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Let $f(x)$ be the continuous probability density function of a random variable $x$, the probability that $a < x \leq b$, is :

  1. $f(b-a)$
  2. $f(b) - f(a)$
  3. $\int\limits_a^b f(x) dx$
  4. $\int\limits_a^b xf (x)dx$
asked in Probability by Boss (18.3k points) | 2.2k views
+2

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2 Answers

+12 votes
Best answer

$A.$ This gives the probability at the point of $b-a$ which is not having any significant w.r.t $a$ and $b.$

$B.$ This gives the difference of the probabilities at $b$ and $a$. Note: This is different from cumulative distribution function $F(b) - F(a).$ Ref: https://en.wikipedia.org/wiki/Cumulative_distribution_function

$C.$ This is Probability Density Function. Ref: https://en.wikipedia.org/wiki/Probability_density_function

$D.$ This is expected value of continuous random variable. Ref: https://en.wikipedia.org/wiki/Expected_value
Answer is $C$.

answered by Boss (43.2k points)
edited by
+1
(b) is not CDF
0
@Bikram sir How is b) part CDF
+6

(B) f(b)−f(a) is not CDF.

Reason: Here function f is pdf not cdf.

 Note: If  X is Discrete RV  then P(a≤X≤b) = F(b) - F(a) ,where F is CDF .Both f and F are diffrent.

+1

@Akash Kanase and @Kapil I think selected answer requires small correction. 

0
What would be the answer if conditions were:

 

1. a≤x≤b

2. a≤x<b

3. a<x<b
0

@anchitjindal07 For all the $3$ intervals, the probability will be the same as for the closed interval. 

$P(a \leq x \leq b ) = \underset{=0}{\underbrace{P(x=a)}}+ \underset{=0}{\underbrace{P(x=b)}} + P(a <  x < b )\\P(a \leq x \leq b ) =P(a <  x < b ) $ 
 

0

 Why P(x=a) and P(x=b)= 0

 

+13 votes
C should be used if prob density function is given B should be used if prob distribution function is given D must be used to calculate expectation when pdf is given
answered by Boss (14.4k points)
Answer:

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