X is a random variable with distribution Po(2.90). A random sample of 100 values of X is taken.

Mark scheme

Marking points from the official mark scheme (May/June 2022). The mark or mark type is shown in brackets.

1. [B1] X̄ ~ N(2.9, 2.9/100) OR Totals method N(290, 290) B1 for N(2.9, ...) OR N(290, ...)
2. [B1] B1 for Var = 2.9/100 OR for var = 290 stated or implied
3. [M1] (2.88-2.90)/√(2.9/100) [= -0.1174] OR (288-290)/√290 Standardising with their values Allow without -ve sign AND/OR with incorrect continuity correction No mixed methods
4. [M1] 1- Φ('0.1174') For area consistent with their values
5. [A1] 0.453 (3 sf) As final answer
6. Alternative method for question 7
7. [B1] X̄ ~ N(2.9, 2.9/100) OR Totals method N(290, 290) B1 for N(2.9, ...) OR N(290, ...)
8. [B1] B1 for Var = 2.9/100 OR Var = 290 stated or implied
9. [M1] (2.88-1/200)-2.90/√(2.9/100) [= -0.1468] OR (287.5 – 290)/√290 Standardising with their values Allow without -ve sign AND/OR with incorrect continuity correction No mixed methods
10. [M1] 1 – Φ('0.1468') For area consistent with their values
11. [A1] 0.442 (3 sf) As final answer