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Cotinine is a chemical that is made by the body from nicotine which is found in cigarette smoke - Edexcel - A-Level Maths Statistics - Question 2 - 2008 - Paper 1

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Cotinine is a chemical that is made by the body from nicotine which is found in cigarette smoke. A doctor tested the blood of 12 patients, who claimed to smoke a pac... show full transcript

Worked Solution & Example Answer:Cotinine is a chemical that is made by the body from nicotine which is found in cigarette smoke - Edexcel - A-Level Maths Statistics - Question 2 - 2008 - Paper 1

Step 1

Find the mean and standard deviation of the level of cotinine in a patient’s blood.

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Answer

To find the mean cotinine level, we sum the cotinine levels and divide by the number of patients:

extMean=xn=160+390+169+175+186+210+243+250+420+258+186+24312=275712=229.75 ext{Mean} = \frac{\sum x}{n} = \frac{160 + 390 + 169 + 175 + 186 + 210 + 243 + 250 + 420 + 258 + 186 + 243}{12} = \frac{2757}{12} = 229.75

To find the standard deviation, we first calculate the variance:

  1. Calculate the mean: 229.75.
  2. Compute the squared differences from the mean: (xiMean)2=724961\sum (x_i - \text{Mean})^2 = 724961
  3. The variance is: Variance=(xiMean)2n=72496112=60413.42\text{Variance} = \frac{\sum (x_i - \text{Mean})^2}{n} = \frac{724961}{12} = 60413.42
  4. Therefore, the standard deviation is: SD=60413.4287.34\text{SD} = \sqrt{60413.42} \approx 87.34

Step 2

Find the median, upper and lower quartiles of these data.

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Answer

First, we order the data:

Ordered List: 125, 160, 169, 171, 175, 186, 210, 243, 250, 258, 390, 420

  • Median: The median is the average of the 6th and 7th values: Q2=186+2102=198Q_2 = \frac{186 + 210}{2} = 198

  • Lower Quartile: The lower quartile is the median of the first half (first 6 values): Q1=160+1692=164.5Q_1 = \frac{160 + 169}{2} = 164.5

  • Upper Quartile: The upper quartile is the median of the second half (last 6 values): Q3=243+2502=246.5Q_3 = \frac{243 + 250}{2} = 246.5

Step 3

Identify which patient(s) may have been smoking more than a packet of cigarettes a day. Show your working clearly.

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Answer

To identify potential outliers using the formula: Q3+1.5(Q3Q1)=246.5+1.5(246.5164.5)Q_3 + 1.5(Q_3 - Q_1) = 246.5 + 1.5(246.5 - 164.5) Calculating: Q3Q1=246.5164.5=82Q_3 - Q_1 = 246.5 - 164.5 = 82 1.5(82)=1231.5(82) = 123 Thus, Q3+123=369.5Q_3 + 123 = 369.5

Patients with cotinine levels above 369.5 are considered outliers. Checking the data:

  • Patient B: 390
  • Patient I: 420

Therefore, Patients B and I may have been smoking more than a packet of cigarettes a day.

Step 4

Evaluate this measure and describe the skewness of these data.

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Answer

To calculate the measure of skewness:

Q12Q2+Q3Q_1 - 2Q_2 + Q_3 Substituting the quartile values: 164.52(198)+246.5=164.5396+246.5=15164.5 - 2(198) + 246.5 = 164.5 - 396 + 246.5 = 15

Since the result is positive, it indicates a positive skew in the distribution of cotinine levels among the patients. This suggests that there are more patients with lower cotinine levels compared to those with higher levels.

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