A large collection of one-digit random numbers should have about 50% odd and 50%

Question

A large collection of one-digit random numbers should have about 50% odd and 50% even digits because five of the ten digits are odd (1 3, 5 7, and 9) and five are even (0, 2, 4, 6, and 8). a. Find the proportion of odd-numbered digits in the following lines from a random number table. Count carefully. 9 3 3 0 8 4 3 5 8 0 3 3 7 2 2 2 9 6 8 6 2 2 4 5 5 5 2 0 9 8 b. Does the proportion found in part (a> represent p (the sample proportion, or p (the Ration proportion)? c. Find the error in this estimate, the difference between p and p (or p- p) a. The given random number table consists of

Explanation / Answer

Total number of digits = 30

No. of odd numbered digits = 9,3,3 - first number, 3,5 - > second number; 3,3 7 -> third number

9->4th number 5,5->5th number , 5,9->6th number

= 3+2+3+1+2+2 = 13

a. proportion of odd-numbered digits = No. of odd numbered digits / Total no. of digits = 13/30 = 0.43333

b. The propotion found in part (a) represent p^ : the sample proportion. (because it's calculated from a sample)

c. Error in estimate = p - p^ = 0.5 -0.4333 = 0.0667 (p: population to be 50% of odd digits i.e 0.5 propotion of odd digits)

a.The given random number table consists of 43.33% odd-numbered digits.

13 odd digits out of 30; conver them into percentage = (13/30)*100 = 43.33%

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.