Edexcel S1 2003 June — Question 2 6 marks

Exam BoardEdexcel
ModuleS1 (Statistics 1)
Year2003
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicNormal Distribution
TypeFind k given probability statement
DifficultyModerate -0.5 This is a straightforward inverse normal problem requiring students to find a value given a probability (P(X > t) = 0.2). It involves standard bookwork: finding the z-score for the 80th percentile and applying the transformation t = μ + zσ. While it requires understanding of the normal distribution, it's a routine S1 exercise with no problem-solving complexity.
Spec2.04e Normal distribution: as model N(mu, sigma^2)2.04f Find normal probabilities: Z transformation
2. The lifetimes of batteries used for a computer game have a mean of 12 hours and a standard deviation of 3 hours. Battery lifetimes may be assumed to be normally distributed. Find the lifetime, \(t\) hours, of a battery such that 1 battery in 5 will have a lifetime longer than \(t\).
AnswerMarks Guidance
SchemeMarks Guidance
\(P(L > t) = 0.2\)M1
\(\therefore \frac{t-12}{3} = 0.8416\)M1 B1 A1
\(\therefore t = 14.5248\)A1 (6 marks)
Alt:
AnswerMarks Guidance
\(P(L > t) = 0.2 \therefore P(L \leq t) = 0.8\)M1
\(\therefore \frac{t-12}{3} = 0.84(16)\)B1 A1
\(\therefore t = 14.52(54)\)M1 A1 (6 marks) 
| Scheme | Marks | Guidance |
|--------|-------|----------|
| $P(L > t) = 0.2$ | M1 | |
| $\therefore \frac{t-12}{3} = 0.8416$ | M1 B1 A1 | |
| $\therefore t = 14.5248$ | A1 | (6 marks) |

**Alt:** 
| $P(L > t) = 0.2 \therefore P(L \leq t) = 0.8$ | M1 | |
| $\therefore \frac{t-12}{3} = 0.84(16)$ | B1 A1 | |
| $\therefore t = 14.52(54)$ | M1 A1 | (6 marks) |

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2. The lifetimes of batteries used for a computer game have a mean of 12 hours and a standard deviation of 3 hours. Battery lifetimes may be assumed to be normally distributed.

Find the lifetime, $t$ hours, of a battery such that 1 battery in 5 will have a lifetime longer than $t$.\\

\hfill \mbox{\textit{Edexcel S1 2003 Q2 [6]}}
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