E9a — AQA GCSE Statistics

Test yourself on E9a with AQA GCSE practice questions.

7 days Premium · Then free forever · No card, no charge

Your focus

Interpret given Spearman’s rank correlation coefficient in the context of the problem.

E9a exam tips

Quick Revision Summary (Key Takeaway)

E9a covers standardised scores in GCSE Statistics, which measure how many standard deviations a raw data value lies above or below the distribution mean. Calculated using (value - mean) / standard deviation, they enable fair comparisons across datasets with different scales and spreads.

How Students Lose Marks (Examiner Pitfalls)

Pitfall: Failing to account for context when evaluating timed sporting events where smaller raw values represent better performances.

❌ Weak Answer (Loses Marks):Athlete A was better because their standardised score was +1.5 which is bigger than Athlete B who scored -0.8.

Example improved answer:Athlete B performed relatively better. In running events, a lower time indicates a faster and superior performance. Athlete B finished 0.8 standard deviations faster than the group mean (z = -0.8), whereas Athlete A finished 1.5 standard deviations slower than the group mean (z = +1.5).

Examiner Tip: Always check the variable context before concluding whether a positive or negative standardised score represents a superior outcome.

Pitfall: Making algebraic sign errors when rearranging the formula to find a raw score from a negative standardised score.

❌ Weak Answer (Loses Marks):x = 60 + 1.2 * 5 = 66, because the minus sign was discarded.

Example improved answer:Rearranging the formula gives x = mean + (z * standard deviation). Substituting z = -1.2, mean = 60, and standard deviation = 5 yields x = 60 + (-1.2 * 5) = 60 - 6 = 54.

Examiner Tip: Whenever the standardised score is negative, carry out an immediate sanity check to ensure your final raw score is strictly lower than the distribution mean.

Step-by-Step Worked Solutions

Question: Maya sits two examinations. In French, she scores 76 where the mean is 68 and the standard deviation is 5. In History, she scores 81 where the mean is 73 and the standard deviation is 6. By calculating standardised scores for both subjects, determine in which examination Maya achieved a relatively better performance.

1.Step 1: State the standardised score formula: z = (x - mean) / standard deviation.
2.Step 2: Calculate the standardised score for French: z_French = (76 - 68) / 5 = 8 / 5 = +1.60.
3.Step 3: Calculate the standardised score for History: z_History = (81 - 73) / 6 = 8 / 6 = +1.33 (to 2 d.p.).
4.Step 4: Compare the two standardised scores and state the final contextual conclusion: +1.60 > +1.33, meaning Maya scored further above the cohort average in French than in History.

Final Answer: Maya performed relatively better in French (z = +1.60) compared to History (z = +1.33).

Question: The heights of adult males in a region are normally distributed with a mean of 175 cm and a standard deviation of 8 cm. A participant in a medical trial has a standardised height score of -1.75. Calculate the actual height of this participant in centimetres.

1.Step 1: Identify the given values from the question: mean = 175, standard deviation = 8, z = -1.75.
2.Step 2: Set up the formula: -1.75 = (x - 175) / 8.
3.Step 3: Multiply both sides by the standard deviation: -1.75 * 8 = -14.
4.Step 4: Solve for the raw score x: x = 175 - 14 = 161 cm.

Final Answer: The participant has an actual height of 161 cm.

Your focus

E9a exam tips

More AQA GCSE Statistics topics

Cookies on MasteryMind