Lengths of a certain species of lizard are known to be normally distributed with standard deviation $$\(3.2 \mathrm{~cm}\)$$. A naturalist measures the lengths of a random sample of 100 lizards of this species and obtains an $$\(\alpha \%\)$$ confidence interval for the population mean. He finds that the total width of this interval is $$\(1.25 \mathrm{~cm}\)$$. Find $$\(\alpha\)$$. ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................ ................................................................................................................................................................
Exam No：9709_m20_qp_62 Year：2020 Question No：2
Answer:
\(2 \times z \times \frac{3.2}{10}=1.25\)
\(z=1.953\)
\(\phi(\) 'their 1.953') (=0.9746)
\(=1-2(1-6.9746\) ' \()\)
\(=0.9492\)
\(\alpha=94.9\) or 95
\(z=1.953\)
\(\phi(\) 'their 1.953') (=0.9746)
\(=1-2(1-6.9746\) ' \()\)
\(=0.9492\)
\(\alpha=94.9\) or 95
Knowledge points:
6.4.7 determine and interpret a confidence interval for a population mean in cases where the population is normally distributed with known variance or where a large sample is used
Solution:
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