A real number X is selected uniformly at random in the conti

A real number X is selected uniformly at random in the continuous interval [0,10]. (For example, X could be 3.87.) Find P(2

Solution

A)

Note that here,          
          
a = lower fence of the distribution =    0      
b = upper fence of the distribution =    10      
          
Thus, the area between the said numbers is          
          
c = lower number =    2      
d = higher number =    5      
          
Thus, the probability between these two values is          
          
P = (d - c)/(b - a) =    0.3   [ANSWER]

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b)

It is like reducing the fences to (0,5).

Note that here,          
          
a = lower fence of the distribution =    0      
b = upper fence of the distribution =    5      
          
Note that P(x<c) = P(a<x<c) = (c-a)/(b-a). Thus, as          
          
c = critical value =    2      
          
Then          
          
P(x<2|x<=5) =    0.4   [ANSWER]  

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c)

Note that

P(3<=x<=8|x>=4) = P(4<=x<=8|x>=4)

which sort of reduces the distribution to (4,10).

Note that here,          
          
a = lower fence of the distribution =    4      
b = upper fence of the distribution =    10      
          
Thus, the area between the said numbers is          
          
c = lower number =    4      
d = higher number =    8      
          
Thus, the probability between these two values is          
          
P(3<=x<=8|x>=4) = (d - c)/(b - a) =    0.666666667   [ANSWER]