 3 } \times { } ^ { 20 } C _ { 5 } } { { } ^ { 28 } C _ { 8 } }\)M1 A13.1b 1.1(Product of two \({ } ^ { n } C _ { r }\) ) ÷ \({ } ^ { n } C _ { r }\) At least two \({ } ^ { n } C _ { r }\) correct\multirow[t]{2}{*}{Or \(\frac { 8 } { 28 } \times \frac { 7 } { 27 } \times \frac { 6 } { 26 } \times \frac { 20 } { 25 } \times \ldots \times \frac { 16 } { 21 } \times { } ^ { 8 } C _ { 3 } = 0.27934 \ldots\)}
\(\frac { 56 \times 15504 } { 3108105 } = 0.27934 \ldots\)A1 [3]1.1Any exact form or awrt 0.279
2(b)
× B × B × B × B × B × B × B × B x
GGG in one \(\mathrm { x } , \mathrm { G }\) in another: \(9 \times 8\) \(\div \frac { 12 ! } { 8 ! \times 4 ! }\) \(= \frac { 72 } { 495 } = \frac { 8 } { 55 } \text { or } 0.145 \ldots\)
M1 A13.1b 2.1
Or e.g. \(\frac { 10 ! } { 8 ! } - 2 \times 9\)
Divide by \({ } _ { 12 } \mathrm { C } _ { 4 }\) oe
Or, e.g. find \({ } _ { 12 } \mathrm { C } _ { 4 }\) - (\# (all separate) +\#(all together) \(+ \# ( 2,1,1 ) \times 3 +\) \#(2,2))
M11.1
A11.1
[4]
QuestionAnswerMarkAOGuidance
\multirow{7}{*}{3}\multirow{7}{*}{(a)}\(\mathrm { H } _ { 0 } : \mu = 700\)B21.1One error, e.g. no or wrongIgnore failure to define \(\mu\)
\(\mathrm { H } _ { 1 } : \mu < 700\) where \(\mu\) is the mean reaction1.1letter, \(\neq\), etc : B1here
\(\bar { x } = 607\)M13.3Find sample mean
\(z = - 1.822\) or \(p = 0.0342\) or \(\mathrm { CV } = 616.05 \ldots\)A13.4Correct \(z , p\) or CV
\(z < - 1.645\) or \(p < 0.05\) or \(607 < \mathrm { CV }\)A11.1Correct comparison
Reject \(\mathrm { H } _ { 0 }\)M1ft1.1Correct first conclusionNeeds correct method, like-
Significant evidence that mean reaction timesA1ft2.2bContext, not too definite (e.g. not "international athletes' reaction times are shorter"ft on their \(z , p\) or CV
3(b)(i)Uses more information (e.g. magnitudes of differences)B1 [1]2.4
\multirow{5}{*}{3}\multirow{5}{*}{(b)}\multirow{5}{*}{(ii)}\(\mathrm { H } _ { 0 } : m = 700 , \mathrm { H } _ { 1 } : m < 700\) where \(m\) is the median reaction time for all international athletesB12.5Same as in (i) but different letter or "median" stated
\(W _ { - } = 18\)
\(W _ { + } = 3\) so \(T = 3\)
For both, and \(T\) correct
\(n = 6 , \mathrm { CV } = 2\)A11.1Correct CV
Do not reject \(\mathrm { H } _ { 0 }\). Insufficient evidence that median reaction times of international athletes are shorterA1ft [6]2.2bIn context, not too definiteFT on their \(T\)
3(c)They use different assumptionsB1 [1]2.3Not "one is more accurate"
QuestionAnswerMarkAOGuidance
4(a)\(\begin{aligned}\int _ { 0 } ^ { a } x \frac { 2 x } { a ^ { 2 } } d x = 4
{ \left[ \frac { 2 x ^ { 3 } } { 3 a ^ { 2 } } \right] = 4 }
\frac { 2 } { 3 } a = 4 \Rightarrow a = 6 \end{aligned}\)
M1
B1
A1 [3]
3.1a
1.1
2.2a
4(b)
\(\mathrm { F } ( x ) = \frac { x ^ { 2 } } { 36 }\)
Let the CDF of \(M\) be \(\mathrm { H } ( m )\). Then \(\mathrm { H } ( m ) = \mathrm { P } (\) all observations less than \(m )\) \(= [ \mathrm { P } ( X \leqslant m ) ] ^ { 5 }\) \(= \left[ \frac { m ^ { 2 } } { 36 } \right] ^ { 5 }\)
\(\mathrm { H } ( m ) = \begin{cases} 0m < 0 ,
\frac { m ^ { 10 } } { 60466176 }0 \leq m \leq 6 ,
1m > 6 . \end{cases}\)
M1 A1ft
M1
M1
A1
A1
A1
A1
[8]
1.1
1.1
2.1
3.1a
2.2a
2.1
2.1
1.2
Find \(\mathrm { F } ( x ) ; = \frac { x ^ { 2 } } { a ^ { 2 } }\)
Correct basis for CDF of \(m\)
Correct function, any letter Range \(0 \leq m \leq 6\)
Letter not \(x\), and 0, 1 present
ft on their \(a\)
Allow