Edexcel A Level Statistics 2021 (Paper 31)

Any of the following are acceptable:

• It is not random.
• It cannot be used to reliably make inferences about the population.
• It may be biased (e.g., interviewer selection bias).

✓ (B1)

Model: \( X \sim B(36, 0.08) \) ✓ (M1)

(i) Find \( P(X=4) \):

Using calculator Binomial PD:

\[ P(X=4) = \binom{36}{4}(0.08)^4(0.92)^{32} \approx 0.167387… \]

Rounding to 3 s.f.:

\[ P(X=4) = 0.167 \] ✓ (A1)

(ii) Find \( P(X \geqslant 7) \):

Calculators typically compute Cumulative Distribution \( P(X \leqslant x) \). We need to rewrite the inequality:

\[ P(X \geqslant 7) = 1 – P(X \leqslant 6) \]

Using calculator Binomial CD with \( x=6 \):

\[ P(X \leqslant 6) = 0.97776… \] \[ P(X \geqslant 7) = 1 – 0.97776… = 0.02223… \]

Rounding to 3 s.f.:

\[ P(X \geqslant 7) = 0.0222 \] ✓ (A1)

\[ P(\text{Club} \cap \text{Tango}) = 0.08 \times 0.40 \] \[ = 0.032 \]

✓ (B1)

Model: \( T \sim B(50, 0.032) \) ✓ (M1)

We need “fewer than 3”, which means \( T < 3 \) or \( T \leqslant 2 \).

\[ P(T < 3) = P(T \leqslant 2) \]

Using calculator Binomial CD with \( x=2 \):

\[ P(T \leqslant 2) = 0.78508… \]

Rounding to 3 s.f.:

\[ 0.785 \] ✓ (A1)

(a) Negative correlation. ✓ (B1)

(b) Marc’s suggestion is compatible because there is negative correlation (or the points lie close to a line with a negative gradient). ✓ (B1)

Using calculator Statistics Mode (2-Var):

Enter the data points into the lists.

Read off \( r \):

\[ r = -0.54458… \]

Rounding to 3 s.f.:

\[ r = -0.545 \] ✓ (B1)

Hypotheses:

\[ H_0: \rho = 0 \] \[ H_1: \rho < 0 \] ✓ (B1)

(Where \( \rho \) is the population correlation coefficient)

Critical Value:

Sample size \( n = 16 \), Significance Level \( 5\% \) (0.05).

From tables, for \( n=16 \), the critical value is \( 0.4259 \).

Since we are testing for negative correlation, the critical region is \( r < -0.4259 \).

CV \( = -0.4259 \) ✓ (M1)

Comparison:

\[ -0.545 < -0.4259 \]

The test statistic falls in the critical region (it is more negative than the critical value).

Conclusion:

Reject \( H_0 \).

There is evidence of a negative correlation between the number of letters in a student’s last name and their first name.

✓ (A1)

Hectopascal (or hPa)

✓ (B1)

Mean of \( y \):

\[ \bar{y} = \frac{\sum y}{n} = \frac{214}{30} \]

Decoded mean \( \bar{x} \):

\[ \bar{x} = \bar{y} + 1010 \] \[ \bar{x} = \frac{214}{30} + 1010 = 1017.133… \]

Answer:

\[ 1017 \text{ hPa} \] (Accept 1017 or 1017.1) ✓ (A1)

Formula for standard deviation of \( y \):

\[ \sigma_y = \sqrt{\frac{\sum y^2}{n} – (\bar{y})^2} \]

Substitute values:

\[ \sigma_y = \sqrt{\frac{5912}{30} – \left(\frac{214}{30}\right)^2} \] \[ \sigma_y = \sqrt{197.066… – (7.133…)^2} \] \[ \sigma_y = \sqrt{197.066… – 50.88…} \] \[ \sigma_y = \sqrt{146.18…} \approx 12.09 \]

Since \( \sigma_x = \sigma_y \):

\[ \sigma_x = 12.1 \] ✓ (A1)

Reasoning:

• The pressures are high (> mean + SD approx), implying a High Pressure system.
• High pressure systems rotate Clockwise.
• Imagine a clock face over the UK. Wind moves from the “top” (North) towards the “right” (East), then “bottom” (South), then “left” (West).
• Actually, wind blows around the center.
• Leuchars (North) -> Wind direction is from the West?
• Let’s map the locations relative to the center or each other.

Correct Allocation:

Given the mark scheme logic: “High pressure… so clockwise”.

Locations North to South: Leuchars, Heathrow, Hurn.

Wind Direction means “Direction wind blows FROM”.

• Heathrow: NE
• Hurn: E
• Leuchars: W

Table:

Heathrow	NE
Hurn	E
Leuchars	W

✓ (B1)

Sum the probabilities in the “only” regions:

\[ P(\text{exactly one}) = 0.08 + 0.09 + 0.36 = 0.53 \] ✓ (B1)

(i) Value of \( p \):

We are told “No students read all three magazines”. This corresponds to the center intersection region \( G \cap E \cap S \).

\[ p = 0 \] ✓ (B1)

(ii) Value of \( q \):

We are told \( P(G) = 0.25 \). The circle G contains four regions summing to 0.25:

\[ 0.08 + 0.05 + p + q = 0.25 \]

Substitute \( p = 0 \):

\[ 0.13 + q = 0.25 \] \[ q = 0.12 \] ✓ (M1 A1)

(i) Find \( r \):

Identify regions in \( E \): \( 0.05 + 0.09 + p + r \)

\[ P(E) = 0.14 + r \quad (\text{since } p=0) \]

