Question

Suppose that 8% of all Canadians are left-handed.

A) If we randomly select 200 Canadians, what is the approximate probability that less than 5% of them are left-handed? (Keep 6 decimal places in intermediate calculations and report your final answer to 4 decimal places.

B) If we randomly select 200 Canadians, what is the approximate probability that more than 20 of them are left-handed? (Keep 6 decimal places in intermediate calculations and report your final answer to 4 decimal places.

Answer 1

Solution:

Given ,

p = 8% = 0.08(population proportion)

1 - p = 1 - 0.08 = 0.92

n = 200 (sample size)

Let be the sample proportion.

The sampling distribution of is approximately normal with

mean = =  p = 0.08

SD =   =     

=  

=  0.01918332609

A)

P( < 5% )

= P( < 0.05)

=  

=  P(Z <(0.05-0.08)/0.01918332609)

= P(Z < -1.56)

= 0.0594 ...use z table

P(Sample proportion is less than 5%) = 0.0594

B)

P(X > 20)

= P( > 20/200)

= P( > 0.1)

=

=  P(Z > (0.1 -0.08)/0.01918332609)

= P(Z > 1.04)

= P(Z < -1.04)

= 0.1492 ...use z table

Required Probability = 0.1492