X + 2y = 53x + 5y = 14Solve the system of equations.(3.1)(7.-1)(2.3/2)
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Answer: (a) Percentage of 25 year old men that are above 6 feet 2 inches is 11.5%.
(b) Percentage of 25 year old men in the 6 footer club that are above 6 feet 5 inches are 2.4%.
Step-by-step explanation:
Given that,
Height (in inches) of a 25 year old man is a normal random variable with mean and variance
.
To find: (a) What percentage of 25 year old men are 6 feet, 2 inches tall
(b) What percentage of 25 year old men in the 6 footer club are over 6 feet. 5 inches.
Now,
(a) To calculate the percentage of men, we have to calculate the probability
P[Height of a 25 year old man is over 6 feet 2 inches]= P[X>]
P[X>74] = P[ >
]
= P[Z > 1.2]
= 1 - P[Z ≤ 1.2]
= 1 - Ф (1.2)
= 1 - 0.8849
= 0.1151
Thus, percentage of 25 year old men that are above 6 feet 2 inches is 11.5%.
(b) P[Height of 25 year old man is above 6 feet 5 inches gives that he is above 6 feet] = P[X,
- X,
]
P[X >
I X >
] = P[X > 77 I X > 72]
=
=
=
=
=
=
= 0.024
Thus, Percentage of 25 year old men in the 6 footer club that are above 6 feet 5 inches are 2.4%.
Let be the randomly selected lengths. Without loss of generality, suppose
where are independent random variables with the same uniform distribution on [0, 1].
By their mutual independence, we have
so that the joint density function is
where is the cube with vertices at (0, 0, 0) and (1, 1, 1).
Consider the plane
with . This plane passes through (0, 0, 0), (1, 0, 1), and (0, 1, 1), and thus splits up the cube into one tetrahedral region above the plane and the rest of the cube under it. (see attached plot)
The point (0, 0, 1) (the vertex of the cube above the plane) does not belong the region , since
. So the probability we want is the volume of the bottom "half" of the cube. We could integrate the joint density over this set, but integrating over the complement is simpler since it's a tetrahedron.
Then we have
