Answer: a) There is a 99% likelihood that they will sell between 127 and 153 units.
b) There is a 95% likelihood that they will sell between 130 and 150 units.
c) There is a 68% likelihood that they will sell between 135 and 145 units.
Step-by-step explanation:
Given : Sample size : n= 100
Sample mean : [tex]\overline{x}=140[/tex]
Standard deviation : [tex]\sigma=50[/tex]
a) Significance level : [tex]\alpha: 1-0.99=0.01[/tex]
Critical value: [tex]z_{\alpha/2}=2.576[/tex]
The formula to find the confidence interval is given by :-
[tex]\overline{x}\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}[/tex]
i.e. [tex]140\pm (2.576)\dfrac{50}{\sqrt{100}}[/tex]
i.e. [tex]140\pm 12.88[/tex]
[tex]=(140-12.88,140+12.88)=(127.12,\ 152.88)\approx(127,\ 153)[/tex]
So, There is a 99% likelihood that they will sell between 127 and 153 units.
b) Significance level : [tex]\alpha: 1-0.95=0.05[/tex]
Critical value: [tex]z_{\alpha/2}=1.96[/tex]
then, 95% confidence interval will be :
i.e. [tex]140\pm (1.96)\dfrac{50}{\sqrt{100}}[/tex]
i.e. [tex]140\pm 12.88[/tex]
[tex]=(140-9.8,140+9.8)=(130.2,\ 149.8)\approx(130,\ 150)[/tex]
So, There is a 95% likelihood that they will sell between 130 and 150 units.
c) Significance level : [tex]\alpha: 1-0.68=0.32[/tex]
Critical value: [tex]z_{\alpha/2}=0.9945[/tex]
then, 95% confidence interval will be :
i.e. [tex]140\pm (0.9945)\dfrac{50}{\sqrt{100}}[/tex]
i.e. [tex]140\pm 4.9725[/tex]
[tex]=(140-4.9725,140+4.9725)=(135.0275,\ 144.9725)\approx(135,\ 145)[/tex]
So, There is a 68% likelihood that they will sell between 135 and 145 units.
