Edexcel A-level Maths · Topic S4
Practice questions covering statistical distributions from this specification. Every question below is the real Pearson question as it is printed on the paper, with its mark scheme.
6 questions on Statistical distributions, each with its mark scheme and worked method. Reveal the answer when you have had a go.
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(3)
(Total for Question 1 is 3 marks)
2.
(3)
(Total for Question 2 is 3 marks)
3.
(4)
(Total for Question 3 is 4 marks)
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(2)
(Total for Question 4 is 2 marks)
5.
(3)
(Total for Question 5 is 3 marks)
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(2)
(Total for Question 6 is 2 marks)
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Printable question pack — S4 Statistical distributionsAll 30 questions across 3 specification points — 24 more than the 6 above — with the full mark scheme. Guest accessible · browser print.14 questions shown as printed, 12 reference-only · newest series first
Question and mark-scheme extracts © Pearson Education Ltd. Reproduced for educational use; the original paper is linked on every question. Contact hello@oscarsongtutoring.co.uk for removal.
Edexcel 9MA0 · June 2024 · Paper 31 · Q1 · 11 marks
11 marks
Examiner report
Almost all students were able to write down or use the correct binomial model in part (a) and usually evaluate P(X = 3) correctly. Some struggled with part (ii) and an answer of 0.930… from P(X ⩽ 3) was often seen suggesting that students are not entirely familiar with how their calculator operates. Others seemed to mis-interpret the inequality and attempted 1 – P(X ⩽ 2). In part (b) a new random variable was needed and students who defined this clearly usually made good progress, but some simply tried to use B(60, 1 6 ). Interpreting the phrase “at least 12 days” caused problems for some students but generally this part was answered well. The final two parts required a third random variable S ~B(600, 1 6 ) and many wrote down the correct mean for part (c) and often stated that this random variable was approximated by N(100, …) to get started in part (d). Many students seemed unsure about finding the standard deviation (the formulae are available in the formula booklet) and there was a great deal of mislabelling or use of the variance instead of the standard deviation. A good number realised that a continuity correction was required, though of course not always in the right direction, but the correct final answer was not uncommon.
Xian rolls a fair die 10 times. The random variable X represents the number of times the die lands on a six.
Edexcel 9MA0 · June 2024 · Paper 31 · Q5 · 10 marks
10 marks
Examiner report
Part (a) was answered very well by most students and in part (b) a good proportion also recognised the need for the events to be independent. Some students failed to score the mark in part (b) because they did not use correct mathematical language: we saw comments such as “they are not affected by each other” or “the events are not mutually exclusive” and a few simply said “it is continuous data” or “they must take part in both events”. Many students answered part (c) successfully, but some made errors when trying to find P(T < 300) either by using a continuity correction and finding P(T < 299.5) or failing to convert minutes to seconds and trying to find P(T < 5). There were many good attempts at part (d). Most students could start the problem and write down a pair of linear equations in and with suitable values of z. Many students showed some working when solving their equations whilst others simply wrote down solutions using their calculators. This is fine if the equations are correct but if there are errors in the equations then the method mark cannot be awarded if no method for solving the equations is seen. The front of the paper indicates that if students use their calculators instead of the tables, then they should use an equivalent degree of accuracy which in this case, for finding the z values, was 4 decimal places. Some failed to do this and lost one of the accuracy marks but there were a good number of students scoring full marks for this part. A common error here was missing the minus sign on the z value for the lower tail and a few students assigned the z values the wrong way around.
For a school athletics club, high-jump height H is normally distributed with mean 1.4 m and standard deviation 0.15 m. The 1500 m time T is normally distributed with mean 330 s and standard deviation 26 s.
