Computers are often used to randomly generate digits of telephone numbers to be called for telemarketing purposes. Can the methods used with normal probability distributions be used to find the probability that when one digit is randomly generated, it is greater than 2? Why or why not? What is the probability of getting a digit greater than 2?

Question

anonymous · Answer

Are you an IITian ?

perl · Answer

you can have 3,4,5,6,7,8,9 , out of 0,1,2,3,4,5,6,7,8,9

perl · Answer

i take that back, the method of normal probability wont really help here. 
this problem has equally likely events, you can have 0 or 1 or 2 or ... 9 
equally likely.

perl · Answer

if you generated a million random numbers, you would get a uniform distribution (not normal)

kropot72 · Answer

The normal distribution is a bell-shaped curve and cannot be used for this purpose. A discrete uniform distribution applies in this case. The probability of generating any of the 10 digits from 0 to 9 is 1/10. The selections of digits are mutually exclusive, therefore the probability of selecting a digit greater than 2 is
P(3) + P(4) + P(5) + P(6) + P(7) + P(8) + P(9) =
$$\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+\frac{1}{10}$$

perl · Answer

let X = digit 0 - 9 
P( X is greater than 2 ) = P ( X = 3 or X = 4 or X = 5 ... or X = 9 or X = 10) 
                                 = P ( X = 3) + P ( X = 4) + .... P ( X = 9 ) + P( X = 10) 
                                  = 1/10    + 1/10  + ... + 1/10  = 7/10