Respuesta :
Solution :
Let [tex]$x_i$[/tex] represents the prices for the digital camera.
[tex]$x_i$[/tex] [tex]$x_i^2$[/tex]
225 50625
236 55696
217 47089
202 40804
231 53361
185 35225
193 37249
185 34225
217 47089
[tex]$\sum x_i = 2136$[/tex] [tex]$\sum x_i^2 = 460388$[/tex]
Mean, [tex]$(\overline X) = \frac{\sum x_i}{n}$[/tex]
[tex]$=\frac{2136}{10}$[/tex]
= 213.6
Variance, [tex]$\sigma^2 = \frac{\sum x_i^2}{n}-(\overline X)^2$[/tex]
[tex]$ = \frac{460388}{10}-(213.6)^2$[/tex]
= 46038.8 - 45624.96
= 413.84
Therefore, standard deviation,
[tex]$\sigma = \sqrt{Var}$[/tex]
[tex]$=\sqrt{413.84}$[/tex]
= 20.343
Now to find the 93% confidence interval, the formula us
[tex]$(\overline X)\pm (z) \times \frac{\sigma}{\sqrt{n}}$[/tex]
For 93%, the value of z is 1.81
∴ Confidence interval = [tex]$(\overline X)\pm (z) \times \frac{\sigma}{\sqrt{n}}$[/tex]
C.I. = [tex]$213.6\pm 1.81 \times \frac{20.343}{\sqrt{10}}$[/tex]
[tex]$=213.6 \pm 11.6438$[/tex]
Therefore, lower bound = 213.6 - 11.6438
= 201.9562
Upper bound = 213.6 + 11.6438
= 225.2438
∴ [tex]$201.9562 < \mu < 225.2438$[/tex]
or [tex]$202< \mu < 225.2$[/tex]
