Answer:
a) 0.0082
b) 0.9987
c) 0.9192
d) 0.5000
e) 1
Step-by-step explanation:
The question is concerned with the mean of a sample.
From the central limit theorem we have the formula:
[tex]z=\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex]
a) [tex]z=\frac{1224-1200}{\frac{60}{\sqrt{36} } }=2.40[/tex]
The area to the left of z=2.40 is 0.9918
The area to the right of z=2.40 is 1-0.9918=0.0082
[tex]\therefore P(\bar X\:>\:1224)=0.0082[/tex]
b) [tex]z=\frac{1230-1200}{\frac{60}{\sqrt{36} } }=3.00[/tex]
The area to the left of z=3.00 is 0.9987
[tex]\therefore P(\bar X\:<\:1230)=0.9987[/tex]
c) The z-value of 1200 is 0
The area to the left of 0 is 0.5
[tex]z=\frac{1214-1200}{\frac{60}{\sqrt{36} } }=1.40[/tex]
The area to the left of z=1.40 is 0.9192
The probability that the sample mean is between 1200 and 1214 is
[tex]P(1200\:<\:\bar X\:<\:1214)=0.9192-0.5000=0.4192[/tex]
d) From c) the probability that the sample mean will be greater than 1200 is 1-0.5000=0.5000
e) [tex]z=\frac{73.46-1200}{\frac{60}{\sqrt{36} } }=-112.65[/tex]
The area to the left of z=-112.65 is 0.
The area to the right of z=-112.65 is 1-0=1