In a shopping survey a random sample of 104 teenagers were asked how many hours, to the nearest hour, they spent shopping in the last month - Edexcel - A-Level Maths Statistics - Question 5 - 2009 - Paper 1
Question 5
In a shopping survey a random sample of 104 teenagers were asked how many hours, to the nearest hour, they spent shopping in the last month. The results are summaris... show full transcript
Worked Solution & Example Answer:In a shopping survey a random sample of 104 teenagers were asked how many hours, to the nearest hour, they spent shopping in the last month - Edexcel - A-Level Maths Statistics - Question 5 - 2009 - Paper 1
Step 1
Calculate the width and height of the rectangle representing the group (16 – 25) hours.
96%
114 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
To calculate the width for the group (16 – 25) hours, we find the difference between 16 and 25:
extWidth=25−16=9exthours
The height represents the frequency for this group shown in the table, which is 15.
Thus,
Width = 9 hours
Height = 15.
Step 2
Use linear interpolation to estimate the median and interquartile range.
99%
104 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
To find the median, we first locate the position using:
extMedianposition=2n+1=2104+1=52.5
The cumulative frequency shows that the median falls in the 11-15 hour group. Through interpolation,
Median ≈ 12.8.
The lower quartile (Q1) is at 26 (which is 25% of 104) and positioned at ... gives:
Q1 ≈ 9.4.
The upper quartile (Q3) at 78 gives:
Q3 ≈ 15.4.
Interquartile Range (IQR) is calculated as:
extIQR=Q3−Q1=15.4−9.4=6.
Step 3
Estimate the mean and standard deviation of the number of hours spent shopping.
96%
101 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
State, giving a reason, the skewness of these data.
98%
120 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
The data is positively skewed because the mean (12.8) is greater than the median (around 10), indicating a longer tail on the right side.
Step 5
State, giving a reason, which average and measure of dispersion you would recommend to use to summarise these data.
97%
117 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
I would recommend using the median and interquartile range (IQR) to summarise the data, as they are less influenced by outliers and better represent the central tendency and spread of the skewed data.