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 packet of cigarettes a day, for cotinine. The results, in appropriate units, are shown below. Patient A B C D E F G H I J K L Cotinine level, x 160 390 169 175 420 171 250 258 186 243 250 243 [You may use $ \sum x^2 = 724961 $] (a) Find the mean and standard deviation of the level of cotinine in a patient’s blood. (b) Find the median, upper and lower quartiles of these data. (c) A doctor suspects that some of his patients have been smoking more than a packet of cigarettes per day. He decides to use $ Q_3 + 1.5(Q_3 - Q_1) $ to determine if any of the cognitive results are far enough away from the upper quartile to be outliers. (d) Identify which patient(s) may have been smoking more than a packet of cigarettes a day. Show your working clearly. Research suggests that cotinine levels in the blood form a skewed distribution. One measure of skewness is found using $ Q_1 - 2Q_2 + Q_3 $ d) Evaluate this measure and describe the skewness of these data.

A-Level Edexcel Maths Statistics Working with Data: Cotinine is a chemical that is m

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

Question 2

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, we sum all cotinine levels and divide by the number of patients:

Mean: $\text{Mean} = \frac{\sum x}{n} = \frac{2757}{12} = 229.75$

Next, we calculate the standard deviation:

First, find the variance: $\text{Variance} = \frac{\sum x^2}{n} - \left( \frac{\sum x}{n} \right)^2$ where ( \sum x^2 = 724961 ).
Therefore, $\text{Variance} = \frac{724961}{12} - (229.75)^2 = \frac{724961}{12} - 52751.5625 = 7645.375$
Taking the square root gives us the standard deviation: $\text{SD} = \sqrt{7645.375} \approx 87.34$ .

Step 2

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

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Answer

First, we arrange the cotinine levels in ascending order: 125, 160, 169, 171, 175, 186, 243, 250, 250, 258, 390, 420.

To find the median (Q2): Since there are 12 data points, the median is the average of the 6th and 7th values:

$Q_2 = \frac{186 + 243}{2} = 214.5$

Next, we find the lower quartile (Q1):

For Q1, we take the lower half (first six values): 125, 160, 169, 171, 175, 186. Thus,

$Q_1 = \frac{169 + 171}{2} = 170$

For the upper quartile (Q3):

For Q3, we take the upper half (last six values): 186, 243, 250, 250, 258, 390. Thus,

$Q_3 = \frac{250 + 258}{2} = 254$ .

Step 3

A doctor suspects that some of his patients have been smoking more than a packet of cigarettes per day. He decides to use Q3 + 1.5(Q3 - Q1) to determine if any of the cognitive results are far enough away from the upper quartile to be outliers.

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Answer

Calculating the outlier threshold:

Find ( Q_3 - Q_1 ): $Q_3 - Q_1 = 254 - 170 = 84$
Then compute the threshold for identifying outliers: $Q_3 + 1.5(Q_3 - Q_1) = 254 + 1.5(84) = 254 + 126 = 380$ .

Any patient with a cotinine level above 380 is considered an outlier.

Step 4

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

The only patient with a cotinine level above 380 is Patient B, with a level of 390. Thus, Patient B may have been smoking more than a packet of cigarettes a day.

Step 5

Evaluate this measure and describe the skewness of these data.

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Answer

Using the measure of skewness, we evaluate: $Q_1 - 2Q_2 + Q_3 = 170 - 2(214.5) + 254 = 170 - 429 + 254 = -5$ .

A negative skewness indicates that the data is skewed to the left, meaning there are a larger number of lower values in the distribution.

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

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

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

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

A doctor suspects that some of his patients have been smoking more than a packet of cigarettes per day. He decides to use Q3 + 1.5(Q3 - Q1) to determine if any of the cognitive results are far enough away from the upper quartile to be outliers.

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

Evaluate this measure and describe the skewness of these data.

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