Revision as of 03:05, 27 June 2024 by Admin (Created page with "'''Solution: A''' The probability that the digit 3 appears as a random digit is 1/10, and the number of digits equalling 3 among 10,000 random digits is approximately normally distributed with mean 10,000 * 1/10 = 1,000 and variance 10,000 * 0.1 * 0.9 = 900. Hence the probaility that we observe the digit 3 more than 931 times is approximately equal to the probability that a standard normal variable exceeds 31/30 and this is approximately equal to 0.1507.")
Exercise
ABy Admin
Jun 27'24
Answer
Solution: A
The probability that the digit 3 appears as a random digit is 1/10, and the number of digits equalling 3 among 10,000 random digits is approximately normally distributed with mean 10,000 * 1/10 = 1,000 and variance 10,000 * 0.1 * 0.9 = 900. Hence the probaility that we observe the digit 3 more than 931 times is approximately equal to the probability that a standard normal variable exceeds 31/30 and this is approximately equal to 0.1507.