Question

Joe is pursuing a major in computer science. He notices a memory chip containing 2¹² = 4,096 bits is full of data. The data seems to be generated, bit-by-bit, at random, with 0’s and 1’s equally likely, and the bits are stored independently. If each bit is equally likely to be a 0 or 1, estimate the probability that there are actually 2,140 or more 1’s stored on the memory chip. Hint, use the normal approximation with continuity correction.

Answer 1

Since 0's and 1's are equally likely, probability of 0 or 1 = 0.5

We have mean of the binomial distribution = 4096*0.5 = 2048

Standard deviation = SQRT(4096*0.5*0.5) = 32

With approximation to normal distribution, and with the continuity correction factor of 0.5, we need to find the probability P (X >= 2140-0.5) = P (X>= 2139.5)

We need to find P (Z >= 2.8594) = 1 - P(Z < 2.8594)

Using a standard normal table, or using the MS Excel function NORM.S.DIST(2.8594,TRUE),

we have P(Z<2.8594) = 0.9979

P (Z >= 2.8594) = 1 - P(Z < 2.8594) = -0.9979 = 0.0021