Question

The next two questions (10 and 11) refer to the following: Suppose that 13% of Canadians are left-handed. Question 10 (1 point) If we randomly select 500 Canadians, what is the approximate probability that less than 10% of them are left-handed? Keep 6 decimal places in intermediate calculations and report your final answer to 4 decimal places. Your Answer: Answer Question 11 (1 point) If we randomly select 500 Canadians, what is the approximate probability that more than 60 of them are left-handed? Keep 6 decimal places in intermediate calculations and report your final answer to 4 decimal places. Your Answer: Answer

Answer 1

Solution(10)
Given in the question
P(Canadians are left handed) = 0.13
Number of sample (n)= 500
we need to calculate that less than 10% are left handed i.e. P(p^<0.10), here we will use standard normal distribution, First we will calculate Z-score which can be calculated as
Z-score = (p^-p)/sqrt(p*(1-p)/n) = (0.10-0.13)/sqrt(0.13*0.87/500) = -2.0
From Z table we found p-value i.e.
P(p^<0.10) = 0.0228
So there is 2.28% probability that less than 10% are left handed.
Solution(11)
Given in the question
P(Canadians are left handed) = 0.13
Number of sample (n)= 500
We need to calculate the probability that more than 60 are left handed
sample proportion = 60/500 = 0.12
here we will use standard normal distribution, First we will calculate Z-score which can be calculated as
Z-score = (p^-p)/sqrt(p*(1-p)/n) = (0.12-0.13)/sqrt(0.13*0.87/500) = -0.66
From Z table we found p-value i.e.
P(p^>0.12) = 0.7454
So there is 74.54% probability that more than 60 are left handed.