Edexcel S3 2005 June — Question 6 10 marks

Exam BoardEdexcel
ModuleS3 (Statistics 3)
Year2005
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConfidence intervals
TypeFind minimum sample size
DifficultyStandard +0.3 This is a straightforward S3 confidence interval question requiring standard formulas: part (a) uses basic sample mean and variance calculations (unbiased estimator with n-1), while part (b) applies the standard sample size formula n = (z*σ/E)² with given values. Both parts are direct applications of textbook methods with no problem-solving or novel insight required, making it slightly easier than average.
Spec5.05b Unbiased estimates: of population mean and variance5.05c Hypothesis test: normal distribution for population mean

A computer company repairs large numbers of PCs and wants to estimate the mean time to repair a particular fault. Five repairs are chosen at random from the company's records and the times taken, in seconds, are 205 \quad 310 \quad 405 \quad 195 \quad 320.
  1. Calculate unbiased estimates of the mean and the variance of the population of repair times from which this sample has been taken. [4]
It is known from previous results that the standard deviation of the repair time for this fault is 100 seconds. The company manager wants to ensure that there is a probability of at least 0.95 that the estimate of the population mean lies within 20 seconds of its true value.
  1. Find the minimum sample size required. [6]
(Total 10 marks)

AnswerMarks Guidance
ContentMarks Guidance
(a) Let \(x\) represent slider time. \(\sum w = 1433\) \(\bar{x} = \frac{1433}{5} = 286.7\) \(\sum x^2 = 442375\) \(S^2 = \frac{1}{4}\left\{442375 - \frac{1433^2}{5}\right\} = 7662.5\)B1, M1, A1 (4)
(b) \(P(μ - \bar{x} < 2σ = 0.95\) \(\therefore \frac{2σ}{\sqrt{n}} = 1.96\) \(\therefore n = 1.96^2 = 1.96 \times 100^2 = 96.04\) Sample size (\(>\))97 required
| Content | Marks | Guidance |
|---------|-------|----------|
| **(a)** Let $x$ represent slider time. $\sum w = 1433$ $\bar{x} = \frac{1433}{5} = 286.7$ $\sum x^2 = 442375$ $S^2 = \frac{1}{4}\left\{442375 - \frac{1433^2}{5}\right\} = 7662.5$ | B1, M1, A1 (4) | |
| **(b)** $P(|μ - \bar{x}| < 2σ = 0.95$ $\therefore \frac{2σ}{\sqrt{n}} = 1.96$ $\therefore n = 1.96^2 = 1.96 \times 100^2 = 96.04$ **Sample size** ($>$)97 required | M1, B1, A1, M1, A1 (6) | Solving for n |

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A computer company repairs large numbers of PCs and wants to estimate the mean time to repair a particular fault. Five repairs are chosen at random from the company's records and the times taken, in seconds, are

205 \quad 310 \quad 405 \quad 195 \quad 320.

\begin{enumerate}[label=(\alph*)]
\item Calculate unbiased estimates of the mean and the variance of the population of repair times from which this sample has been taken. [4]
\end{enumerate}

It is known from previous results that the standard deviation of the repair time for this fault is 100 seconds. The company manager wants to ensure that there is a probability of at least 0.95 that the estimate of the population mean lies within 20 seconds of its true value.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the minimum sample size required. [6]
\end{enumerate}
(Total 10 marks)

\hfill \mbox{\textit{Edexcel S3 2005 Q6 [10]}}