Answer:
a) x = 80.40
s = 29.54
b) lower limit = 69.53
upper limit = 91.27
Step-by-step explanation:
The mean price is:
[tex]\bar X=\frac{1}{n}\sum x_i=\frac{1}{20}(50+80+...+70)=\frac{1608}{20} =80.40[/tex]
The standard deviation is
[tex]s=\sqrt{\frac{1}{n-1}\sum(x_i-\bar X)^2}=\sqrt{\frac{1}{19}[(50-80.4)^2+...+(70-80.4)^2]} \\\\s=\sqrt{\frac{15,471}{19} }=\sqrt{814.25} =29.54[/tex]
We have to calculate a 90% CI for the mean.
The z-value for a 90% confidence interval is z=1.645.
The margin of error is:
[tex]E=z\cdot \sigma / \sqrt{n}=1.645*29.54/\sqrt{20}=48.59/4.47=10.87[/tex]
The 90% CI is
[tex]\bar X-z\sigma/\sqrt{n} \leq \mu \leq \bar X-z\sigma/\sqrt{n}\\\\80.40-10.87 \leq \mu \leq 80.40+10.87\\\\69.53\leq \mu \leq 91.27[/tex]
