Pearson Edexcel GCE A-Level Statistics (Paper 31) – June 2019

Why we do this: A tree diagram helps visualize the sequential events. We start at Bag A. If we get Green, we move to Bag B. If we get Green again, we move to Bag C. If we get Red at any point, we stop.

How:

• Bag A: Total 10 marbles (1 Red, 9 Green). \( P(R) = \frac{1}{10} \), \( P(G) = \frac{9}{10} \).
• Bag B: Total 5 marbles (1 Red, 4 Green). \( P(R) = \frac{1}{5} \), \( P(G) = \frac{4}{5} \).
• Bag C: Total 3 marbles (1 Red, 2 Green). \( P(R) = \frac{1}{3} \), \( P(G) = \frac{2}{3} \).

Strategy: Follow the path G \( \to \) G \( \to \) G on the tree diagram.

Why we do this: The question asks for “at least 1 marble of each colour”. Since we stop on Red, we only pick multiple marbles if we pick Greens first.

However, re-reading the question: “Three bags… Sasha selects from A…”.

Wait, if we select Red immediately, we have 1 Red, 0 Green. (Not 1 of each).

If we select Green then Red, we have 1 Green, 1 Red. (1 of each).

If we select Green then Green then Red, we have 2 Green, 1 Red. (1 of each).

If we select Green then Green then Green, we have 3 Green, 0 Red. (Not 1 of each).

So the valid outcomes are:

• Green from A, Red from B (End). Outcome: {G, R}
• Green from A, Green from B, Red from C (End). Outcome: {G, G, R}

Why we do this: We need \( P(\text{Red from B} \mid \text{Red selected}) \).

Formula: \( P(A \mid B) = \frac{P(A \cap B)}{P(B)} \).

Here: \( \frac{P(\text{Red from B})}{P(\text{Any Red selected})} \).

Calculate Denominator (Total Probability of Red):

• Red from A: \( \frac{1}{10} \)
• Red from B: \( \frac{9}{10} \times \frac{1}{5} = \frac{9}{50} \)
• Red from C: \( \frac{9}{10} \times \frac{4}{5} \times \frac{1}{3} = \frac{6}{25} \) (Wait, calculated above as 36/150)

Sum of all probabilities involving Red:

Alternatively: \( P(\text{Red}) = 1 – P(\text{No Red}) = 1 – P(G,G,G) = 1 – \frac{12}{25} = \frac{13}{25} = \frac{26}{50} \). This matches.

Calculate Numerator (Red from B):

Strategy: Calculate the Interquartile Range (IQR) and the outlier boundaries (fences). Determine which points are outliers and draw the whiskers to the last non-outlier values.

Values from graph/text:

• \( Q_1 = 19.4 \) (Read from left of box)
• \( Q_3 = 26.6 \) (Read from right of box)
• \( \text{IQR} = 26.6 – 19.4 = 7.2 \)

Outlier Fences:

• Lower Fence: \( Q_1 – 1.5 \times \text{IQR} = 19.4 – 1.5(7.2) = 19.4 – 10.8 = 8.6 \)
• Upper Fence: \( Q_3 + 1.5 \times \text{IQR} = 26.6 + 1.5(7.2) = 26.6 + 10.8 = 37.4 \)

Analyzing Data Points:

• Lowest values: 7.6, 8.1, 9.1.
• Are they below 8.6? Yes: 7.6 and 8.1 are outliers. 9.1 is inside the fence.
• Lower whisker goes to 9.1.
• Highest value: 32.5.
• Is it above 37.4? No.
• Upper whisker goes to 32.5.

(b) Month of outliers: The outliers are very low temperatures (7.6, 8.1). In Beijing (Northern Hemisphere), the coldest months are in winter. The data covers May to October. The coldest month in this range would be October.

(c) Standard Deviation:

Use the formula \( \sigma = \sqrt{\frac{S_{xx}}{n}} \).

Why we do this: We need the difference between the 90th percentile and the 10th percentile of the normal distribution \( N(22.6, 5.19^2) \).

How: Find the \( z \)-score for 90% (Area = 0.9) and 10% (Area = 0.1).

From tables or calculator, \( z \approx 1.2816 \) for 90%.

The 10th percentile is symmetrical, so \( z \approx -1.2816 \).

