You plan to conduct a survey to find what proportion of the workforce has two or more jobs. You decide on the 98% confidence level and a margin of error of 2%. A pilot survey reveals that 7 of the 50 sampled hold two or more jobs.
1. How many in the workforce should be interviewed to meet your requirement?

1635 people

Step-by-step explanation:

The proportion of people with two or more jobs in the sample  is  

7/50 = 0.14

The z-value that gives us a 98% confidence level is 2.33 (this means that 98% of the area under the normal curve N(0;1) is between -2.33 and 2.33)

If we use Simple Random sampling, the size of the sample should be

where z is the z-value which gives us the desired confidence level, p the proportion and e the error. In our case

z = 2.33

p = 0.14

e = 2% = 0.02

So, the sample size should be

rounding up to the nearest integer.

amount of oatypop that contains the prize = 40%

amount   of oatypop which doesn't contain the prize = (100-40)=60%

probability that hannah will win a prize

    = s +f s +f f s+ff   f s+ff f f s+ infinity.where f= failure,s= success

=   or 60%

probability that hannah only has to buy 3 or less boxes before getting a prize= (s) and ( f s) and (ff s) and (f f f s)

        = = 87.04 %  

step-by-step explanation: