Question

A corporation is considering a new issue of convertible bonds. Management believes that the offer terms will be found attractive by 20% of all its current stockholders. Suppose that the belief is correct. A random sample of 130 stockholders is taken. What is the probability that the sample proportion of those stockholders that find the terms attractive will be between .18 and .23? 1. In a small town, households recycle on average 30% of their waste. The new recycling committee wants to increase this proportion and study the relationship between recycling and income. They select 25 households from the two wealthier neighborhoods and estimate a 92% confidence interval for the true proportion of recycling. What is the error term of this interval? 2. The same recycling committee also focuses on poor neighborhoods. How many households do they need to sample to get a 95% confidence interval, with an error of +/- 0.09? 3. Finally, the same recycling committee wants to know the probability of 12 random households recycling more than 45% of their waste. This probability is: 1. Plastic bags used for packaging produce are manufactured so that the breaking strengths of the bags are normally distributed with a standard deviation of 1.8 pounds per square inch. A random sample of 16 bags is selected. The probability is .15 that the sample mean exceeds the population mean by more than or equal to how much? 2. Plastic bags used for packaging produce are manufactured so that the breaking strengths of the bags are normally distributed with a standard deviation of 1.8 pounds per square inch. What proportion of bags will exceed the average breaking strength by more than 3 pounds?

Answer 1

1)

                       population proportion ,p=   0.2                      
                       n=   130                      
                                                  
                       std error , SE = √( p(1-p)/n ) =    0.0351                      
                                                  
                       we need to compute probability for                           
                       0.18   < p̂ <   0.23                  
                                                  
                       Z1 =( p̂1 - p )/SE= (   0.18   -   0.2   ) /    0.0351   =   -0.570
                       Z2 =( p̂2 - p )/SE= (   0.23   -   0.2   ) /    0.0351   =   0.855
P(   0.18   < p̂ <   0.23   ) =    P[( p̂1-p )/SE< Z <(p̂2-p)/SE ]    =P(    -0.570   < Z <   0.855   )          
                                                  
= P ( Z <   0.855   ) - P (    -0.570   ) =    0.8038   -   0.2843   =   0.5195 (answer)