A recent study found that 42% of college students engage In binge drinking (5 dr
ID: 3253341 • Letter: A
Question
A recent study found that 42% of college students engage In binge drinking (5 drinks at a sitting for men, 4 for women). Use the 88-95-99.7 Rule to describe the sampling distribution model for the proportion of students in a randomly selected group of 269 college students who engage in binge drinking. Do you think the appropriate conditions are met? Which graph shows the distribution of the proportion with the intervals for 68% 95% and 99 7%? Are the conditions necessary to use a Normal model met? A. Yes all the conditions are met. B. No the success/failure condition is not met. C. No the randomization and 10% conditions are not met. D. No, the 10% condition is not met. E. No, the 10% and success/failure conditions are not met. F. No, the randomization and success/failure conditions are not met. G. No, the randomization condition is not met. H. No, none of the conditions are met.Explanation / Answer
Sol:
mean=p^=sample proportion=x/n=42/100=0.42
stddev=sqrt[p(1-p)/n]=sqrt[0.42(1-0.42)/260]=0.03
68% values lies in between mean-stddev and mean+stddev=0.42-0.03 and 0.42+0.03
=0.39 and 0.45
95% values lies inbetween mean-2stddev and mean+2stddev=0.42-2(0.03) and 0.42+2(0.03)
=0.36 and 0.48
99.7% values lies in between mean-3stddev and mean+3stddev=0.42-3(0.03) and 0.42+3(0.03)
=0.33 and 0.51
center line is mean =42%
ANSWER C
n is large enough=n=260
atleat 10 success and 10 failures(10% condition)
successes=np=260(0.42=109.2>10
failures=n(1-p)=260*(1-0.42)=150.8>10
conditions for normality met
ANSWER A
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