Edexcel 9MA0 · June 2023 · Paper 31 · Q2 · 9 marks
9 marks
Examiner report
This question was also answered well. In part (a) students were often aware of the conditions needed for a binomial distribution but did not always present their answers in context with the precision required. Some focussed on the need for a fixed number of trials or that packets either did or did not contain a prize but these features were clearly identified in the question. Those scoring the mark either identified the need for prizes to be placed in packets at random or that the probability of a packet containing a prize needed to be constant. There was some confusion between “events”, such as a prize being placed in a packet, being independent and the “probability” of such an event being constant. We sometimes saw a statement such as “the probability of a packet containing a prize is independent” which didn’t score this first mark. Part (b) was generally answered well though some gave answers to 2dp or 2sf rather than the 3sf as instructed on the front of the paper. Most answered (i) correctly but in (ii) common incorrect answers were from those who found P(T = 3) or P(T ⩽ 3). In part (c) many were able to identify the correct distribution and those who stated this were guaranteed the method mark even if their answer to (b)(ii) was incorrect. Whilst many did obtain the correct answer here those with an incorrect answer to (b)(ii) who simply wrote down a probability could not be given either mark. Some students in part (c) didn’t identify the binomial model but wrote an expression for the probability of the form 5 2 3 2 C (1 ) p p − but a few forgot the 5 2 C term. The hypothesis test in part (d) was answered well with correct notation used for the hypotheses and the appropriate one-tailed test identified. Most selected the correct binomial model but it was disappointing to see a number finding P(X = 9) rather than P(X ⩽ 9). A correct conclusion in context was often given: some using the simple response that there is evidence to support Kamil’s claim and others stating that there was evidence that the proportion of packets containing a prize is less than 1 7. It was encouraging to see that most of those using the second approach were referring to “proportion” rather than “number of” this time.
A machine fills packets with sweets, and 1/7 of the packets also contain a prize. The packets are placed in boxes containing 40 packets each. The random variable T is the number of packets containing a prize in a box.
Edexcel 9MA0 · June 2023 · Paper 31 · Q4 · 6 marks
6 marks
Examiner report
Whilst it was rare to see an incorrect answer to part (a) a minority of students were unable to use their calculator to evaluate this simple probability using a normal distribution. In part (b) most students stated their hypotheses correctly in terms of with very few using a one-tailed test. To score the method mark in this type of question the students simply need to select and write down the correct model (distribution). A number seemed to be listing numbers they needed for their calculator, but we expect to see the appropriate mathematical notation used. Fortunately, many obtained a correct probability and would therefore receive the method mark by implication. Some chose to standardise and find a z value (1.9088… which should be compared with 1.96) or a critical region (𝑆̄ ⩾ 177.248. . . ) but given that a p-value was required in part (c) this was not the most efficient route. A common error in stating the model is to use an incorrect value for the mean. On this occasion we condoned 2 6.8 ~ N 177.2, 52 S for the M1 and first A1 but penalised the final accuracy mark. In a hypothesis test the probability (or critical region) is always calculated assuming H0 is true and so the mean should be 175.4 and the probability calculated is ( ) P 177.2 . S Some students stopped after stating that the result was not significant or not rejecting H0 but a good number gave a correct conclusion in context. Part (c) was not answered well with only a few scoring this mark. The common error was to use 0.028 or 0.025 but others used values like 175.4
The heights of adult men from region A are normally distributed with mean 175.4 cm and standard deviation 6.8 cm. A student claims that the mean height in region B is different from the mean in region A. A random sample of 52 adult men from region B has mean height 177.2 cm. The student assumes that the standard deviation is 6.8 cm in both regions.
Edexcel 9MA0 · June 2023 · Paper 31 · Q5 · 8 marks
8 marks
Examiner report
This question proved to be the most challenging on the paper for students. In part (a) few could see how to use the two bullet points to reach the required result. A good number realised that P(S |{X = x}) = k x but couldn’t always use the conditional probability formula to reach ( ) P 50 50 k S X b = = . Those that did get this far were often able to give a convincing proof of the given result. A common misconception was believing that , etc 20 50 k k a b = = and then using the sum of the probabilities equalling 1 but this scored no marks. It was good to see a number of students using the result from part (a) to form similar relationships between the probabilities, such as d = 2b and going on to solve their equations to find the probability distribution. The final part was particularly challenging with many misinterpreting the 30% probability given in the stem to part (c). Many confused this 0.3 [= P(S | X = 100) ] with the 0.4 [= P(X = 100)] from the probability distribution and simply compared these two values. Those that did realise that Nav’s 30% meant that k = 30 often went on to show that when x = 20 this would give a probability [P(S | X = 20)] > 1 and so the model was not suitable.
Tisam plays a game using a spinner, a ball and a cup. The random variable X has values 20, 50, 80 and 100 with probabilities a, b, c and d respectively. After spinning, Tisam stands X cm from the cup and tries to throw the ball into it. Let S be the event that the throw is successful. Her model assumes P(S | X = x) = k/x, where k is constant, and that P(S intersect {X = x}) is the same for every possible x.
