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

To calculate the standard deviation, we first need the variance. The formula for variance is:

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

Using the values:

• $\sum X^2 = 261.8$
• $n = 120$

We first calculate:

$s^2 = \frac{261.8}{120} - \left(\frac{165.6}{120}\right)^2$

Now, calculating:

$s^2 = 2.18167 - 1.9044 = 0.27727$

Then the standard deviation $s$ is:

$s = \sqrt{0.27727} \approx 0.526$

To find the probability $P(0.5 < X < 1.5)$, we standardize the values:

$Z = \frac{X - \bar{X}}{s}$

Where:

• $\bar{X} = 1.38$
• $s \approx 0.526$

For $X = 0.5$:

$Z_1 = \frac{0.5 - 1.38}{0.526} \approx -1.672$

For $X = 1.5$:

$Z_2 = \frac{1.5 - 1.38}{0.526} \approx 0.227$

Using the Z-table:

• $P(Z < 0.227) \approx 0.591$$
• $P(Z < -1.672) \approx 0.047$$

Therefore,

$P(0.5 < X < 1.5) = P(Z < 0.227) - P(Z < -1.672) \approx 0.591 - 0.047 = 0.544$

In a continuous distribution, the probability of a specific value is zero:

$P(X = 1) = 0$

To determine if a normal distribution is suitable, we must assess the mean and standard deviation:

• The calculated mean is $1.38$ and the standard deviation is $0.526$. Given the range of values observed and the nature of the data, we can conclude:

1. If values lie mostly within three standard deviations from the mean, it indicates suitability.
2. If the values extend beyond reasonable ranges or show significant skewness, normality may not hold.

In this case, checking the computed Z-scores shows acceptable values, suggesting normal distribution could be appropriate.

We use the cumulative probability approach:

$P(Y > 0.75) = 1 - P(Y \leq 0.75) = 0.10$

Thus,

$P(Y \leq 0.75) = 0.90$

Find the Z-score for $0.90$:

$Z \approx 1.281$

We standardize this value considering:

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

Substituting:

$1.281 = \frac{0.75 - \mu}{0.21}$

Solving for $\mu$ yields:

$0.75 - \mu = 1.281 \times 0.21 \approx 0.268$

Thus,

$\mu = 0.75 - 0.268 = 0.482$

Finally, rounding to three significant figures gives:

$\mu \approx 0.481$