Edexcel A level Maths (9MA0) · Statistics › Data presentation and interpretation
Practise Measures of location and spread. unlimited generated questions on this subtopic, at up to four difficulty levels, with full mark schemes and a progress tracker. Free, no account needed.
Written for this site in the style of Edexcel exam questions. They are not taken from real past papers.
Question 1Easy4 marks
The numbers of goals scored in a random sample of football matches are: 4, 2, 2, 3, 3, 5, 4, 3, 2, 3
(a) Find the mean, giving your answer to 3 significant figures.[2]
(b) Find the median and the mode.[2]
Show the answer and mark scheme
(a)Answer: 3.10
M1 for \(\dfrac{31}{10}\)
A1 for awrt 3.10
(b)Answer: Median 3, mode 3
B1 for median 3
B1 for mode 3
Question 2Medium4 marks
The table shows the number of cups of coffee drunk yesterday by each of a random sample of office workers.
[object Object]
(a) Calculate the mean and the standard deviation of these data, giving your answers to 3 significant figures.[3]
(b) One more office worker, who drank 4 cups of coffee yesterday, is added to the sample. Without calculating the new mean, state whether the mean number of cups of coffee increases, decreases or stays the same. Give a reason for your answer.[1]
Show the answer and mark scheme
(a)Answer: Mean 1.73, standard deviation 1.15
M1 for \(\dfrac{\Sigma fx}{\Sigma f} = \dfrac{64}{37}\) or a correct method for the standard deviation, e.g. \(\sqrt{\dfrac{160}{37} - \bar{x}^2}\)
A1 for mean awrt 1.73
A1 for standard deviation awrt 1.15 (allow 1.17 using \(n - 1\))
(b)Answer: The mean increases, because the new value, 4, is greater than the current mean of 1.73.
B1 for “increases” with a reason: 4 is greater than the mean (1.73)
Question 3Hard4 marks
The table shows the reaction times for a random sample of students. One of the frequencies is unknown. Using linear interpolation, with the median at position \(\tfrac{n}{2}\), where \(n\) is the total frequency, an estimate of the median reaction time is 282.0 ms.
[object Object]
Find the value of \(x\).[4]
Show the answer and mark scheme
Answer: \(x = 13\)
M1 for a total frequency of \(91 + x\) and the median position \(\tfrac{91 + x}{2}\)
M1 for a correct interpolation equation, e.g. \(250 + \dfrac{\tfrac{1}{2}(91 + x) - 36}{25} \times 50 = 282.0\)
dM1 for solving their linear equation
A1 for \(x = 13\)
Worked solution: The median lies in the class 250 ≤ t < 300, so \(250 + \dfrac{\tfrac{1}{2}(91 + x) - 36}{25} \times 50 = 282.0\), which gives \(x = 13\).