Question

1. A string of binary values (bits) is generated randomly such that the probability of any bit being a 1 is 0.15 and the values of any two bits are independent of one another. What is the probability that a randomly chosen 15-bit sub-string contains more than three 1s?

2. Quality control testing on a manufactured shaft indicates that the finished diameter follows a Normal distribution with a mean of 19.96mm and a standard deviation of 0.025mm. If the user’s quality control department rejects any shaft with a diameter smaller than 19.92mm or larger than 20.02mm, what proportion of the manufactured shafts are accepted by the user?

Answer 1

1) n = 15

P = 0.15

It is a binomial distribution.

P(X = x) = nCx * p^x * (1 - p)^{n - x}

P(X > 3) = 1 - P(X < 3)

= 1 - (P(X = 0) + P(X = 1) + P(X = 2))

= 1 - (15C0 * (0.15)^0 * (0.85)^15 + 15C1 * (0.15)^1 * (0.85)^14 + 15C2 * (0.15)^2 * (0.85)^13)

= 1 - 0.8227 = 0.1773

2) P(19.92 < X < 20.02)

= P((19.92 - )/ < (X - )/ < (20.02 - )/)

= P((19.92 - 19.96)/0.025 < Z < (20.02 - 19.96)/0.025)

= P(-1.6 < Z < 2.4)

= P(Z < 2.4) - P(Z < -1.6)

= 0.9918 - 0.0548

= 0.9370

The proportion of the manufacturer shafts are accepted by the user is 0.9370.