Practise the method.
Four linked problems drawn from this topic’s subtopics. 12 parts · untimed · answers and worked solutions at the end.
Example 1
Arrival time X, in minutes after opening, has density f(x)= kx for 0≤x≤3,f(x)=k(6−x) for 3<x≤6, and 0 otherwise. A visitor is on time if 1.5<X<3.75. The graph gives the unnormalised shape kf(x).
Find k exactly.
Show worked answer
Answer: 91
- The area under f is a triangle of base 6 and height 3k.
- 1=(21)×6×3k=9k, so k=91.
Example 2
Fill volume X, in millilitres, is normally distributed. Its mean μ and standard deviation σ are unknown. The calibration table gives observed quantile values and their exact corresponding standard-normal scores z=σx−μ.
a Find σ, in millilitres.
Find P(X>50). Give your answer to 4 decimal places.
| Volume x(mL) | Corresponding z |
|---|---|
| 42.5 | −1.5 |
| 55 | 1 |
Show worked answer
Answer: 0.5000
- Using μ=50 and σ=5,z=550−50=0.
- P(X>50)=1−Φ(0)=0.5.
Example 3
A component dimension X is normal with mean 140mm and unknown standard deviation σ. The calibration statement is P(135<X<145)=2Φ(1)−1(approximately 0.682689), where Φ is the standard-normal CDF. The specification interval is 130<X<147.5mm. Treat the expression using Φ as exact.
a Find σ, in millimetres.
b Find the probability that a component is outside the specification interval. Give your answer to 4 decimal places.
A new process keeps mean 140mm. Find the largest standard deviation s for which at least 90% of components meet the same specification interval. Give s in millimetres. Give your answer to 3 decimal places.
| Quantity | Value (mm) |
|---|---|
| Mean | 140 |
| Calibration lower bound | 135 |
| Calibration upper bound | 145 |
| Specification lower bound | 130 |
| Specification upper bound | 147.5 |
Show worked answer
Answer: 5.170
- For spread s, the probability inside the fixed limits is Φ(s7.5)−Φ(−s10).
- As s increases, this probability decreases. Solve Φ(s7.5)−Φ(−s10)=0.9.
- The largest spread is s=5.1703713878mm(before rounding).