Edexcel 9MA0 · June 2023 · Paper 31 · Q6 · 14 marks
14 marks
Examiner report
This question also proved to be quite challenging. In part (a) most students could use the histogram to calculate the number of patients spending between 10 and 30 hours in hospital but many failed to convert this to a probability by dividing by 90. Some students tried using small squares, but errors were often made and a divisor of 1600 not 1800 was often seen. Part (b) though was not answered so well. Some did explain that the histogram was not symmetric or was skewed and so a normal distribution wouldn’t be suitable. Some focussed on the continuous nature of the variable and thought it was therefore suitable whilst others thought that T was discrete and therefore a normal wasn’t suitable. The question wanted reference to the histogram and there was no expectation or requirement to find probabilities from the normal distribution nor compare the mean of the data with 14.9 If further calculations were expected these would be asked for and reflected in the mark allocation. Many students were able to use integration by parts correctly to establish the result in part (c). Some started the wrong way around (using u = e – x) and soon realised they were not heading for the given answer and there was some muddling over the signs when applying the limits. Part (d) was not answered so well with only a minority of students realising that the area under the curve should equal 90. Those who did realise this were often able to apply the answer to part (c) though some used n = 40 rather than n = 4. The first part of (e) was often answered well and some realised that the answer to part (c) could be used to answer (e)(ii) but some tried using n = 10 and 30 and others stopped after find an estimate for the number of patients as 53.1 and failed to complete the question to find the required probability. In part (f) only a small number of candidates realised that the fact that some patients might stay longer than 40 hours was a limitation of the model. Others did not give any context in their answers or discussed the shape of the curve without considering the stated upper limit.
A medical researcher studies the number of hours T that a patient stays in hospital after an operation, using a random sample of 90 patients. A histogram has the following class intervals and frequency densities: 0-4: 2.5; 4-7: 3.0; 7-12: 4.2; 12-16: 4.0; 16-20: 3.5; 20-40: 1.0. The sample mean is 14.9 hours and the standard deviation is 9.3 hours. Tomas suggests T ~ N(14.9, 9.3^2). Xiang suggests modelling the frequency polygon by y = kx e^(-x) for 0 <= x <= 4, where x is measured in tens of hours and k is constant.
Edexcel 9MA0 · June 2022 · Paper 31 · Q1 · 6 marks
6 marks
Examiner report
This was generally a good start to the paper for many students with almost all being able to score some of the marks on this question. In part (a) students were usually able to find at least one of the two required probabilities although occasionally they miscopied from their calculator with 0.197 being written down rather than 0.0197. The notation demonstrating use of a binomial distribution was varied. Some students wrote down answers without stating the distribution that was being used, in this instance this was condoned provided that a correct probability was seen, but students should be aware that method marks are often awarded for sight of the correct model. Where full marks were not awarded the most common reasons were, rounding answers to fewer than 3 significant figures, not knowing whether to use Binomial PD or CD on their calculator, or for not knowing how to rewrite ( ) P 5 X in a form that allowed use of the binomial CD function. Part (b) caused more problems for students. The question clearly asked the students to use a normal approximation and most found the correct mean and variance for the Normal distribution and went on to find a probability. There is some confusion about whether it is the variance or the standard deviation which is a parameter of the normal distribution and too many failed to summarise their working with a simple statement N(120, 64.2). Almost all students correctly interpreted “more than 110” as requiring P(X > 110) but a minority of students remembered to use a continuity correction, required when approximating a discrete distribution with the normal distribution, and those that did attempt to use a continuity correction often used an incorrect one, most commonly 109.5. Despite being directed to use a normal approximation by the question some students used the binomial distribution. This gave an answer very close to the one obtained from the correct distribution, so it was very important to write down the distribution used, the probability statement with its continuity correction and give the answer to a sufficient degree of accuracy before rounding.
George throws a ball at a target 15 times. On each throw, the probability of hitting the target is 0.48. The random variable X is the number of hits in 15 throws.
