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Question: If the c.d.f (cumulative distribution function) is given by \(F\left( x \right)=\dfrac{x-25}{10}\) ,...

If the c.d.f (cumulative distribution function) is given by F(x)=x−2510F\left( x \right)=\dfrac{x-25}{10} , then P(27≤x≤33)=?P\left( 27\le x\le 33 \right)=?
(a)35\dfrac{3}{5}
(b)310\dfrac{3}{10}
(c)15\dfrac{1}{5}
(d)110\dfrac{1}{10}

Explanation

Solution

Hint: First look at definition of cumulative distribution function. Relate this function to probability by using this relation to find the probability in the given range of X by using the function value in the range. This probability in particular range is the required result.

Complete step-by-step answer:
Cumulative distribution function: In probability and theory of statistics the cumulative distribution function of a real valued random variable or just distribution function of evaluated at is the probability that will take less than or equal to X.
The value of F(x) at X is
P(x≤x).............(i)P\left( x\le x \right).............\left( i \right)
So, we will do 2 cases to find two values for F(x).
Case (i): Find F(x) at x = 33. So, we need F(33).
The cdf is given in the question in terms of x as
F(x)=x−2510F\left( x \right)=\dfrac{x-25}{10}
By substituting x = 33, we get value of function as
F(33)=33−2510F\left( 33 \right)=\dfrac{33-25}{10}
By simplifying the equation, we can write it in form of:
F(33)=810F\left( 33 \right)=\dfrac{8}{10}
By simplifying we can write the value of F(33) as F(33) = 0.8
From the statement (i), we can say value of P(x≤33)P\left( x\le 33 \right) as:
P(x≤33)=0.8................(ii)P\left( x\le 33 \right)=0.8................(ii)
Case (ii): Find value of F(x) at the value of x being 27. By substituting value of x as 27, we can write F(27) as
F(27)=27−2510F\left( 27 \right)=\dfrac{27-25}{10}
By simplifying this, we can write the value of F(x) as:
F(27)=210F\left( 27 \right)=\dfrac{2}{10}
By simplifying, we can write final value of F(27) as:
F(27) = 0.2
From statement (I) we can write value of P(x≤27)P\left( x\le 27 \right) as:
P(x≤27)=0.2.................(iii)P\left( x\le 27 \right)=0.2.................(iii)
By subtracting equation (iii) from equation (ii) we get
P(x≤33)−P(x≤27)=0.8−0.2P\left( x\le 33 \right)-P\left( x\le 27 \right)=0.8-0.2
So, all the values of x≤27x\le 27 will get cancel, we get
P(27≤x≤33)=0.6=610P\left( 27\le x\le 33 \right)=0.6=\dfrac{6}{10}
By simplifying the above equation, we get it as
P(27≤x≤33)=35P\left( 27\le x\le 33 \right)=\dfrac{3}{5}
Therefore option (a) is the correct answer.

Note: Be careful while relating F, P as it is the point which forms the base to our solution, so if you do wrong in that step the whole answer might go wrong. While subtracting, also be careful to subtract the lesser term greater one to get the result exactly.