Question

there are a total of 38 macaw parrots at a Disney park they weigh on average u= 2.3 pounds with Q= .20 pounds assume the weight of macaw parrots is normally distributed Construct a 90% confidence interval for the macaw parrots

Answer 1

Given data

*The given mean is u = 2.3 pounds

*The total number of macaw parts is n = 38

*The given standard deviation is Q = 0.20 Pounds

Z-score for 90% confidence interval is Z = 1.645

The formula for the confidence interval is given as

$s=u\pm\frac{zQ}{\sqrt[]{n}}$

Substitute the values in the above expression as

$\begin{gathered} s=2.3\pm\frac{1.645\times0.20}{\sqrt[]{38}} \\ =2.3\pm0.05337 \\ =2.24,2.35\text{ pounds} \end{gathered}$