Edexcel Maths A Level Paper 31 Statistics QP
Course
Edexcel Maths A Level Paper 31 Statistics QP
Question 1
Q: A survey of 100 students found that the mean number of hours they spent studying each
week was 15 hours. What is the total number of hours studied by all students?
A: 1500 hours
Rationale: To find the total number of hours, multiply the mean by the number of students:
Total Hours=Mean×Number of Students=15×100=1500.\text{Total Hours} = \text{Mean} \
times \text{Number of Students} = 15 \times 100 =
1500.Total Hours=Mean×Number of Students=15×100=1500.
Question 2
Q: The heights of students in a class are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. What proportion of students are taller than 180 cm?
A: Approximately 0.1587 (15.87%)
Rationale: To find this proportion, use the z-score formula:
z=(X−μ)σ=(180−170)10=1.z = \frac{(X - \mu)}{\sigma} = \frac{(180 - 170)}{10} =
1.z=σ(X−μ)=10(180−170)=1.
Using z-tables, the area to the left of z = 1 is approximately 0.8413. Therefore, the area to the
right is:
1−0.8413=0.1587.1 - 0.8413 = 0.1587.1−0.8413=0.1587.
Question 3
Q: A box contains 4 red balls and 6 blue balls. If a ball is picked at random, what is the
probability that it is red?
A: 0.4
Rationale: The probability of picking a red ball is calculated as:
P(Red)=Number of Red BallsTotal Number of Balls=44+6=410=0.4.P(\text{Red}) = \frac{\
text{Number of Red Balls}}{\text{Total Number of Balls}} = \frac{4}{4 + 6} = \frac{4}
{10} = 0.4.P(Red)=Total Number of BallsNumber of Red Balls=4+64=104=0.4.
Question 4
,Q: The following data shows the number of books read by a group of students in a month: 3,
5, 2, 4, 6, 3, 7, 2. What is the median number of books read?
A: 3.5
Rationale: To find the median, arrange the data in ascending order: 2, 2, 3, 3, 4, 5, 6, 7. Since
there are 8 numbers (an even set), the median is the average of the 4th and 5th values:
Median=3+42=3.5.\text{Median} = \frac{3 + 4}{2} = 3.5.Median=23+4=3.5.
Question 5
Q: A researcher finds that the correlation coefficient between hours studied and exam scores
is 0.8. What does this imply about the relationship between these two variables?
A: There is a strong positive correlation.
Rationale: A correlation coefficient (r) close to 1 indicates a strong positive linear
relationship, meaning that as one variable increases, the other tends to increase as well.
Question 6
Q: A dataset has a mean of 50 and a standard deviation of 5. If one data point is removed
from the dataset, causing the new mean to be 48, what can be inferred about the removed data
point?
A: The removed data point was greater than the mean.
Rationale: The mean decreased when a value was removed, indicating that the removed
value was greater than the original mean of 50. Removing a larger value would lower the
mean.
Question 7
Q: A student scores 45 out of 60 on a test. What percentage does this score represent?
A: 75%
Rationale: To calculate the percentage, divide the score by the total possible score and
multiply by 100:
Percentage=(4560)×100=75%.\text{Percentage} = \left(\frac{45}{60}\right) \times 100 =
75\%.Percentage=(6045)×100=75%.
,Question 8
Q: In a probability distribution, the probability of event A is 0.3 and the probability of event
B is 0.5. If A and B are mutually exclusive, what is the probability of either A or B
occurring?
A: 0.8
Rationale: For mutually exclusive events, the probability of either occurring is the sum of
their probabilities:
P(A or B)=P(A)+P(B)=0.3+0.5=0.8.P(A \text{ or } B) = P(A) + P(B) = 0.3 + 0.5 =
0.8.P(A or B)=P(A)+P(B)=0.3+0.5=0.8.
Question 9
Q: A die is rolled. What is the probability of rolling an even number?
A: 0.5
Rationale: There are three even numbers on a standard six-sided die (2, 4, 6). Therefore, the
probability is:
P(Even)=Number of Even OutcomesTotal Outcomes=36=0.5.P(\text{Even}) = \frac{\
text{Number of Even Outcomes}}{\text{Total Outcomes}} = \frac{3}{6} =
0.5.P(Even)=Total OutcomesNumber of Even Outcomes=63=0.5.
Question 10
Q: A school’s math test scores are normally distributed with a mean of 75 and a standard
deviation of 10. What score corresponds to the 90th percentile?
