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 Mechanics - Question 16 - 2018 - Paper 3

The standard deviation is calculated using the formula:

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

Here:

• $\sum X^2 = 261.8$
• $\sum X = 165.6$
• $n = 120$

First, we calculate:

$\frac{\sum X^2}{n} = \frac{261.8}{120} = 2.1816667$

Then we calculate:

$\left(\frac{\sum X}{n}\right)^2 = \left(1.38\right)^2 = 1.9044$

Now, substituting into the standard deviation formula:

$\sigma = \sqrt{2.1816667 - 1.9044} = \sqrt{0.2772667} \approx 0.526$

Using the Z-score formula, we transform the values:

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

Where:

• $\mu = 1.38$
• $\sigma \approx 0.526$

Calculating for $0.5$ and $1.5$:

1. For $0.5$:

$Z_{0.5} = \frac{0.5 - 1.38}{0.526} \approx -1.6697$

2. For $1.5$:

$Z_{1.5} = \frac{1.5 - 1.38}{0.526} \approx 0.2277$

Now we find the probabilities using the Z-table:

• $P(Z < -1.6697) \approx 0.0478$
• $P(Z < 0.2277) \approx 0.5910$

Thus,

$P(0.5 < X < 1.5) = P(Z < 0.2277) - P(Z < -1.6697) \approx 0.5910 - 0.0478 = 0.5432$

For a continuous distribution like the normal distribution, the probability of a specific value is always:

$P(X = 1) = 0$

To determine if a normal distribution is suitable, we compare the calculated values and consider:

• The calculated mean is $1.38$, and for the standard deviation is approximately $0.526$.
• Any probability calculations yield positive values. However, since $P(0.5 < X < 1.5)$ yielded usable results without negative probabilities, we can conclude that a normal model seems reasonable, but a check on skewness or kurtosis may be necessary to be definitive.

Given $P(Y > 0.75) = 0.10$, converting this into a Z-score:

$P(Z > z) = 0.10 \Rightarrow z \approx 1.2816$

Using the Z-score formula:

$z = \frac{Y - \mu}{\sigma}$

Substituting values:

• $Y = 0.75$
• $\sigma = 0.21$

Therefore:

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

Rearranging gives:

$0.75 - \mu = 1.2816 \times 0.21 \Rightarrow 0.75 - \mu \approx 0.2681$

Thus,

$\mu = 0.75 - 0.2681 \approx 0.4819$

To three significant figures:

$\mu \approx 0.482$