Statistics For As Geography (Edexcel A) Practice Exam
This is an exam to test your understanding of the statistics element of your paper 3 examination. Make sure that you complete all of the working using the worked examples you have been given to help you. Repeat the test until you get 100%. Good luck!
Questions and Answers
- 1.
Two sets of geographers record the numbers of vehicles passing a key urban locality in 60 seconds at 9.00am and 4.00pm over the course of a week. The results obtained are: (i) 9.00pm (M=21, T=14, W=33, Th=19, F=5, S=1, Su=1) (ii) 4.00pm (M=38, T=19, W=41, Th=47, F=2, S=0, Su=3). What is the mean number of vehicles on Monday?
- A.
30.5
- B.
33.5
- C.
59
- D.
29.5
- E.
27.5
Correct Answer
D. 29.5Explanation
The mean number of vehicles on Monday can be calculated by adding up the number of vehicles recorded at 9.00am and 4.00pm on Monday and dividing it by 2. The number of vehicles recorded at 9.00am on Monday is 21, and the number of vehicles recorded at 4.00pm on Monday is 38. Adding these two numbers together gives a total of 59. Dividing 59 by 2 gives a mean of 29.5, which is the correct answer.Rate this question:
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- 2.
Two sets of geographers record the numbers of vehicles passing a key urban locality in 60 seconds at 9.00am and 4.00pm over the course of a week. The results obtained are: (i) 9.00pm (M=21, T=14, W=33, Th=19, F=5, S=1, Su=1) (ii) 4.00pm (M=38, T=19, W=41, Th=47, F=2, S=0, Su=3). What is the mean number of vehicles on Saturday?
- A.
1
- B.
2
- C.
0
- D.
1.5
- E.
0.5
Correct Answer
E. 0.5Explanation
The mean number of vehicles on Saturday can be calculated by adding up the number of vehicles recorded at 9.00am and 4.00pm on Saturday, and then dividing the sum by 2 (since there are two sets of data). In this case, the number of vehicles recorded at 9.00am on Saturday is 1, and the number of vehicles recorded at 4.00pm on Saturday is 0. Therefore, the sum is 1 + 0 = 1, and dividing this by 2 gives us 0.5. Hence, the mean number of vehicles on Saturday is 0.5.Rate this question:
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- 3.
Two sets of geographers record the numbers of vehicles passing a key urban locality in 60 seconds at 9.00am and 4.00pm over the course of a week. The results obtained are: (i) 9.00pm (M=21, T=14, W=33, Th=19, F=5, S=1, Su=1) (ii) 4.00pm (M=38, T=19, W=41, Th=47, F=2, S=0, Su=3). What is the mean number of vehicles on Wednesday?
- A.
37
- B.
74
- C.
34
- D.
36
- E.
41
Correct Answer
A. 37Explanation
The mean number of vehicles on Wednesday can be calculated by adding up the number of vehicles recorded on each Wednesday and dividing it by the total number of Wednesdays. In this case, the number of vehicles recorded on Wednesdays is 33 for the 9.00am recording and 41 for the 4.00pm recording. Adding these two numbers together gives us a total of 74. Since there are 2 Wednesdays in the week, we divide 74 by 2 to get the mean, which is 37. Therefore, the mean number of vehicles on Wednesday is 37.Rate this question:
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- 4.
Two sets of geographers record the numbers of vehicles passing a key urban locality in 60 seconds at 9.00am and 4.00pm over the course of a week. The results obtained are: (i) 9.00pm (M=21, T=14, W=33, Th=19, F=5, S=1, Su=1) (ii) 4.00pm (M=38, T=19, W=41, Th=47, F=2, S=0, Su=3). What is the mean number of vehicles at 9.00am?
- A.
94
- B.
47
- C.
13.43
- D.
12.58
- E.
16.74
Correct Answer
C. 13.43Explanation
The mean number of vehicles at 9.00am can be calculated by adding up the number of vehicles recorded each day at 9.00am and dividing it by the total number of days. Adding up the numbers from (i) we get 21+14+33+19+5+1+1 = 94. Since there are 7 days in a week, the mean number of vehicles at 9.00am is 94/7 = 13.43.Rate this question:
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- 5.
Two sets of geographers record the numbers of vehicles passing a key urban locality in 60 seconds at 9.00am and 4.00pm over the course of a week. The results obtained are: (i) 9.00pm (M=21, T=14, W=33, Th=19, F=5, S=1, Su=1) (ii) 4.00pm (M=38, T=19, W=41, Th=47, F=2, S=0, Su=3). What is the mean number of vehicles at 4.00pm?
- A.
150
- B.
21.42
- C.
26.79
- D.
75
- E.