A: 84
Rationale: To find the 90th percentile, use the z-score for 90% (which is approximately
1.28). Calculate the score using the formula:
X=μ+(z⋅σ)=75+(1.28⋅10)=75+12.8=87.8.X = \mu + (z \cdot \sigma) = 75 + (1.28 \cdot 10) =
75 + 12.8 = 87.8.X=μ+(z⋅σ)=75+(1.28⋅10)=75+12.8=87.8.
Rounding gives a score of approximately 84.
Question 11
Q: A bag contains 5 red marbles and 3 green marbles. If one marble is drawn at random,
what is the probability that it is green?
A: 0.375
, Rationale: The probability of drawing a green marble is calculated as:
P(Green)=Number of Green MarblesTotal Number of Marbles=35+3=38=0.375.P(\
text{Green}) = \frac{\text{Number of Green Marbles}}{\text{Total Number of Marbles}}
= \frac{3}{5 + 3} = \frac{3}{8} =
0.375.P(Green)=Total Number of MarblesNumber of Green Marbles=5+33=83=0.375.
Question 12
Q: The following dataset shows the ages (in years) of a group of friends: 22, 25, 25, 23, 24,
21, 23. What is the mode of the ages?
A: 25
Rationale: The mode is the value that appears most frequently in the dataset. Here, 25
appears twice, which is more than any other age.
Question 13
Q: A student scores in the 85th percentile on an exam. What does this indicate about their
performance compared to their peers?
A: They scored better than 85% of the students.
Rationale: Being in the 85th percentile means that the student scored higher than 85% of all
test-takers, indicating strong performance.
Question 14
Q: In a normal distribution, if the mean is 100 and the standard deviation is 15, what is the z-
score for a value of 130?
A: 2.0
Rationale: The z-score is calculated using the formula:
z=(X−μ)σ=(130−100)15=3015=2.z = \frac{(X - \mu)}{\sigma} = \frac{(130 - 100)}{15} = \
frac{30}{15} = 2.z=σ(X−μ)=15(130−100)=1530=2.
Question 15
Q: A survey reveals that 70% of students prefer online classes to in-person classes. If 200
students are surveyed, how many students prefer online classes?
Course
Edexcel Maths A Level Paper 31 Statistics QP
Question 1
Q: A survey of 100 students found that the mean number of hours they spent studying each
week was 15 hours. What is the total number of hours studied by all students?
A: 1500 hours
Rationale: To find the total number of hours, multiply the mean by the number of students:
Total Hours=Mean×Number of Students=15×100=1500.\text{Total Hours} = \text{Mean} \
times \text{Number of Students} = 15 \times 100 =
1500.Total Hours=Mean×Number of Students=15×100=1500.
Question 2
Q: The heights of students in a class are normally distributed with a mean of 170 cm and a
standard deviation of 10 cm. What proportion of students are taller than 180 cm?
A: Approximately 0.1587 (15.87%)
Rationale: To find this proportion, use the z-score formula:
z=(X−μ)σ=(180−170)10=1.z = \frac{(X - \mu)}{\sigma} = \frac{(180 - 170)}{10} =
1.z=σ(X−μ)=10(180−170)=1.
Using z-tables, the area to the left of z = 1 is approximately 0.8413. Therefore, the area to the
right is:
1−0.8413=0.1587.1 - 0.8413 = 0.1587.1−0.8413=0.1587.
Question 3
Q: A box contains 4 red balls and 6 blue balls. If a ball is picked at random, what is the
probability that it is red?
A: 0.4
Rationale: The probability of picking a red ball is calculated as:
P(Red)=Number of Red BallsTotal Number of Balls=44+6=410=0.4.P(\text{Red}) = \frac{\
text{Number of Red Balls}}{\text{Total Number of Balls}} = \frac{4}{4 + 6} = \frac{4}
{10} = 0.4.P(Red)=Total Number of BallsNumber of Red Balls=4+64=104=0.4.
Question 4
,Q: The following data shows the number of books read by a group of students in a month: 3,
5, 2, 4, 6, 3, 7, 2. What is the median number of books read?
A: 3.5
Rationale: To find the median, arrange the data in ascending order: 2, 2, 3, 3, 4, 5, 6, 7. Since
there are 8 numbers (an even set), the median is the average of the 4th and 5th values:
Median=3+42=3.5.\text{Median} = \frac{3 + 4}{2} = 3.5.Median=23+4=3.5.
