Q171 · Practice questionTechnology-activeComplex familiar6 marks
QUESTION 171 (6 marks)
Measurements are normally distributed with unknown mean and known standard deviation mm. Independent measurements are selected at random. A sample of size has mean .
a)[3 marks]
For , state the distribution of using variance notation and determine to four decimal places.
b)[3 marks]
Determine the least sample size for which . Justify why it is the least integer.
WORKED SOLUTION
6 marksPractice marking scheme
ANSWER
(a) For , and . (b) Least sample size .
Worked solution
(a) The sample mean is normal with mean and variance , hence standard deviation 2 mm. Standardising,
(b) For a general , the probability is
For this to be at least 0.95, . Thus
The least integer is 139. The achieved probabilities at and are approximately 0.9498 and 0.9506 respectively, confirming minimality. The population mean need not be known to design the precision about it. Normality of the individual measurements makes the sample-mean distribution exact.
Equivalent mathematically justified methods accepted.
Exact answers unless a decimal accuracy is specified.
Part a: State a normal distribution with mean mu and variance 4.
Part a: Standardise the two-sided event.
Part a: Obtain 0.6827.
Part b: Use the 0.975 normal quantile in the precision inequality.
Part b: Obtain the threshold approximately 138.2925 and round up to 139.
Part b: Justify minimality using the threshold or adjacent probabilities.
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