Continuous random variable p.d.f. question
The continuous random variable $X$ has p.d.f $f(x)$ given by
$$f(x) = begin{cases}
k (9 - x^2) &text{for } 0 leq x leq 3 \
0 &text{otherwise}
end{cases}$$
Where $k$ is a constant
Show that $k = 1/18$, and find the value of $E(X)$
The textbook says the answer is $3/4$, however, the only answer I can get is $1.125$
My workings:
begin{align*}E(X) &= int_0^3 0.5x - frac{x^3}{18} dx \
&= [0.25x^2 - x^4/72]_3^0 \
&= frac{1}{4} (3)^2 - frac{1}{72} (3)^4 \
&= frac{9}{4} - frac{81}{72} \
&= 1.125
end{align*}
Apologies, I am not sure how to use LaTeX. Any help appreciated.
continuity random-variables expected-value
edited Dec 3 '18 at 0:34
platty
3,370320
add a comment |
The continuous random variable $X$ has p.d.f $f(x)$ given by
$$f(x) = begin{cases}
k (9 - x^2) &text{for } 0 leq x leq 3 \
0 &text{otherwise}
end{cases}$$
Where $k$ is a constant
Show that $k = 1/18$, and find the value of $E(X)$
The textbook says the answer is $3/4$, however, the only answer I can get is $1.125$
My workings:
begin{align*}E(X) &= int_0^3 0.5x - frac{x^3}{18} dx \
&= [0.25x^2 - x^4/72]_3^0 \
&= frac{1}{4} (3)^2 - frac{1}{72} (3)^4 \
&= frac{9}{4} - frac{81}{72} \
&= 1.125
end{align*}
Apologies, I am not sure how to use LaTeX. Any help appreciated.
continuity random-variables expected-value
edited Dec 3 '18 at 0:34
platty
3,370320
$1.125$ is the correct answer.
– Kavi Rama Murthy
Dec 3 '18 at 0:34
add a comment |
The continuous random variable $X$ has p.d.f $f(x)$ given by
$$f(x) = begin{cases}
k (9 - x^2) &text{for } 0 leq x leq 3 \
0 &text{otherwise}
end{cases}$$
Where $k$ is a constant
Show that $k = 1/18$, and find the value of $E(X)$
The textbook says the answer is $3/4$, however, the only answer I can get is $1.125$
My workings:
begin{align*}E(X) &= int_0^3 0.5x - frac{x^3}{18} dx \
&= [0.25x^2 - x^4/72]_3^0 \
&= frac{1}{4} (3)^2 - frac{1}{72} (3)^4 \
&= frac{9}{4} - frac{81}{72} \
&= 1.125
end{align*}
Apologies, I am not sure how to use LaTeX. Any help appreciated.
continuity random-variables expected-value
edited Dec 3 '18 at 0:34
platty
3,370320
The continuous random variable $X$ has p.d.f $f(x)$ given by
$$f(x) = begin{cases}
k (9 - x^2) &text{for } 0 leq x leq 3 \
0 &text{otherwise}
end{cases}$$
Where $k$ is a constant
Show that $k = 1/18$, and find the value of $E(X)$
The textbook says the answer is $3/4$, however, the only answer I can get is $1.125$
My workings:
begin{align*}E(X) &= int_0^3 0.5x - frac{x^3}{18} dx \
&= [0.25x^2 - x^4/72]_3^0 \
&= frac{1}{4} (3)^2 - frac{1}{72} (3)^4 \
&= frac{9}{4} - frac{81}{72} \
&= 1.125
end{align*}
Apologies, I am not sure how to use LaTeX. Any help appreciated.
continuity random-variables expected-value
continuity random-variables expected-value
edited Dec 3 '18 at 0:34
platty
3,370320
edited Dec 3 '18 at 0:34
platty
3,370320
edited Dec 3 '18 at 0:34
platty
3,370320
edited Dec 3 '18 at 0:34
platty
3,370320
edited Dec 3 '18 at 0:34
platty
3,370320
3,370320
asked Dec 3 '18 at 0:29
user546944
$1.125$ is the correct answer.
– Kavi Rama Murthy
Dec 3 '18 at 0:34
add a comment |
$1.125$ is the correct answer.
– Kavi Rama Murthy
Dec 3 '18 at 0:34
$1.125$ is the correct answer.
– Kavi Rama Murthy
Dec 3 '18 at 0:34
$1.125$ is the correct answer.
– Kavi Rama Murthy
Dec 3 '18 at 0:34
add a comment |
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$1.125$ is the correct answer.
– Kavi Rama Murthy
Dec 3 '18 at 0:34