since 211 were minivans, then 474 must have been either sedans or sport cars.
so the probability of choosing either a sedan or sports car would be 474/685 = .692
I'm having trouble with estimating probability:
In a study of 685 cars in a parking lot, there were only 3 different types: sedan, sports car and mini-van. 211 of the cars were mini-vans. Use this information to estimate the probability of choosing a car that is a sedan or sports car.
I figure n=685 but after that I'm stuck!
Any help would be great! Thanks!
5 answers
Thank you.
a recent poll of 700 peole who work indoors found that 278 of the smoke. If the researchers want to be 98% confident of thier results to within 3.5% how large a sample is necessary?
how do i work this problem?
how do i work this problem?
Already answered in another post. :D
Listed below are the number of homeruns for the National League leader over the last 20 years, through 2006. Assuming that number of homeruns is normally distributed, if this is sample data collected from a population of all past and future homerun leaders, test the claim that the mean homerun leader has less than 47 homeruns, where α=.05. Set up and complete the appropriate hypothesis test. For this data, also compute the p-value and describe how you could have used this information to complete the analysis. Finally, compute 85% and 98% Confidence Intervals for this data.
Year Homeruns Year Homeruns
2006 58 1996 47
2005 51 1995 40
2004 48 1994 43
2003 47 1993 46
2002 49 1992 35
2001 73 1991 38
2000 50 1990 40
1999 65 1989 47
1998 70 1988 39
1997 49 1987 49
Year Homeruns Year Homeruns
2006 58 1996 47
2005 51 1995 40
2004 48 1994 43
2003 47 1993 46
2002 49 1992 35
2001 73 1991 38
2000 50 1990 40
1999 65 1989 47
1998 70 1988 39
1997 49 1987 49
1 answer