Statistical Reasoning Flashcards
6 cards from real Civil Service Numerical Reasoning Test practice questions. Tap to flip, then mark Knew It or Still Learning — missed cards come back until you master them.
Read the first 6 Statistical Reasoning flashcards as text
A dataset shows annual staff turnover rates for 8 years: 5.2%, 6.1%, 4.8%, 7.3%, 5.5%, 6.8%, 5.0%, 6.3%. What is the interquartile range (IQR)?
Answer: 1.45 percentage points
Sorted: 4.8, 5.0, 5.2, 5.5, 6.1, 6.3, 6.8, 7.3. Q1 = average of 2nd and 3rd values: (5.0+5.2)/2 = 5.1. Q3 = average of 6th and 7th values: (6.3+6.8)/2 = 6.55. IQR = 6.55 minus 5.1 = 1.45 percentage points.
A correlation coefficient of r = -0.82 is found between staff absence rates and team productivity scores. Which statement best describes this relationship?
Answer: Strong negative correlation: as absence rises, productivity falls
An r value of -0.82 indicates a strong negative correlation. The negative sign means the variables move in opposite directions: as absence increases, productivity decreases. Values close to -1 indicate strong negative relationships.
A policy proposal is tested with two groups. Group A (n=200) shows a 15% improvement. Group B (n=800) shows a 3% improvement. What is the overall weighted improvement across both groups?
Answer: 5.4%
Weighted average = [(200 x 15%) + (800 x 3%)] / (200 + 800) = (30 + 24) / 1000 = 54 / 1000 = 5.4%. The larger group's lower improvement pulls the overall figure down significantly.
A moving average of quarterly benefit claims uses a 4-quarter window. The claims (in thousands) are: Q1: 320, Q2: 345, Q3: 380, Q4: 310, Q5: 355. What is the 4-quarter moving average for the period Q1 to Q4?
Answer: 338.75
The 4-quarter moving average for Q1 to Q4: (320 + 345 + 380 + 310) / 4 = 1,355 / 4 = 338.75 thousand.
A department's error rate follows a Poisson distribution with a mean of 3 errors per day. What is the probability of exactly 0 errors on a given day? (Use e to the power of -3 = 0.0498)
Answer: 4.98%
For a Poisson distribution, P(X=0) = e^(-lambda) x lambda^0 / 0! = e^(-3) x 1 / 1 = 0.0498 = 4.98%. When X equals zero, the formula simplifies to just e raised to the negative mean.
A time series shows seasonal indices for passport applications: Spring 1.15, Summer 1.30, Autumn 0.85, Winter 0.70. The deseasonalised figure for Summer is 450,000. What is the actual (raw) figure?
Answer: 585,000
Actual = Deseasonalised x Seasonal Index = 450,000 x 1.30 = 585,000. To reverse deseasonalisation, multiply by the seasonal index.