Answer :
P(X ≤ 5) is 0.5102, b) P(4.5 ≤ X ≤ 5) is 0.09694, c) P(X>5) is 0.3827, d) the median is 4.9497, e) the density function is 0 , x<0 f(x) = 2x/49 , 0 ≤ x ≤7 0, 7≤ x f) the variance and standard deviation are 2.7222 and 1.64991 respectively, while h) E(h(x)) is 24.5.
Here it is given that the CDF of the distribution is
0 , x<0
F(x) = x²/49 , 0 ≤ x ≤ 7
1, 7≤ x
a)
We need to find P(X ≤ 5)
Since we will have to sum up all values from (X=0) to P(X ≤ 5)
This implies we need to find F(X=5)
This can be easily found in the formulae where we get
F(X=5) = 5²/49
= 0.5102
b) To calculate P(4.5 ≤ X ≤ 5)
We can calculate the following by subtracting F(X=4.5) from F(X=5)
= 25/29 - 4.5²/49
= 4.75/49
= 0.09694
c)
P(X>5.5)
= 1 - P(X≤5.5)
= 1 - F(X = 5.5)
= 1 - 5.5²/49
= 0.3827
d)
Let the median be x.
A median divides the cumulative frequency into 2 equal halves.
Therefore, the value of F(X=x) = 0.5
=> x²/49 = 0.5
or, x² = 24.5
or, x = 4.9497
e)
To obtain the density function, f(x0 we must differentiate the cumulative distribution function with respect to x.
Doing so will give us
0 , x<0
f(x) = 2x/49 , 0 ≤ x ≤ 7
0, 7≤ x
f)
Since f(x) is clearly a continuous probability distribution function
E(X) = ∫xf(x)dx with limits 0 to 7
= ∫2x²/49dx with limits 0 to 7
= 2x³/147 with limits 0 to 7
Applying the limits gives us
686/147
= 4.6667
g)
V(X) = variance of f(x)
= E(X²) - E²(X)
E(X²)
= ∫2x³/49dx with limits 0 and 7
= x⁴/98 with limits 0 to 7
= 24.5
Hence
V(X) = 24.5 - 21.7778
= 2.7222
σ(x) = √V(X)
= 1.64991
h)
Now here instead of being charged X amount i.e x²/49, the borrower is being charged X² amount which is (x²/49)².
We need to compute the expected charge E[h(X)]
= E(X²)
This we already know is 24.5
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