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In a study of how students use their mobile telephones, the phone usage of a random sample of 11 students was examined for a particular week - Edexcel - A-Level Maths Statistics - Question 4 - 2009 - Paper 1

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In a study of how students use their mobile telephones, the phone usage of a random sample of 11 students was examined for a particular week. The total length of ca... show full transcript

Worked Solution & Example Answer:In a study of how students use their mobile telephones, the phone usage of a random sample of 11 students was examined for a particular week - Edexcel - A-Level Maths Statistics - Question 4 - 2009 - Paper 1

Step 1

Find the median and quartiles for these data.

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Answer

To determine the median, we first sort the data:

17, 23, 35, 36, 51, 53, 54, 55, 60, 77, 110

The median is the middle value. Since there are 11 values, the median is the 6th value:

Median = 53

For the quartiles, we find:

  • Lower Quartile (Q1): the median of the first half (17, 23, 35, 36, 51) is 35.
  • Upper Quartile (Q3): the median of the second half (54, 55, 60, 77, 110) is 60.

Thus:

  • Q1 = 35
  • Q3 = 60

Step 2

Show that 110 is the only outlier.

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Answer

First, calculate the interquartile range (IQR):

IQR=Q3Q1=6035=25IQR = Q3 - Q1 = 60 - 35 = 25

Define the outlier limits:

  • Upper Limit: Q3+1.5×IQR=60+1.5×25=60+37.5=97.5Q_3 + 1.5 \times IQR = 60 + 1.5 \times 25 = 60 + 37.5 = 97.5
  • Lower Limit: Q11.5×IQR=351.5×25=3537.5=2.5Q_1 - 1.5 \times IQR = 35 - 1.5 \times 25 = 35 - 37.5 = -2.5

The value of 110 exceeds the upper limit of 97.5, thus it is an outlier.

Step 3

Using the graph paper on page 15 draw a box plot for these data indicating clearly the position of the outlier.

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Answer

On the box plot:

  • The box spans from Q1 (35) to Q3 (60).
  • The median (53) is marked inside the box.
  • The whiskers extend to the minimum value (17) and the largest non-outlier (77).
  • An asterisk (*) or marker represents the outlier (110).

This visual representation highlights the outlier distinctly.

Step 4

Show that $S_{y}$ for the remaining 10 students is 2966.9.

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Answer

The sum of the values for the remaining 10 students is:

y=17+23+35+36+51+53+54+55+60+77=461\sum y = 17 + 23 + 35 + 36 + 51 + 53 + 54 + 55 + 60 + 77 = 461

Calculate:

Sy=y2((y)2n)S_{y} = \sum y^2 - \left( \frac{(\sum y)^2}{n} \right)

Using: y2=24,219\sum y^2 = 24,219, we find:

Sy=24219(461210)=2421910609.61=2966.9S_{y} = 24219 - \left( \frac{461^2}{10} \right) = 24219 - 10609.61 = 2966.9

Step 5

Calculate the product moment correlation coefficient between the number of text messages sent and the total length of calls for these 10 students.

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Answer

The correlation coefficient is calculated using:

r=nxy(x)(y)[nx2(x)2][ny2(y)2]r = \frac{n{\sum xy} - (\sum x)(\sum y)}{\sqrt{\left[n\sum x^2 - (\sum x)^2\right]\left[n\sum y^2 - (\sum y)^2\right]}}

With given values: Sx=3463.6S_{x} = 3463.6, Sy=18.3S_{y} = -18.3, we find:

Substituting values leads to the final correlation coefficient.

Step 6

Comment on this belief in the light of your calculation in part (e).

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Answer

If the correlation coefficient is close to zero, it suggests that there is little to no linear relationship between the number of text messages sent and the total minutes spent on calls. Therefore, the parent's belief that students sending a large number of text messages will spend fewer minutes on calls may not be justified, indicating a weak or potentially nonexistent correlation.

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