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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.
Answer
To calculate the mean ( \bar{x} ), sum the cotinine levels and divide by the number of patients:
[ \bar{x} = \frac{\sum x}{n} = \frac{160 + 390 + 169 + 175 + 186 + 210 + 243 + 250 + 420 + 258 + 186 + 243}{12} = \frac{2757}{12} \approx 229.75 ]\
To find the standard deviation (sd), we first calculate the variance:
[ \sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n} ]\
Using ( \sum x^2 = 724961 ), we get:
[ \sigma^2 = \frac{724961 - (2757^2 / 12)}{12} \approx 87.34 ]
Thus, the standard deviation, ( \sigma ), is:
[ sd = \sqrt{\sigma^2} \approx 9.32 ]
Step 2
Find the median, upper and lower quartiles of these data.
Answer
First, order the cotinine levels:
Ordered List: 125, 160, 169, 171, 175, 186, 186, 210, 243, 250, 258, 420
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Median (Q2): The median is the average of the 6th and 7th numbers: [ Q_2 = \frac{186 + 186}{2} = 186 ]
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Lower quartile (Q1): The median of the first half (first 6 numbers) is: [ Q_1 = \frac{160 + 169}{2} = 164.5 ]
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Upper quartile (Q3): The median of the second half (last 6 numbers) is: [ Q_3 = \frac{250 + 258}{2} = 254 ]
Step 3
Identify which patient(s) may have been smoking more than a packet of cigarettes a day. Show your working clearly.
Answer
To determine if any patient is an outlier, calculate: [ Q_3 + 1.5(Q_3 - Q_1) = 254 + 1.5(254 - 164.5) = 254 + 1.5(89.5) = 254 + 134.25 = 388.25 ]
Patients with cotinine levels exceeding 388.25 are considered outliers. Therefore, patients I (420) and B (390) may be smoking more than a packet of cigarettes a day.
Step 4
Evaluate this measure and describe the skewness of these data.
Answer
Using the measure of skewness: [ Q_1 - 2Q_2 + Q_3 ] Substituting the values: [ 164.5 - 2(186) + 254 = 164.5 - 372 + 254 = 46.5 ]
A positive skewness value indicates that the distribution of cotinine levels is positively skewed, meaning that there are more low values and a tail extending towards higher values.