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June 2012 p61 q6
3259
The lengths of body feathers of a particular species of bird are modelled by a normal distribution. A researcher measures the lengths of a random sample of 600 body feathers from birds of this species and finds that 63 are less than 6 cm long and 155 are more than 12 cm long.
(i) Find estimates of the mean and standard deviation of the lengths of body feathers of birds of this species.
(ii) In a random sample of 1000 body feathers from birds of this species, how many would the researcher expect to find with lengths more than 1 standard deviation from the mean?
Solution
(i) We use the standard normal distribution to find the mean \(\mu\) and standard deviation \(\sigma\).
For lengths less than 6 cm: \(Z = \frac{6 - \mu}{\sigma} = -1.253\)
For lengths more than 12 cm: \(Z = \frac{12 - \mu}{\sigma} = 0.648\)
Solving these equations simultaneously gives:
\(\mu = 9.9\)
\(\sigma = 3.15 \text{ or } 3.16\)
(ii) We need \(P(Z < -1 \text{ or } Z > 1)\).
This is equivalent to \(1 - \Phi(1) + \Phi(-1)\).
\(= 2 - 2 \times 0.8413\)
\(= 0.3174\)
In a sample of 1000, the expected number is \(0.3174 \times 1000 = 317\).