Question 5
Q: A researcher finds that the correlation coefficient between hours studied and exam scores
is 0.8. What does this imply about the relationship between these two variables?
A: There is a strong positive correlation.
Rationale: A correlation coefficient (r) close to 1 indicates a strong positive linear
relationship, meaning that as one variable increases, the other tends to increase as well.
Question 6
Q: A dataset has a mean of 50 and a standard deviation of 5. If one data point is removed
from the dataset, causing the new mean to be 48, what can be inferred about the removed data
point?
A: The removed data point was greater than the mean.
Rationale: The mean decreased when a value was removed, indicating that the removed
value was greater than the original mean of 50. Removing a larger value would lower the
mean.
Question 7
Q: A student scores 45 out of 60 on a test. What percentage does this score represent?
A: 75%
Rationale: To calculate the percentage, divide the score by the total possible score and
multiply by 100:
Percentage=(4560)×100=75%.\text{Percentage} = \left(\frac{45}{60}\right) \times 100 =
75\%.Percentage=(6045)×100=75%.
,Question 8
Q: In a probability distribution, the probability of event A is 0.3 and the probability of event
B is 0.5. If A and B are mutually exclusive, what is the probability of either A or B
occurring?
A: 0.8
Rationale: For mutually exclusive events, the probability of either occurring is the sum of
their probabilities:
P(A or B)=P(A)+P(B)=0.3+0.5=0.8.P(A \text{ or } B) = P(A) + P(B) = 0.3 + 0.5 =
0.8.P(A or B)=P(A)+P(B)=0.3+0.5=0.8.
Question 9
Q: A die is rolled. What is the probability of rolling an even number?
A: 0.5
Rationale: There are three even numbers on a standard six-sided die (2, 4, 6). Therefore, the
probability is:
P(Even)=Number of Even OutcomesTotal Outcomes=36=0.5.P(\text{Even}) = \frac{\
text{Number of Even Outcomes}}{\text{Total Outcomes}} = \frac{3}{6} =
0.5.P(Even)=Total OutcomesNumber of Even Outcomes=63=0.5.
Question 10
Q: A school’s math test scores are normally distributed with a mean of 75 and a standard
deviation of 10. What score corresponds to the 90th percentile?
A: 84
Rationale: To find the 90th percentile, use the z-score for 90% (which is approximately
1.28). Calculate the score using the formula:
X=μ+(z⋅σ)=75+(1.28⋅10)=75+12.8=87.8.X = \mu + (z \cdot \sigma) = 75 + (1.28 \cdot 10) =
75 + 12.8 = 87.8.X=μ+(z⋅σ)=75+(1.28⋅10)=75+12.8=87.8.
Rounding gives a score of approximately 84.
Question 11
Q: A bag contains 5 red marbles and 3 green marbles. If one marble is drawn at random,
what is the probability that it is green?
A: 0.375
, Rationale: The probability of drawing a green marble is calculated as:
P(Green)=Number of Green MarblesTotal Number of Marbles=35+3=38=0.375.P(\
text{Green}) = \frac{\text{Number of Green Marbles}}{\text{Total Number of Marbles}}
= \frac{3}{5 + 3} = \frac{3}{8} =
0.375.P(Green)=Total Number of MarblesNumber of Green Marbles=5+33=83=0.375.
Question 12
Q: The following dataset shows the ages (in years) of a group of friends: 22, 25, 25, 23, 24,
21, 23. What is the mode of the ages?
A: 25
Rationale: The mode is the value that appears most frequently in the dataset. Here, 25
appears twice, which is more than any other age.
Question 13
Q: A student scores in the 85th percentile on an exam. What does this indicate about their
performance compared to their peers?
A: They scored better than 85% of the students.
Rationale: Being in the 85th percentile means that the student scored higher than 85% of all
test-takers, indicating strong performance.
Question 14
Q: In a normal distribution, if the mean is 100 and the standard deviation is 15, what is the z-
score for a value of 130?
A: 2.0
Rationale: The z-score is calculated using the formula:
z=(X−μ)σ=(130−100)15=3015=2.z = \frac{(X - \mu)}{\sigma} = \frac{(130 - 100)}{15} = \
frac{30}{15} = 2.z=σ(X−μ)=15(130−100)=1530=2.
Question 15
Q: A survey reveals that 70% of students prefer online classes to in-person classes. If 200
students are surveyed, how many students prefer online classes?
