A cable provider wants to contact customers in a particular telephone exchange t

Question

A cable provider wants to contact customers in a particular telephone exchange to see how satisfied they are with the new digital TV service the company has provided. All numbers are in the 153 exchanges, so there are 10,000 possible numbers from 153-0000 to 153-9999. If they select the numbers with equal probability:

a) What distribution would they use to model the selection?

b) What is the probability the number selected will be an odd number?

c) What is the probability the number selected will end in 555?

A)

a.uniform

b. binomial

c. Poisson

d. geometric

distribution should be used to model the selection.

B) The probability that the number selected will be an odd number is ----------

(Round to three decimal places as needed.)

c) The probability that the number selected will end in 555 is ---------

(Round to three decimal places as needed.)

Explanation / Answer

All the numbers from 1530000 to 1539999 are equally probable

This follows a uniform distribution

a) So, to model this we use the uniform distribution

In this list of 10,000 numbers, we have an equal number of Odd and Even numbers- 5000 each

b) The probability of selecting an Odd number = 0.5

c) The probability that the number will end in 555

We have 153 X555

X can have a total of 10 values from 0 to 9

So, required probability = 10/10000 = 0.001

Hope this helps and Please don't forget to rate the answer :)

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