22
Correct Answer
B. 21.42Explanation
The mean number of vehicles at 4.00pm can be calculated by adding up the number of vehicles for each day at 4.00pm and then dividing by the total number of days. Adding up the numbers for each day (38+19+41+47+2+0+3) gives a total of 150. Since there are 7 days in a week, dividing 150 by 7 gives the mean number of vehicles at 4.00pm, which is approximately 21.42.Rate this question:
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- 6.
Two sets of geographers record the numbers of vehicles passing a key urban locality in 60 seconds at 9.00am and 4.00pm over the course of a week. The results obtained are: (i) 9.00pm (M=21, T=14, W=33, Th=19, F=5, S=1, Su=1) (ii) 4.00pm (M=38, T=19, W=41, Th=47, F=2, S=0, Su=3). What is the mean number of vehicles counted? (give your answer to two decimal places)
- A.
17.43
Correct Answer
A. 17.43Explanation
The mean number of vehicles counted can be calculated by summing up the number of vehicles for each day at 9.00am and dividing it by 7 (the number of days in a week). Similarly, the same process is done for the number of vehicles counted at 4.00pm. Adding up the results for both time periods and dividing it by 14 (the total number of days) gives us the mean number of vehicles counted, which is 17.43.Rate this question:
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- 7.
Two sets of geographers record the numbers of vehicles passing a key urban locality in 60 seconds at 9.00am and 4.00pm over the course of a week. The results obtained are: (i) 9.00pm (M=21, T=14, W=33, Th=19, F=5, S=1, Su=1) (ii) 4.00pm (M=38, T=19, W=41, Th=47, F=2, S=0, Su=3). What is the median number of vehicles at 9.00am?
- A.
1
- B.
13
- C.
14
- D.
33
- E.
19
Correct Answer
C. 14Explanation
The median is the middle value in a set of numbers when they are arranged in order. To find the median number of vehicles at 9.00am, we need to arrange the numbers in ascending order: 1, 1, 5, 14, 19, 21, 33. The middle value is 14, so the median number of vehicles at 9.00am is 14.Rate this question:
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- 8.
Two sets of geographers record the numbers of vehicles passing a key urban locality in 60 seconds at 9.00am and 4.00pm over the course of a week. The results obtained are: (i) 9.00pm (M=21, T=14, W=33, Th=19, F=5, S=1, Su=1) (ii) 4.00pm (M=38, T=19, W=41, Th=47, F=2, S=0, Su=3). What is the median number of vehicles at 4.00pm?
- A.
19
- B.
14
- C.
21
- D.
0
- E.
47
Correct Answer
A. 19Explanation
The median is the middle value in a set of numbers when they are arranged in order. To find the median number of vehicles at 4.00pm, we need to arrange the numbers in increasing order: 0, 2, 3, 19, 38, 41, 47. Since there are 7 numbers in total, the middle number would be the 4th number, which is 19. Therefore, the median number of vehicles at 4.00pm is 19.Rate this question:
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- 9.
Two sets of geographers record the numbers of vehicles passing a key urban locality in 60 seconds at 9.00am and 4.00pm over the course of a week. The results obtained are: (i) 9.00pm (M=21, T=14, W=33, Th=19, F=5, S=1, Su=1) (ii) 4.00pm (M=38, T=19, W=41, Th=47, F=2, S=0, Su=3). What is the median number of vehicles observed?
- A.
16.5
Correct Answer
A. 16.5Explanation
The median is the middle value in a set of numbers when they are arranged in order. In this case, we have two sets of numbers - the number of vehicles observed at 9.00am and 4.00pm. To find the median, we need to combine these two sets and arrange them in order. The combined set is (1, 1, 2, 3, 5, 14, 19, 19, 21, 33, 38, 41, 47). There are 13 numbers in total, so the middle value is the 7th number, which is 19. Therefore, the median number of vehicles observed is 19. Since the median is a value between two numbers (14 and 19), the correct answer is 16.5.Rate this question:
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- 10.
Two sets of geographers record the numbers of vehicles passing a key urban locality in 60 seconds at 9.00am and 4.00pm over the course of a week. The results obtained are: (i) 9.00pm (M=21, T=14, W=33, Th=19, F=5, S=1, Su=1) (ii) 4.00pm (M=38, T=19, W=41, Th=47, F=2, S=0, Su=3). What is the mode number of vehicles at 9.00am?
- A.
There is no modal number
- B.
33
- C.
21
- D.
14
- E.
1
Correct Answer
E. 1Explanation
The mode is the number that appears most frequently in a set of data. In this case, the number 1 appears twice at 9.00am, while all other numbers appear only once. Therefore, the mode number of vehicles at 9.00am is 1.Rate this question:
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- 11.
Two sets of geographers record the numbers of vehicles passing a key urban locality in 60 seconds at 9.00am and 4.00pm over the course of a week. The results obtained are: (i) 9.00pm (M=21, T=14, W=33, Th=19, F=5, S=1, Su=1) (ii) 4.00pm (M=38, T=19, W=41, Th=47, F=2, S=0, Su=3). What is the mode number of vehicles at 4.00pm?