Edexcel 9MA0 · June 2022 · Paper 31 · Q2 · 12 marks
12 marks
Examiner report
Students had mixed success with part (a) of this question. Most students who recognised that they needed to use standardisation and found a z-value of ± 1.96 completed the proof without error scoring both marks. A few students who used + 1.96 then “fudged” the signs part way through their solution losing the A mark. Part (b) was a good introduction to part (c) with the majority of students scoring the mark although there were some who didn’t read the question carefully enough and used the 7.902 from (a) as the lower limit rather than 7.94 The majority of marks lost were due to rounding to 2 s.f Most students were able to make a credible attempt at answering part (c). However, there was a lot of information for students to process and it was common to see errors or omissions at some point in their solution. There were several correct ways of approaching this question. All approaches required the calculation of the probability of a rod lying in each of the categories, some students mistakenly used P(L < 7.94) = 0.025 presumably assuming that 7.94 was the value given in the question (7.902). Another common error was assuming that 7.94 and 8.09 were evenly spaced about the mean so the probability from (b) was subtracted from 1 and then the result halved. Once the students had found the probabilities a common approach was to start with calculating the number of rods in each category, and then calculating the income/profit. Many students calculated the total income and went no further. Whether this was because of a misunderstanding of the term “profit” or whether having got that far they simply forgot to subtract the costs was unclear. The ability for students to apply their skills in “real life” situations needs further development including training in setting out working methodically Common errors included rounding the probabilities inappropriately, rounding the number of rods to integer values and some occasionally being uncertain whether they were working in pounds or pence. The most successful answers were those where the calculation was combined into a single expression. Students who attempted part (d) usually selected the correct Binomial model. A few students, recognising that n was large, used a normal approximation which was not appropriate in this case with very small p. It was common for students to find ( ) P 6 X rather than ( ) P 6 X < . Those who made this error were however still able to access the mark available for a correct interpretation. A minority of the students who gave a correct interpretation did not make the required comparison of their probability with the given 95% aim thus losing the final A mark. Students need to be aware that in questions such as this they should be using the values calculated to justify and explain any conclusions they make. A reasonable proportion of students were confused about what the question was asking and tried to set up a hypothesis test, making a conclusion based on whether they were accepting or rejecting the null hypothesis rather than judging the likelihood of the manufacturer meeting his aim by comparing their probability with the given 95% aim. Some Weaker students simply worked out 6 out of 200 as a percentage and compared this with 5%.
A machine makes metal rods whose length L cm is normally distributed with mean 8 cm and standard deviation x cm. The proportion of rods shorter than 7.902 cm is 2.5%.
Edexcel 9MA0 · June 2022 · Paper 31 · Q3 · 7 marks
7 marks
Examiner report
Part (a) tested knowledge of the data set. Some students were able to identify tr or trace as the required answer, but many displayed their lack of knowledge of the data set, with n/a being a typical wrong answer, or 0 or 0.025. Many students gave no answer of any sort and there were a few nonsensical or confused answers such as explaining how the rainfall might be collected (a particular favourite being ‘with a bucket and ruler’) or how it might be recorded (in an excel spreadsheet). The vast majority of students were able to find the mean correctly in Part (b). Mistakes if any in (i) tended to be rounding the answer to 5.6 without previously showing a more accurate answer, hence losing the single mark. A few students showed the calculation without evaluating it. Disturbingly there was a tiny minority who seemed to have formulae involving square root signs etc. It is disappointing to see evidence of students studying statistics at advanced level who still do not know how to calculate a mean. There were more errors (ii) although well answered by the majority. Errors tended to be missing the square root in their expression for the standard deviation giving the variance as the final answer, with students believing this to be the standard deviation; subtracting the mean rather than the square of the mean premature approximation of the mean leading to inaccurate final answers and giving the answer to 2sf or an even lesser degree of accuracy. A small number of students gave an answer with no working and were then at the mercy of having obtained a sufficiently accurate answer or risk losing these relatively easy marks. The calculation and use of 𝑆𝑆𝑥𝑥𝑥𝑥 was seen occasionally. Part (c) proved to be more problematic. Many students lacked knowledge of the LDS and did not know the relative geographical locations of Camborne and Leuchars within the UK. As such, a significant number of students made the incorrect assumption that Leuchars is in the South and Camborne is in the North. However, when students were able to locate the relative positions of the two places correctly, they were generally able to draw the correct conclusion, resulting from comparing the mean rainfall between the two places. The main errors included • not making reference to the relative positions of the two locations in their answer e.g. stating that Camborne was in the South with no reference to the relative position of Leuchars, • not making any statements about either of the geographical locations at all and only compared means, with responses such as: no, as the mean is less in Leuchars than in Camborne. • not drawing the correct conclusion after correctly comparing locations and means • not stating any conclusion. • incorrectly comparing standard deviations rather than means. There was a wide variety of responses to part (d). Successful students often stated one or more of the conditions of the Binomial Distribution, with reasons why this distribution was not suitable in this case. e.g. • the probability needs to be independent but the probability of rainfall is not independent as it often depends on the weather from the day before. • the probability of rain needs to be constant but the probability was taken from May to October and summer is June to September and therefore the probability is likely to be higher than in the summer when there is less rain Errors included • referring to independence without context • commenting on the great storm of 1987 or global warming • stating that the data is out of date • the sample size is too small • some values are trace values and/or rainfall is influenced by other factors.
