4 David, a zoologist, is investigating a particular species of monitor lizard. He measures the lengths, in centimetres, of a random sample of this particular species of lizard. His measured lengths are $$\begin{array} { l l l l l l l l l l } 53.2 & 57.8 & 55.3 & 58.9 & 59.0 & 60.2 & 61.8 & 62.3 & 65.4 & 66.5 \end{array}$$ The lengths may be assumed to be normally distributed.
David correctly constructed a 90\% confidence interval for the mean length of lizard using the measured lengths given and the formula \(\bar { x } \pm \left( b \times \frac { s } { \sqrt { n } } \right)\) This interval had limits of 57.63 and 62.45, correct to two decimal places.
4

1. State the value for \(b\) used in David's formula. 4
2. David interprets his interval and states,
   "My confidence interval indicates that exactly 90\% of the population of lizard lengths for this particular species lies between 57.63 cm and \(62.45 \mathrm {~cm} ^ { \prime \prime }\). Do you think David's statement is true? Explain your reasoning. 4
3. David's assistant, Amina, correctly constructs a \(\beta \%\) confidence interval from David's random sample of measured lengths. Amina informs David that the width of her confidence interval is 8.54 .
   Find the value of \(\beta\).
   [0pt] [3 marks]
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