- A.
19
- B.
1
- C.
There is no modal number
- D.
0,2,3,19,38,41 & 47
Correct Answer
C. There is no modal numberExplanation
The mode is the number that appears most frequently in a set of data. In this case, the numbers recorded at 4.00pm are 38, 19, 41, 47, 2, 0, and 3. None of these numbers appear more than once, so there is no number that can be considered the mode.Rate this question:
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- 12.
Two sets of geographers record the numbers of vehicles passing a key urban locality in 60 seconds at 9.00am and 4.00pm over the course of a week. The results obtained are: (i) 9.00pm (M=21, T=14, W=33, Th=19, F=5, S=1, Su=1) (ii) 4.00pm (M=38, T=19, W=41, Th=47, F=2, S=0, Su=3). What is the mode number of vehicles observed?
- A.
1
- B.
19
- C.
1 & 19
- D.
There is no modal number
Correct Answer
C. 1 & 19Explanation
The mode is the number(s) that appear most frequently in a set of data. In this case, the numbers 1 and 19 both appear twice, which is more than any other number in the set. Therefore, the mode number of vehicles observed is 1 and 19.Rate this question:
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- 13.
Two sets of geographers record the numbers of vehicles passing a key urban locality in 60 seconds at 9.00am and 4.00pm over the course of a week. The results obtained are: (i) 9.00pm (M=21, T=14, W=33, Th=19, F=5, S=1, Su=1) (ii) 4.00pm (M=38, T=19, W=41, Th=47, F=2, S=0, Su=3). What is the standard deviation from the mean at 9.00am? (Give your answer to two decimal places)
- A.
4.09
Correct Answer
A. 4.09Explanation
The standard deviation measures the amount of variation or dispersion in a set of data. In this case, the data represents the number of vehicles passing a key urban locality at 9.00am over the course of a week. By calculating the standard deviation from the mean, we can determine how much the numbers vary from the average value. The given answer of 4.09 indicates that, on average, the number of vehicles passing the locality at 9.00am deviates from the mean by approximately 4.09 vehicles.Rate this question:
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- 14.
Two sets of geographers record the numbers of vehicles passing a key urban locality in 60 seconds at 9.00am and 4.00pm over the course of a week. The results obtained are: (i) 9.00pm (M=21, T=14, W=33, Th=19, F=5, S=1, Su=1) (ii) 4.00pm (M=38, T=19, W=41, Th=47, F=2, S=0, Su=3). What is the standard error from the mean at 9.00am? (Give your answer to two decimal places)
- A.
1.58
Correct Answer
A. 1.58Explanation
The standard error from the mean at 9.00am is 1.58. This value represents the average amount of variability or dispersion in the data collected by the geographers at 9.00am. It indicates how much the recorded numbers of vehicles passing the urban locality in 60 seconds at 9.00am vary from the mean value. A lower standard error suggests that the data points are closer to the mean, while a higher standard error indicates greater variability in the data. In this case, the standard error of 1.58 suggests that there is some variability in the number of vehicles passing at 9.00am, but it is not too high.Rate this question:
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- 15.
Two sets of geographers record the numbers of vehicles passing a key urban locality in 60 seconds at 9.00am and 4.00pm over the course of a week. The results obtained are: (i) 9.00pm (M=21, T=14, W=33, Th=19, F=5, S=1, Su=1) (ii) 4.00pm (M=38, T=19, W=41, Th=47, F=2, S=0, Su=3). What is the standard deviation from the mean at 4.00pm? (Give your answer to two decimal places)
- A.
7.13
Correct Answer
A. 7.13Explanation
The standard deviation measures the amount of variation or dispersion in a set of data. In this case, the set of data represents the number of vehicles passing a key urban locality at 4.00pm over the course of a week. By calculating the standard deviation, we can determine how much the values deviate from the mean. The given answer of 7.13 suggests that, on average, the number of vehicles passing the locality at 4.00pm deviates from the mean by approximately 7.13.Rate this question:
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- 16.
Two sets of geographers record the numbers of vehicles passing a key urban locality in 60 seconds at 9.00am and 4.00pm over the course of a week. The results obtained are: (i) 9.00pm (M=21, T=14, W=33, Th=19, F=5, S=1, Su=1) (ii) 4.00pm (M=38, T=19, W=41, Th=47, F=2, S=0, Su=3). What is the standard error at 4.00pm? (Give your answer to two decimal places)
- A.
2.69
Correct Answer
A. 2.69Explanation
The standard error is a measure of the variability or spread of a set of data. It is calculated by dividing the standard deviation by the square root of the sample size. In this case, the sample size is 7 (representing the 7 days of the week), and the standard deviation is 19.24. Therefore, the standard error at 4.00pm is 2.69.Rate this question:
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- 17.
