Question

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Q1 ) Males from Saskatchewan account for 1,350 of the 22,400 total deaths from the four provinces. An estimate for the proportion of males from Saskatchewan who die as a result of cancer is given by 1,350 22,400 ≈ 0.0602 (or 6.0%). Suppose that you randomly sample ten medical records of deaths from western Canada due to cancer. Determine the probability that out of the ten selected records:

a) None are males from Saskatchewan. [2 Marks]

b) All are males from Saskatchewan.

c) At most one are males from Saskatchewan.

d) At least two are males from Saskatchewan.

Answer 1

(a)

Binomial Distribution

n = 10

p = 0.06

q = 1 - p = 0.94

) × 0.94 10 × 0.060 = 1 × 0.5386 × 1 P(X 0) 0.5386

So,

Answer is:

0.5386

(b)

× 0.940 × 0.0610-1 × 1 × 0.0000 0.0000 P(X 10)

(c)

P(X$\leq$1) = P(X=0) + P(X=1)

) × 0.94 10 × 0.060 = 1 × 0.5386 × 1 P(X 0) 0.5386

PlX-1)-(#9 × 0.949 × 0.061 = 10 × 0.5730 × 0.06 = 0.3438

So,

P(X$\leq$1) = 0.8824

So,

Answer is:

0.8824

(d)

P(X$\geq$2) = 1 - [P(X=0) + P(X=1)]

      = 1 - (0.5386 + 0.3438)

       = 1 - 0.8824

         = 0.1176

So,

Answer is:

0.1176