In the South West region of England, 100 households were randomly selected and, for each household, the weekly expenditure, £X, per person on food and drink was recorded - AQA - A-Level Maths Mechanics - Question 13 - 2019 - Paper 3

The standard deviation is calculated using the formula:

S = rac{ ext{Square root}igg( ext{Sum of squared deviations} igg)}{n - 1} = rac{ ext{Square root}(1746.29)}{99} \\ \\ S = rac{41.8}{99} \\ \\ S ext{ is approximately } 4.20

Thus, the standard deviation of X is approximately 4.20.

The normal distribution can be utilized for modeling X due to several reasons:

1. The sample size is sufficiently large (n = 100), which helps in achieving normality according to the Central Limit Theorem.
2. The data is assumed to be continuous and the values of expenditure typically vary throughout a range, supporting the use of a bell-shaped curve.

To find this probability, we first need to compute the Z-score:

Z = rac{X - ar{X}}{S} = rac{25 - 30.46}{4.20} \ \\ \ Z ext{ is approximately } -1.29

Using Z-tables or calculator, we find that:

$$P(Z < -1.29) = 0.097$$

Thus, the probability that a household spends less than £25.00 is approximately 0.097.

Given that P(Y < 30) = 0.55, we first find the Z-score corresponding to this probability:

Assuming standard normal distribution, we find:

• Z ≈ 0.125 (from standard normal distribution tables)

Using the Z-score formula for Y:

Z = rac{30 - 29.55}{ ext{SD}_Y} = 0.125

Solving for SD_Y:

ext{SD}_Y = rac{30 - 29.55}{0.125} = 3.6

Thus, the standard deviation of Y is approximately 3.6.