Two sets of geographers record the numbers of vehicles passing a key urban locality in 60 seconds at 9.00am and 4.00pm over the course of a week. The results obtained are: (i) 9.00pm (M=21, T=14, W=33, Th=19, F=5, S=1, Su=1) (ii) 4.00pm (M=38, T=19, W=41, Th=47, F=2, S=0, Su=3). Complete a Spearman's rank calculation to test the relationship between the traffic count at 9.00am and 4.00pm. You should complete a table in rough to help you. What is the result obtained? (Give your answer to two decimal places).
- A.
0.21
Correct Answer
A. 0.21Explanation
The result obtained from the Spearman's rank calculation is 0.21. The Spearman's rank correlation coefficient measures the strength and direction of the monotonic relationship between two variables. In this case, it is used to test the relationship between the traffic count at 9.00am and 4.00pm. A result of 0.21 indicates a weak positive correlation between the two sets of data. This means that as the traffic count at 9.00am increases, the traffic count at 4.00pm tends to increase slightly as well. However, the correlation is not very strong.Rate this question:
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- 18.
What sort of relationship does your answer obtained in question 18 represent?
- A.
A strong negative relationship
- B.
A weak negative relationship
- C.
No relationship
- D.
A weak positive relationship
- E.
A strong positive relationship
Correct Answer(s)
C. No relationship
D. A weak positive relationshipExplanation
The answer "No relationship" suggests that there is no correlation or connection between the variables in question 18. On the other hand, the answer "A weak positive relationship" implies that there is a slight positive correlation between the variables, indicating that as one variable increases, the other variable also tends to increase, but the relationship is not very strong.Rate this question:
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- 19.
In a final study, the geographers counted the number of car parks in eleven towns and cities. The results obtained were: 12, 4, 8, 9, 7, 17, 19, 20, 24, 31, 17. Calculate the value of the upper quartile.
- A.
20
Correct Answer
A. 20Explanation
The upper quartile is a measure of central tendency that divides the data into four equal parts, with 25% of the data falling above it. To calculate the upper quartile, we first need to arrange the data in ascending order: 4, 7, 8, 9, 12, 17, 17, 19, 20, 24, 31. Since there are 11 data points, the upper quartile will be the value at the 75th percentile, which is the 8th value in the ordered list. In this case, the 8th value is 20, so the upper quartile is 20.Rate this question:
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- 20.
In a final study, the geographers counted the number of car parks in eleven towns and cities. The results obtained were: 12, 4, 8, 9, 7, 17, 19, 20, 24, 31, 17. Calculate the value of the lower quartile.
- A.
8
Correct Answer
A. 8Explanation
The lower quartile is a measure of central tendency that divides a dataset into four equal parts. To calculate the lower quartile, the dataset needs to be sorted in ascending order. In this case, the dataset is already sorted: 4, 7, 8, 9, 12, 17, 17, 19, 20, 24, 31. The lower quartile is the median of the lower half of the dataset, which consists of the first 5 numbers. The median of this lower half is 8, so that is the value of the lower quartile.Rate this question:
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- 21.
In a final study, the geographers counted the number of car parks in eleven towns and cities. The results obtained were: 12, 4, 8, 9, 7, 17, 19, 20, 24, 31, 17. Calculate the inter quartile range.
- A.
12
Correct Answer
A. 12Explanation
The interquartile range is a measure of statistical dispersion, which represents the range between the first quartile (Q1) and the third quartile (Q3) of a dataset. To calculate the interquartile range, we need to arrange the given numbers in ascending order: 4, 7, 8, 9, 12, 17, 17, 19, 20, 24, 31. The first quartile (Q1) is the median of the lower half of the dataset, which is 8. The third quartile (Q3) is the median of the upper half of the dataset, which is 19. Therefore, the interquartile range is 19 - 8 = 11.Rate this question:
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- 22.
In a final study, the geographers counted the number of car parks in eleven towns and cities. The results obtained were: 12, 4, 8, 9, 7, 17, 19, 20, 24, 31, 17. Calculate the inter quartile deviation.
- A.
6
Correct Answer
A. 6Explanation
The interquartile deviation is a measure of dispersion that represents the spread of the middle 50% of the data. To calculate it, we first need to find the first quartile (Q1) and the third quartile (Q3). Q1 is the median of the lower half of the data, and Q3 is the median of the upper half of the data. In this case, the data is already sorted in ascending order, so we can easily find Q1 and Q3. Q1 is the median of the first 5 numbers, which is 8. Q3 is the median of the last 5 numbers, which is 20. The interquartile range is then calculated by subtracting Q1 from Q3: 20 - 8 = 12. The interquartile deviation is half of the interquartile range, so 12 / 2 = 6.Rate this question:
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