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

To find the standard deviation ( s ), we use the formula:

$s = \sqrt{\frac{\sum (X - \bar{X})^2}{n - 1}}$

Substituting the values, we have:

$s = \sqrt{\frac{1746.29}{99}} \approx 4.20$

Thus, the standard deviation is approximately ( 4.20 ).

The normal distribution can be used to model ( X ) because the data obtained from the selected households can be regarded as continuous and approximately symmetrically distributed around the mean. Given that the sample mean is ( \bar{X} = 30.46 ) and the standard deviation is relatively small compared to the range of values, we can conclude that the expenditure data falls within the range defined by ( \bar{X} \pm 3s ), allowing us to model it as a normal distribution.

To find this probability, we first standardize the value:

$Z = \frac{X - \bar{X}}{s} = \frac{25 - 30.46}{4.20} \approx -1.29$

Using the standard normal distribution table, we find that:

$P(Z < -1.29) \approx 0.1003$

Thus, the probability that a household spends less than £25.00 is approximately ( 0.1003 ).

For households in the North West, the distribution of ( Y ) is given with a mean of £29.55. The problem statement does not provide direct information about the standard deviation for ( Y ). However, if we assume the standard deviation to be similar to the data given for the South West, and given the model is normal, we can derive it from data points if available. In this case, we will need additional information to calculate it accurately. If assumed similar to ( X ), we could use the standard deviation found previously as an approximation.