Identify regions in \( S \cap E \): \( p + r = r \)

Set up equation:

\[ \frac{r}{0.14 + r} = \frac{5}{12} \] \[ 12r = 5(0.14 + r) \] \[ 12r = 0.7 + 5r \] \[ 7r = 0.7 \] \[ r = 0.1 \] ✓ (A1)

(ii) Find \( t \):

The sum of all probabilities must be 1.

\[ 0.08 + 0.09 + 0.36 + 0.05 + q + r + p + t = 1 \]

Substitute known values (\( q=0.12, r=0.1, p=0 \)):

\[ 0.08 + 0.09 + 0.36 + 0.05 + 0.12 + 0.1 + 0 + t = 1 \] \[ 0.8 + t = 1 \] \[ t = 0.2 \] ✓ (B1ft)

Let Event A be \( S \cap E’ \) and Event B be \( G \).

1. Find \( P(S \cap E’) \):

This is the part of S that is NOT in E. In the diagram: \( 0.36 + q \).

\[ P(S \cap E’) = 0.36 + 0.12 = 0.48 \]

2. Find \( P(G) \):

\[ P(G) = 0.25 \]

3. Calculate Product \( P(S \cap E’) \times P(G) \):

\[ 0.48 \times 0.25 = 0.12 \]

4. Find Intersection \( P((S \cap E’) \cap G) \):

This is the region inside S, outside E, and inside G.

Looking at the diagram, this is exactly region \( q \).

\[ P((S \cap E’) \cap G) = q = 0.12 \]

Conclusion:

Since \( 0.12 = 0.12 \), the condition \( P(A \cap B) = P(A) \times P(B) \) is satisfied.

The events are independent.

✓ (A1)

Let \( F \sim N(166.5, 6.1^2) \).

We need \( k \) such that \( P(F < k) = 0.01 \).

Using Inverse Normal on calculator:

• Area = 0.01
• \( \mu = 166.5 \)
• \( \sigma = 6.1 \)

\[ k = 152.309… \]

Answer:

\[ k = 152 \text{ cm} \text{ (or 152.3)} \] ✓ (A1)

We calculate \( P(150 < F < 175) \).

Using Normal CD on calculator:

• Lower = 150
• Upper = 175
• \( \mu = 166.5, \sigma = 6.1 \)

\[ P = 0.9148… \] \[ \text{Answer } = 0.915 \] ✓ (B1)

Numerator \( P(160 < F < 175) \):

Using calculator: \( 0.7749… \)

Denominator \( P(150 < F < 175) \):

From part (b): \( 0.9148… \)

Calculate Fraction:

\[ \frac{0.7749…}{0.9148…} = 0.84708… \] \[ = 0.847 \] ✓ (A1)

Hypotheses:

\[ H_0: \mu = 166.5 \] \[ H_1: \mu < 166.5 \] ✓ (B1)

Test Statistic \( Z \):

Sample mean \( \bar{x} = 164.6 \), \( n=50 \), \( \sigma=7.4 \).

\[ Z = \frac{\bar{x} – \mu}{\frac{\sigma}{\sqrt{n}}} = \frac{164.6 – 166.5}{\frac{7.4}{\sqrt{50}}} \] \[ Z = \frac{-1.9}{1.046…} = -1.815… \]

Critical Region:

For 5% significance level (one-tail), \( Z_{crit} = -1.645 \).

Since \( -1.815 < -1.645 \), the result is significant.

Or using p-value:

\[ P(\bar{X} < 164.6) = 0.0347... \]

Since \( 0.0347 < 0.05 \), reject \( H_0 \).

Conclusion:

Reject \( H_0 \).

There is sufficient evidence to support Mia’s belief that the mean height is less than 166.5 cm.

✓ (A1)

Step 1: Use the sum of probabilities

\[ \sum P(X=x) = 1 \] \[ \log_{36} a + \log_{36} b + \log_{36} c = 1 \]

Using log laws (\( \log x + \log y = \log xy \)):

\[ \log_{36} (abc) = 1 \]

Convert from log form to index form:

\[ abc = 36^1 \] \[ abc = 36 \] ✓ (M1 A1)

Step 2: Find distinct integers

We need three distinct integers \( a < b < c \) that multiply to 36.

Also, since \( P(X=x) > 0 \), we know \( \log_{36}x > 0 \Rightarrow x > 1 \).

Let’s list factor combinations of 36 (excluding 1):

• 2, 2, 9 -> Not distinct.
• 2, 3, 6 -> Distinct! Product = 36.
• 3, 3, 4 -> Not distinct.

The only valid set is \( \{2, 3, 6\} \).

Since \( a < b < c \):

\[ a = 2, \quad b = 3, \quad c = 6 \] ✓ (A1)

Calculate individual probabilities:

\[ P(X=2) = \log_{36} 2 \] \[ P(X=3) = \log_{36} 3 \] \[ P(X=6) = \log_{36} 6 = 0.5 \]

Sum of squares:

\[ P(X_1 = X_2) = (\log_{36} 2)^2 + (\log_{36} 3)^2 + (\log_{36} 6)^2 \]

Using calculator:

\[ (\log_{36} 2)^2 \approx (0.193…)^2 \approx 0.0374… \] \[ (\log_{36} 3)^2 \approx (0.306…)^2 \approx 0.0939… \] \[ (\log_{36} 6)^2 = (0.5)^2 = 0.25 \]

Total:

\[ 0.0374… + 0.0939… + 0.25 = 0.3814… \]

Rounding to 3 s.f.:

\[ 0.381 \] ✓ (M1 A1)