Respuesta :
Answer:
(1) The mean (or expected value) of the sample mean `X = 13.35
(2) The standard deviation of the sample mean `X = .02
(3) The probability that the sample mean `X is less than or equal to 13.38 OZ = .9332
(4) The probability that the sample mean `X is less than or equal to 13.32 OZ = .0668
(5) The probability that the sample mean `X is between 13.30 and 13.36 OZ = .6853
Step-by-step explanation:
mean [tex](\nu)[/tex] =13.35
Standard deviation [tex](\sigma )[/tex] = .1200
n = 36
The mean (or expected value) of the sample mean `X
[tex]E(\overline{X}) = \nu _{\overline{X}} = \nu[/tex] = 13.35
The standard deviation of the sample mean `X =
[tex]\sigma _{\overline{X}} = \frac{\sigma}{\sqrt{n}}[/tex] = [tex]\frac{.1200}{\sqrt{36}}[/tex] = .02
The probability that the sample mean `X is less than or equal to 13.38 OZ=
[tex]P(\overline{X} \leq 13.38 )[/tex] = [tex]P(\frac{\overline{X} - \nu }{\frac{\sigma}{\sqrt{n}}}\leq \frac{13.38 - 13.35 }{\frac{.1200}{\sqrt{36}}})[/tex] [ Z = [tex]\frac{\overline{X} - \nu }{\frac{\sigma}{\sqrt{n}}}[/tex] ]
= [tex]P(Z\leq 1.5)[/tex] Using Z table
= .9332
. The probability that the sample mean `X is less than or equal to 13.32 OZ =
[tex]P(\overline{X} \leq 13.32)[/tex] = [tex]P(\frac{\overline{X} - \nu }{\frac{\sigma}{\sqrt{n}}}\leq \frac{13.32 - 13.35 }{\frac{.1200}{\sqrt{36}}})[/tex]
= [tex]P(Z\leq -1.5)[/tex]
= .0668
The probability that the sample mean `X is between 13.30 and 13.36 OZ =
[tex]P(13.30\leq \overline{X} \leq 13.36)[/tex] = [tex]P(\frac{13.30 - 13.35 }{\frac{.1200}{\sqrt{36}}})\leq \frac{\overline{X} - \nu }{\frac{\sigma}{\sqrt{n}}}\leq \frac{13.36 - 13.35 }{\frac{.1200}{\sqrt{36}}})[/tex]
= [tex]P(- 2.5\leq Z\leq 0.5)[/tex]
= .6915 - .0062
= .6853
