MAT 2379, Introduction to biostatistics, Assignment 1
1
MAT 2379, Introduction to biostatistics
Assignment 2
Solution to Calculator Problems
Due date: Wednesday October 6, 2010
T otal = 80 marks
Problem 3.7. (10 marks) Let D be the event that the child will have the disease. Let M be the event that
the child is a male, and F be the event that the child is a female. We know that P (D|M ) = 0.5, P (D|F ) = 0 and
P (M ) = 0.513, P (F ) = 0.497. By the total probability rule,
P (D) = P (D|M )P (M ) + P (D|F )P (F ) = (0.5)(0.513)+)(0.497) = 0.2565.
Problem 3.12. (10 marks) (a) The probability that someone smokes is P (S) = 1213/6549 = 0.185.
(b) The conditional probability that someone smokes given that the person has high income is P (S|H) =
247/2115 = 0.1168.
(c) The event S that the person smokes is not independent of the event H that the person has high income since
P (S) 6= P (S|H).
Problem 3.14. (5 marks) Let A be the event that the husband smokes, and B be the event that the wife smokes.
We know that P (A) = 0.3, P (B) = 0.2 and P (A and B) = 0.08. Since P (A)P (B) = 0.06 6= P (A and B), A and B
are not independent.
Problem 3.15. (10 marks) (a) The percentage of trees with diameters between 4 inches and 10 inches is
0.33+0.25+0.12=0.70.
(b) The percentage of trees with diameters less than 4 inches is 0.03+0.20=0.23.
(c) The percentage of trees with diameters more than 6 inches is 0.25+0.12+0.07=0.44.
Problem 3.18. (10 marks) (a) P (Y = 3) = 610/5000 = 0.122.
(b) P (Y ≥ 7) = P (Y = 7) + P (Y = 8) + P (Y = 9) + P (Y = 10) = (130 + 26 + 3 + 1)/5000 = 160/5000 = 0.032.
(C) P (4 ≤ Y ≤ 6) = P (Y = 4) + P (Y = 5) + P (Y = 6) = (1400 + 1760 + 750)/5000 = 3910/5000 = 0.782
Problem 3.24 (5 marks) E(Y ) = 0(0.15) + 1(0.5) + 2(0.35) = 1.2
Problem 3.25 (10 marks) The variance of Y is:
Var(Y ) = 02 (0.15) + 12 (0.5) + 22 (0.35) − (1.2)2 = 0.46
√
The standard deviation of Y is σY = 0.46 = 0.678
Problem 3.26 (10 marks) The probability of a yellow pea is 0.75. Let X be the number of yellow peas in a
sample of size 4. X has a binomial distribution with n = 4 and p = 0.75.
(a) P (X = 3) = 4(0.75)3 (0.25) = 0.4219.
(b) P (X = 4) = (0.75)4 = 0.3164.
(c) P (all 4 are yellow)+P (all 4 are green) = P (X = 4)+P (X = 0) = (0.75)4 +(0.25)4 = 0.3164+0.0039 = 0.3203.
Problem 3.38 (10 marks) (a) Let A be the event that the first girl has iron deficiency and B be the event that
the second girl has iron deficiency. Since A and B are independent, P (A and B) = P (A)P (B) = (0.1)(0.1) = 0.01.
(b) The desired probability is:
P (A and B c ) + P (Ac and B) = P (A)P (B c ) + P (Ac )P (B) = (0.1)(0.9) + (0.9)(0.1) = 0.18.