Photo AI

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 icon

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-Edexcel-A-Level Maths Statistics-Question 5-2009-Paper 1.png

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

Answer

To calculate the width for the group (16 – 25) hours, we find the difference between 16 and 25:

extWidth=2516=9exthours ext{Width} = 25 - 16 = 9 ext{ hours}

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

Answer

To find the median, we first locate the position using:

extMedianposition=n+12=104+12=52.5 ext{Median position} = \frac{n + 1}{2} = \frac{104 + 1}{2} = 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=Q3Q1=15.49.4=6. ext{IQR} = 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

Answer

To calculate the mean, we use:

extMean=(fx)n=(202.75)+(166.5)+(189)+(2513)+(1520.5)+(1038)104=12.8. ext{Mean} = \frac{\sum (f \cdot x)}{n} = \frac{(20 \cdot 2.75) + (16 \cdot 6.5) + (18 \cdot 9) + (25 \cdot 13) + (15 \cdot 20.5) + (10 \cdot 38)}{104} = 12.8.

For standard deviation, we follow:

σ=f(xxˉ)2n.\sigma = \sqrt{\frac{\sum f(x - \bar{x})^2}{n}}.

This results in: Standard Deviation ≈ 9.88.

Step 4

State, giving a reason, the skewness of these data.

98%

120 rated

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

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.

Join the A-Level students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;