In a survey, a random sample of 250 adults in Fromleigh were asked to fill in a questionnaire about their travel. The survey included a question about the amount, $$$x$$$ dollars, spent on travel per year. The results are summarised as follows. $$\[ n=250 \quad \Sigma x=50460 \quad \Sigma x^{2}=19854200 \]$$ Find unbiased estimates of the population mean and variance of the amount spent per year on travel. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mathematics

IGCSE&ALevel

CAIE

Exam No：9709_w20_qp_63 Year：2020 Question No：2(b)

Answer:

Estimate of mean $\left(\frac{50460}{250}\right)=\$ 201.84$
$\frac{250}{249}\left(\frac{19854200}{250}-\left(\frac{50460}{250}\right)^{2}\right) \text { or } \frac{1}{249}\left(19854200-\frac{50460^{2}}{250}\right)$
Estimate of variance $=38832.75$ dollars $^{2}$ or $38800(3 \mathrm{sf})$

Knowledge points:

6.4.6 calculate unbiased estimates of the population mean and variance from a sample, using either raw or summarised data (Only a simple understanding of the term ‘unbiased’ is required, e.g. that although individual estimates will vary the process gives an accurate result ‘on average’.)

Answer:

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