The continuous random variable X has probability density function f given by
$$
f(x) = \begin{cases}
\frac{x}{2} & 2 \leq x \leq 4 \\
0 & \text{otherwise}
\end{cases}$$
The random variable Y is defined by Y = X^3 - CIE - A-Level Further Maths - Question 8 - 2013 - Paper 1
Question 8
The continuous random variable X has probability density function f given by
$$
f(x) = \begin{cases}
\frac{x}{2} & 2 \leq x \leq 4 \\
0 & \text{otherwise}
\end{case... show full transcript
Worked Solution & Example Answer:The continuous random variable X has probability density function f given by
$$
f(x) = \begin{cases}
\frac{x}{2} & 2 \leq x \leq 4 \\
0 & \text{otherwise}
\end{cases}$$
The random variable Y is defined by Y = X^3 - CIE - A-Level Further Maths - Question 8 - 2013 - Paper 1
Step 1
Show that Y has probability density function g given by
96%
114 rated
Only available for registered users.
Sign up now to view full answer, or log in if you already have an account!
Answer
To find the probability density function g(y), we first need to identify the cumulative distribution function F(y) of Y. Since Y = X^3, we want to express the variables in terms of Y:
Change of Variables: For a given value of y, if Y = X^3, then we have X = Y^{1/3}.
This change of variable leads to:
The bounds for Y are 8 to 64 since if X is between 2 and 4, then Y must be between (2^3 = 8) and (4^3 = 64).
Find the cumulative distribution function F(y):
Thus,
F(y)=P(Y≤y)=P(X3≤y)=P(X≤y1/3)
We can calculate this using the given density function of X:
Finding g(y):
The probability density function is given by differentiating F(y):
g(y)=dyd​F(y)=dyd​(4y2/3​−1)=41​⋅32​y−1/3=61​y−1/3
This shows that g(y) is valid for 8 \leq y \leq 64 and is 0 otherwise. Thus, we get:
Sign up now to view full answer, or log in if you already have an account!
Answer
To find E(Y), we can utilize the formula for expected value based on the density function g(y):
E(Y)=∫ab​yg(y)dy
for the limits (a = 8) and (b = 64).
Substitute g(y) into the expected value equation:
E(Y)=∫864​y⋅181​y31​dy
which simplifies to:
E(Y)=181​∫864​y34​dy
Evaluate the integral:
E(Y)=181​[37​y37​​]864​=1263​[6437​−837​]
where, (64^{\frac{7}{3}} = 64^2 = 4096) and (8^{\frac{7}{3}} = 8^{2+\frac{1}{3}} = 64 \cdot 2 = 128).
Substitute these values back:
E(Y)=1263​(4096−128)=1263​⋅3968=12611904​=3992​=330.67
Therefore, the expected value is ( E(Y) \approx 330.67 ).