Range = \( 2 \times z \times \sigma \).

We need variables from the Large Data Set (LDS) for Beijing that aren’t Normal.

Common Answers:

• Rainfall: It is not symmetric (skewed). There are many days with 0 rainfall, so it’s not bell-shaped.
• Wind Speed / Beaufort conversion: Wind speed is skewed. Beaufort scale is discrete/qualitative (ordinal), not continuous.

Why we do this: We are testing if the population correlation coefficient (\( \rho \)) is positive (“greater than zero”).

Hypotheses:

• \( H_0: \rho = 0 \) (No correlation)
• \( H_1: \rho > 0 \) (Positive correlation)

Critical Value: Sample size \( n = 24 \), Significance level 5% (0.05). One-tailed test.

From tables for PMCC with \( n=24 \): Critical Value \( r_{cv} = 0.3438 \).

The new correlation coefficient for the coded data is \( 0.882 \).

This value is closer to 1 than the original correlation of \( 0.446 \). This indicates a stronger linear relationship between \( \log x \) and \( \log y \), which implies the non-linear relationship (power law) fits the original data better than a linear one.

Why we do this: We need to convert the linear log-log model back to the original variables \( y \) and \( x \).

Model: \( c = -1.82 + 0.89m \)

Substitutions: \( c = \log_{10} y \) and \( m = \log_{10} x \)

We need the proportion of days with cloud cover 6, 7, or 8.

Frequencies: 6 (52 days), 7 (52 days), 8 (28 days).

Model: \( X \sim B(8, 0.76) \)

(i) Find \( P(X \ge 6) \):

Using calculator (Cumulative Binomial CD) or formula.

Or sum individual probabilities \( P(X=6) + P(X=7) + P(X=8) \).

(ii) Expected number of days with 7:

First find \( P(X = 7) \).

Expected Number = \( n \times p = 184 \times 0.2811 \)

Compare the data results with the model results.

• Data \( P(\ge 6) \approx 0.717 \). Model \( P(\ge 6) \approx 0.703 \). (Close)
• Data observed days with 7: 52. Model expected days: 51.7. (Very close)

(d) Proportion for following days:

Table shows days following an 8. Total = 28.

Days with 6, 7, or 8: \( 5 + 9 + 9 = 23 \).

(e) Comment on Model:

The overall probability of \( \ge 6 \) is \( 0.717 \). However, given the previous day was an 8, the probability rises to \( 0.821 \).

This suggests that the cloud cover on one day is dependent on the previous day (persistence). The Binomial model assumes independent trials. Therefore, the Binomial model may not be suitable because independence does not hold.

Given: \( D \sim N(25, \sigma^2) \).

\( P(D < 24.63) = 0.15 \).

Find \( \sigma \): Standardize.

From tables/calculator, \( P(Z < z) = 0.15 \Rightarrow z \approx -1.0364 \).

Find \( k \):

We are told \( P(24.63 < D < k) = 0.45 \).

We know \( P(D < 24.63) = 0.15 \).

So \( P(D < k) = P(D < 24.63) + 0.45 = 0.15 + 0.45 = 0.60 \).

Find \( z \) such that \( P(Z < z) = 0.60 \). From tables \( z \approx 0.2533 \).

Define Variable: Let \( Y \) be the number of bottles between 24.63 and \( k \).

Probability \( p = 0.45 \) (from part a). Sample \( n = 200 \).

\( Y \sim B(200, 0.45) \).

Question: “Fewer than half”. So \( P(Y < 100) \).

Approximation: \( n \) is large, so approximate with Normal.

Mean \( \mu = np = 200 \times 0.45 = 90 \).

Variance \( \sigma^2 = np(1-p) = 90 \times 0.55 = 49.5 \).

So \( W \sim N(90, 49.5) \).

Continuity Correction: \( P(Y < 100) \) becomes \( P(W < 99.5) \).

Hypotheses:

• \( H_0: \mu = 25 \)
• \( H_1: \mu < 25 \) (Hannah's belief)

Distribution of Sample Mean: \( \bar{X} \sim N(\mu, \frac{\sigma^2}{n}) \).

\( \sigma = 0.16 \), \( n = 20 \).

Standard Error \( = \frac{0.16}{\sqrt{20}} \).