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Question 64 - Week 5 Practice | Prasnya
Q64
00:00
2 Apr 2023
Q64.
Find the PDF of the marks of a candidate chosen uniformly at random.
@passage
A
$f(y)=\frac{1}{20\sqrt{2\pi}}\left[e^{-(y-60)^2/72}+2e^{-(y-55)^2/98} ight]$
B
$f(y)=\frac{1}{20\sqrt{2\pi}}\left[2e^{-(y-60)^2/36}+e^{-(y-55)^2/49} ight]$
C
f
(
y
)
=
7
20
2
π
e
−
(
y
−
60
)
2
/
72
+
3
20
2
π
e
−
(
y
−
55
)
2
/
98
f(y)=\frac{7}{20\sqrt{2\pi}}e^{-(y-60)^2/72}+\frac{3}{20\sqrt{2\pi}}e^{-(y-55)^2/98}
f
(
y
)
=
20
2
π
7
e
−
(
y
−
60
)
2
/72
+
20
2
π
3
e
−
(
y
−
55
)
2
/98
D
$f(y)=\frac{1}{\sqrt{2\pi}}\left[\frac16e^{-(y-60)^2/72}+\frac17e^{-(y-55)^2/98} ight]$
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Q64
00:00
2 Apr 2023
Q64.
Find the PDF of the marks of a candidate chosen uniformly at random.
@passage
A
$f(y)=\frac{1}{20\sqrt{2\pi}}\left[e^{-(y-60)^2/72}+2e^{-(y-55)^2/98} ight]$
B
$f(y)=\frac{1}{20\sqrt{2\pi}}\left[2e^{-(y-60)^2/36}+e^{-(y-55)^2/49} ight]$
C
f
(
y
)
=
7
20
2
π
e
−
(
y
−
60
)
2
/
72
+
3
20
2
π
e
−
(
y
−
55
)
2
/
98
f(y)=\frac{7}{20\sqrt{2\pi}}e^{-(y-60)^2/72}+\frac{3}{20\sqrt{2\pi}}e^{-(y-55)^2/98}
f
(
y
)
=
20
2
π
7
e
−
(
y
−
60
)
2
/72
+
20
2
π
3
e
−
(
y
−
55
)
2
/98
D
$f(y)=\frac{1}{\sqrt{2\pi}}\left[\frac16e^{-(y-60)^2/72}+\frac17e^{-(y-55)^2/98} ight]$
Save
Read
Check
Details