Dian uses the large data set to investigate Daily Total Rainfall r mm for Camborne. For the 31 days of August 2015, n = 31, sum(r) = 174.9 and sum(r^2) = 3523.283.
Edexcel 9MA0 · October 2021 · Paper 31 · Q1 · 7 marks
7 marks
Examiner report
In part (a) many scored the mark by referring to the possibility of bias being introduced by the interviewer or the lack of randomness when using quota sampling. The most common answers that did not score the mark referred to cost or time failing to appreciate that quota sampling is generally quicker and therefore cheaper to implement than simple random sampling. Some thought that quota sampling required a sampling frame which suggested they were confused about the different types of sampling. Part (b) was generally answered very well with most stating the distribution, and scoring the method mark, and then evaluating part (i) correctly on their calculator. Part (ii) met with less success as many could not deal with the inequality or were unsure how to use their calculator correctly. Some seemed to confuse significant figures with decimal places: the front of the exam paper instructs then to use 3 significant figures but some only gave an answer of 0.022 here and lost this mark. Part (c) was usually answered correctly and many went on to use the correct binomial distribution in part (d) though some found interpreting “fewer than 3” a problem and a common incorrect answer was 0.924 (from P( 3 X ). Overall though this proved a good start to the paper and over 40% scored 6 or 7 marks here.
In a university, 8% of students are members of the university dance club. A random sample of 36 students is taken, and X is the number who are members of the dance club. Only 40% of dance club members can dance the tango.
Edexcel 9MA0 · October 2021 · Paper 31 · Q5 · 11 marks
11 marks
Examiner report
Most students used their calculators in part (a) with reasonable success but a frequent error was to use 0.1 instead of 0.01 and without any working being shown there was no opportunity to award the method mark. Part (b) was a straightforward probability calculation designed to be completed on a calculator as the specification requires and so it was disappointing that nearly 25% of the candidates scored no marks at all on this question. Identifying the conditional probability in part (c) defeated many students and some of those who seemed to be attempting to use a conditional probability were unable to use the correct notation to write their expression. Those who did start with a correct conditional probability and moved onto a ratio of probabilities often had a correct denominator but a common error was to have P(F > 160) as the numerator rather than P(160 < F < 175). Part (d) was a familiar hypothesis test and many students scored something here. The hypotheses were usually stated in terms of µ and most of the time 166.5 rather than 164.6 was used. Selecting the correct model caused a number of problems: some used 164.6 as the mean and others used 7.4 (or occasionally 6.1) as the standard deviation, simply stating the correct model will score the method mark here but a number of students didn’t show us the model they were using and we had to award this mark by implication from correct working. Most students successfully used the probability approach (as we would recommend) with a few attempting a critical region or using a z value and comparing with 1.6449 from the tables. The final mark required a correct interpretation of their calculation in context and some students simply stopped after stating that the result was significant and failed to score this final mark.
The heights F cm of females from a country are normally distributed with mean 166.5 cm and standard deviation 6.1 cm. For a different country, female heights are normally distributed with standard deviation 7.4 cm. Mia believes the mean height in the different country is less than 166.5 cm. She takes a random sample of 50 females there and obtains a sample mean of 164.6 cm.
Edexcel 9MA0 · October 2021 · Paper 31 · Q6 · 7 marks
7 marks
Examiner report
There is a requirement in the specification for questions occasionally to draw together knowledge from different areas of 9MA0 and this was a question aiming to do that. Many students realised that to start the problem they could use the fact that the sum of the probabilities must equal 1 but fewer than half of the students could move beyond that mark. Those who realised they had a sum of logarithms were often able to arrive at the equation abc = 36 but few could devise a strategy to deduce the values of a, b and c. Some stated that all of the values had to be greater than 1 though it was not always clear that they deduced this from the given bullet points in the question. Around a quarter of the students found the correct set of values with varying degrees of explanation on the way. Part (b) was often left blank though the first mark was available without values for a, b and c provided brackets were used correctly for the expression. Nearly 10% managed to score full marks for the question.
