Question

Question 1)

Tay-Sachs disease is a genetic disorder that is usually fatal in young children. If both parents are carriers of the disease, the probability that each of their offspring will develop the disease is approximately 0.25. Suppose a husband and wife are both carriers of the disease and the wife is pregnant on three different occasions. If the occurrence of Tay-Sachs in any one offspring is independent of the occurrence in any other, what are the probabilities of these events? (Round your answers to six decimal places.)

(a)

all three children will develop Tay-Sachs disease

(b)

only one child will develop Tay-Sachs disease

Question 2)

Let x have a uniform distribution on the interval 0 to 10. Find the probability.

P(2 < x < 7)

Question 3)

Let z be a standard normal random variable with mean μ = 0 and standard deviation σ = 1. Use Table 3 in Appendix I to find the probability. (Round your answer to four decimal places.)

P(−1.645 < z < 1.645) =

Question 4)

Let z be a standard normal random variable with mean μ = 0 and standard deviation σ = 1. Use Table 3 in Appendix I to find the probability. (Round your answer to four decimal places.)

to the left of 1.1

Question 5)

Let z be a standard normal random variable with mean μ = 0 and standard deviation σ = 1. Use Table 3 in Appendix I to find the probability. (Round your answer to four decimal places.)

to the right of −1.85

Answer 1

1) n = 3

    p = 0.25

It is a binomial distribution.

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

a) P(all three will develop) = P(X = 3) = 3C3 * (0.25)^3 * (0.75)^0 = 0.015625

b) P(only one children will develop) = P(X = 1) = 3C1 * (0.25)^1 * (0.75)^2 = 0.421875

2) P(2 < X < 7)

= (7 - 2)/(10 - 0)

= 5/10 = 0.5

3) P(-1.645 < Z < 1.645)

= P(Z < 1.645) - P(Z < -1.645)

= 0.9500 - 0.0500

= 0.9000

4) P(Z < 1.1) = 0.8643

5) P(Z > -1.85)

= 1 - P(Z < -1.85)

= 1 - 0.0322

= 0.9678