A survey of 100 households gave the following results for weekly income y £: Income y (£) Mid-point Frequency f 0 ≤ y < 200 100 12 200 ≤ y < 240 220 22 240 ≤ y < 320 280 22 320 ≤ y < 400 360 18 400 ≤ y < 600 500 18 600 ≤ y < 800 700 8 (You may use Σf y² = 12 452 800) A histogram was drawn and the class 200 ≤ y < 240 was represented by a rectangle of width 2 cm and height 7 cm - Edexcel - A-Level Maths Statistics - Question 5 - 2013 - Paper 1

To estimate the median, we first calculate the cumulative frequencies and locate where the median lies:

Median Position = (n + 1) / 2 = (100 + 1) / 2 = 50.5.

The cumulative frequency corresponding to the income class where the 50.5 position lies must be identified. The cumulative frequencies add up as:

• 0 ≤ y < 200: 12
• 200 ≤ y < 240: 34
• 240 ≤ y < 320: 56 (median falls here)

Using linear interpolation:

Median = L + [(N/2 - F) / f] * w Where:

• L = lower boundary of the median class (240)
• N/2 = 50.5
• F = cumulative frequency before the median class (34)
• f = frequency of the median class (22)
• w = class width (80)

Calculating:

Median = 240 + [(50.5 - 34) / 22] * 80 ≈ 240 + 60.73 = 300.73 ≈ 301 (rounded to nearest pound).

To estimate the mean income, the following formula is used:

Mean (ar{y}) = Σ(f * mid-point) / Σf.

Calculating:

Mean = (12 * 100 + 22 * 220 + 22 * 280 + 18 * 360 + 18 * 500 + 8 * 700) / 100 = 316.

For the standard deviation, we utilize:

Standard Deviation (σ) = √(Σ(f * (mid-point - mean)²) / Σf).

Calculating step-by-step:

1. Calculate (mid-point - mean) for each class, square it, and multiply by the frequency:

Variance = Σ(f * (mid-point - 316)²) = 1,245,2800 (from the provided Σf y²).

Standard Deviation = √(12452800 / 100) = √(124528) ≈ 157.

Thus, mean = 316 and standard deviation ≈ 157.

Skewness can be calculated using the formula:

Skewness = 3(Mean - Median) / Standard Deviation.

Here:

• Mean = 316
• Median = 301
• Standard Deviation ≈ 157.

Substituting:

Skewness = 3(316 - 301) / 157 = 3 * 15 / 157 ≈ 0.286.

This value indicates a positive skew, suggesting that there are more low-income households than high-income households.

To find P(240 < X < 400) for a normal distribution defined by:

• Mean = 320
• Standard Deviation = 150,

We use the z-score formula for finding probabilities:

Z = (X - mean) / standard deviation.

Calculating for both limits:

1. For X = 240: Z_1 = (240 - 320) / 150 = -0.53.
2. For X = 400: Z_2 = (400 - 320) / 150 = 0.53.

Using a standard normal distribution table:

P(240 < X < 400) = P(Z < 0.53) - P(Z < -0.53) ≈ 0.703 - 0.298 = 0.405.

Thus, P(240 < X < 400) ≈ 0.40.

Considering the skewness calculated (approximately 0.286, which is positive), it implies that the income distribution is not symmetric and leans toward lower incomes.

Additionally, the calculated probability P(240 < X < 400) ≈ 0.40 shows a significant portion of households earning between these amounts. However, modeling the income with a normal distribution could overlook the true behavior of the skewed data.

Katie's suggestion might provide a general idea, but the underlying income distribution should be analyzed more deeply to capture the true characteristics of the data. Hence, while using a normal distribution offers insights, it may not accurately reflect the skew observed in the raw data.