A survey of 120 adults found that the volume, $X$ litres per person, of carbonated drinks they consumed in a week had the following results: $$\sum X = 165.6$$ $$\sum X^2 = 261.8$$ 16 (a) (i) Calculate the mean of $X$. 16 (a) (ii) Calculate the standard deviation of $X$. 16 (b) Assuming that $X$ can be modelled by a normal distribution find 16 (b) (i) $P(0.5 0.75) = 0.10$, find the value of $\mu$, correct to three significant figures.

A-Level AQA Maths: Pure 2.13 Further Modelling with Functions: A survey of 120 adults found tha

A survey of 120 adults found that the volume, $X$ litres per person, of carbonated drinks they consumed in a week had the following results: $$\sum X = 165.6$$ $$\sum X^2 = 261.8$$ 16 (a) (i) Calculate the mean of $X$ - AQA - A-Level Maths: Pure - Question 16 - 2018 - Paper 3

Question 16

$A-survey-of-120-adults-found-that-the-volume,-$X$-litres-per-person,-of-carbonated-drinks-they-consumed-in-a-week-had-the-following-results:-$$\sum-X-=-165.6$$-$$\sum-X^2-=-261.8$$--16-(a)-(i)-Calculate-the-mean-of-$X$-AQA-A-Level Maths: Pure-Question 16-2018-Paper 3.png$

A survey of 120 adults found that the volume, $X$ litres per person, of carbonated drinks they consumed in a week had the following results: $$\sum X = 165.6$$ $$\su... show full transcript

Worked Solution & Example Answer:A survey of 120 adults found that the volume, $X$ litres per person, of carbonated drinks they consumed in a week had the following results: $$\sum X = 165.6$$ $$\sum X^2 = 261.8$$ 16 (a) (i) Calculate the mean of $X$ - AQA - A-Level Maths: Pure - Question 16 - 2018 - Paper 3

Step 1

Calculate the mean of $X$.

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Answer

To calculate the mean of $X$ , we use the formula:

$\mu = \frac{\sum X}{n}$

where $\sum X = 165.6$ and $n = 120$ .

Thus, $\mu = \frac{165.6}{120} = 1.38$

Step 2

Calculate the standard deviation of $X$.

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Answer

The standard deviation $\sigma$ is calculated using the formula:

$\sigma = \sqrt{\frac{\sum X^2}{n} - (\mu)^2}$

Substituting the values:

$\sum X^2 = 261.8$ , so $\frac{\sum X^2}{n} = \frac{261.8}{120} \approx 2.18167$
Already obtained mean $\mu = 1.38$ , thus $\mu^2 = (1.38)^2 = 1.9044$

Now, calculating:

$\sigma = \sqrt{2.18167 - 1.9044} \approx \sqrt{0.27727} \approx 0.526$

Step 3

Assuming that $X$ can be modelled by a normal distribution find $P(0.5 < X < 1.5)$.

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Answer

To find $P(0.5 < X < 1.5)$ , we first standardize the variable:

$Z = \frac{X - \mu}{\sigma}$

For $X = 0.5$ :

$Z = \frac{0.5 - 1.38}{0.526} \approx -1.677 \implies P(Z < -1.677)$

For $X = 1.5$ :

$Z = \frac{1.5 - 1.38}{0.526} \approx 0.227 \implies P(Z < 0.227)$

Using Z-tables or calculators:

$P(0.5 < X < 1.5) = P(Z < 0.227) - P(Z < -1.677) \approx 0.5910 - 0.0468 = 0.5442$

Step 4

Find $P(X = 1)$.

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Answer

In a continuous distribution, the probability of $P(X = 1)$ for any single point is 0. Therefore,

$P(X = 1) = 0$

Step 5

Determine whether a normal distribution is suitable to model this data.

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To determine if a normal distribution is appropriate, we should consider:

The calculated mean $\mu = 1.38$ and standard deviation $\sigma \approx 0.526$ .
Check for any significant skewness or outliers in the data distribution (this may require graphical analysis, like a histogram). If the values remain within reasonable limits and there are no outliers, it can be considered normal, otherwise it might not be suitable depending on the skewness observed.

Step 6

Find the value of $\mu$, correct to three significant figures.

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Answer

Given $P(Y > 0.75) = 0.10$ , we will use the Z-score formula:

$P(Z \geq z) = 0.10 \implies z \approx 1.2816$

Thus, substituting into the Z-score formula:

$\frac{0.75 - \mu}{0.21} = -1.2816$

Solving for $\mu$ gives:

$0.75 - \mu = -1.2816 \cdot 0.21$

Calculating:

$-\mu = -0.2681 + 0.75$

Thus,

$\mu \approx 0.4819, \text{ correct to three significant figures gives } \mu = 0.482$

A survey of 120 adults found that the volume, $X$ litres per person, of carbonated drinks they consumed in a week had the following results: $$\sum X = 165.6$$ $$\sum X^2 = 261.8$$ 16 (a) (i) Calculate the mean of $X$ - AQA - A-Level Maths: Pure - Question 16 - 2018 - Paper 3

Calculate the mean of $X$.

Calculate the standard deviation of $X$.

Assuming that $X$ can be modelled by a normal distribution find $P(0.5 < X < 1.5)$.

Find $P(X = 1)$.

Determine whether a normal distribution is suitable to model this data.

Find the value of $\mu$, correct to three significant figures.

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