The discrete random variable X has the probability distribution P(X = a) = log_36(a), P(X = b) = log_36(b), P(X = c) = log_36(c), where a, b and c are distinct integers with a < b < c and all three probabilities are greater than zero.
Edexcel 9MA0 · October 2020 · Paper 31 · Q4 · 10 marks
10 marks
Examiner report
Part (a) was generally answered very well. Being a “show that” question meant that we needed to see a full explanation and most clearly showed the sum of the probabilities equal to 1 and then solved the resulting equation to find k. Failing to show this key step usually meant that no marks could be awarded. A significant number of students left the rest of the question blank but those who did make an attempt were usually able to identify some, though rarely all, the cases in (b). Finding the probabilities was more of a problem with some using P(D = 50) + P(D = 30) rather than the product and those having correct probabilities for all 3 cases were quite rare. It was good to see many students who had faltered on (b) still making a start on part (c). Many identified the 4 angles as terms of an arithmetic series and some were able to reach the condition 2a + 3d = 180 (or equivalent) but many could not see how to complete the problem. Some used the information that a > 50 to reach d < 80 3 but were unable to make the connection that this meant that only the two cases d = 10 and d = 20 were required and so the probability could simply be written down.
The discrete random variable D takes values 10, 20, 30, 40 and 50, with respective probabilities k/10, k/20, k/30, k/40 and k/50, where k is a constant.
Edexcel 9MA0 · October 2020 · Paper 31 · Q5 · 15 marks
15 marks
Examiner report
It was disappointing that over 20% of candidates scored no marks here as there were some straightforward marks available to those who had studied this part of the specification. Part (a) was a very straightforward opening part and the vast majority of students could use their calculator efficiently to score this mark. Part (b) was a fairly standard hypothesis test and a good number made progress here. Some still did not use µ for the hypotheses and a null hypothesis of 10 T = was all too common. Most attempts used a normal distribution with a correct variance but a significant number of students used a mean of 11.5 rather than 10. This is not a correct method for a hypothesis test (the probabilities should be calculated on the assumption that H0 is true) but, on this occasion, for those who still managed to reach a correct conclusion a special case enabled them to secure one mark. A variety of different approaches were used in part (b), some finding the critical value for X and others calculating the test statistic of 1.677… but some of these candidates compared this value to the significance level rather than the critical value of 1.6669 from the tables. The first part of (c) was another fairly straightforward calculation which many scoring this mark, sometimes including those who failed to score the mark in part (a). Students need to be aware of which “tail” their calculator gives them. Beyond (c) (i) the question was more challenging but it is encouraging that over 50% of students scored some of these first 6 marks. In (c)(ii) although a good number of students recognised that a conditional probability was required, few could find a correct expression for the numerator and a common error was to see (c)(i) P( 0) T > used rather than a correct numerator of P(0 < T < 2). In part (iii) many realised that the model would assign some probability to negative values of T but this will always be true when a normal distribution is used as a model. The point here was that a significant proportion of the values for T were negative and so this means that the normal model is not very good. Most responses seemed to believe that a model must match reality exactly but a few did comment that P(T > 0) should be closer to 1 or that the difference between (i) and (ii) was too big and therefore the model was not a good one to use. Fully correct answers to part (d) were rare and this part gave the A* students an opportunity to shine. Successful students realised that they needed to look at P(T > 2) and often found this by taking 1 – (c)(i) but they rarely realised that the median splits this probability into 2 so they were seeking the value of t such that P(T > t) = 0.40…Some who reached this stage tried, incorrectly to use their calculator, failing to realise that their calculator gave t for probabilities of the form P(T < t), a few used the tables provided to find z = 0.2533 and some did prepare correctly to use their calculator and found P(T < t ) = 0.5978… (the 0.402… + (c)(i)) from whence the correct value of t = 5.9 could be found. It was encouraging to see some good use of diagrams in this question being used to help students identify which probabilities they needed.
A health centre claims that the time X minutes a doctor spends with a patient is modelled by N(10, 4^2). A receptionist samples 20 patients and obtains a mean time of 11.5 minutes. The centre also claims that the time T minutes a dentist spends on a routine appointment is modelled by T ~ N(5, 3.5^2).
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Everything here stays free. If statistical distributions is where the marks keep going, I teach it one to one from the mark schemes. Free intro call, then a free first lesson.