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In a survey of 300 people, approximately 15% of the people identified themselves as being left-handed....

  1. In a survey of 300 people, approximately 15% of the people identified themselves as being left-handed.
    1. What are the mean, standard deviation, and shape of the distribution of thesample proportion p-hat of respondents who say they are left-handed? (3 points)



    2. Using the normal approximation without the continuity correction, sketch theprobability distribution curve for the distribution of p-hat. Shade equal areas onboth sides of the mean to show an area that represents a probability of .95,and label the upper and lower bounds of the shaded area as values of p-hat(not z-scores). Show yourcalculations for the upper and lower bounds. (3 points)



    3. Considering the sketch in part B, the shaded area shows a .95 probability ofwhat happening? In other words, what does the probability of .95 represent? (2 points)



    4. Using the normal approximation, what's the probability a randomly drawn sample of parents of size 300 will have a sample proportion between 13% and 17%? Draw a sketch of the probability curve, shade the area representing theprobability you're finding, and label the z-scores that represent the upper andlower bounds of the probability you're finding. Don't use the continuitycorrection. (4 points)



    5. Now, use the exact binomial calculation to find the probability of gettingbetween, but not including, 13% and 17% of the respondents in a sample of 300 who indicate they are left-handed. To usethe exact binomial, you'll need to convert the proportions to counts by multiplying each proportion by 300. (2 points)



    6. Now try it again, but this time find the probability of getting at least 13% but no more than 17%. Use the exact binomial calculation. (2 points)



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Answer #1

Sol 令 О.IS ( Sayts(e 110 otha 300 VAc) 300 3 bd o,6826 300 2 16% of 300 48 083)+-- O IS orss 261 soo ) [0-1ず10.8s) 2u9

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