About 72% of young adult Internet users (ages 18 to 29) use social-networking si

Question

About 72% of young adult Internet users (ages 18 to 29) use social-networking sites. Suppose that a sample survey contacts an SRS of 1600 young adult Internet users and calculates the proportion p in this sample who use social-networking sites. (Enter your means to two decimal places and round your standard deviations to four decimal places.)

(a) What is the approximate distribution of p? Express responses in decimal form (proportion not percent) and use two decimal places for the mean and four decimal places for the standard deviation.

mean

standard deviation    

(b) If the sample size were 6000 rather than 1600, what would be the approximate distribution of p? Express your responses in the same form as requsted above.

mean

standard deviation    

mean

standard deviation    

Explanation / Answer

a) Mean is n*p which is 1600*0.72=1152

and standard deviation is sqrt(n*p*(1-p))=sqrt(1600*0.72*(1-0.72))=17.9599

b) Mean is 6000*0.72=4320

standard deviation is sqrt(6000*0.72*(1-0.72)) =34.779

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