for normal distribution z score =(p̂-p)/σ_{p} | |
here population proportion= μp= | 0.4800 |
sample size =n= | 60 |
std error of proportion=σ_{p}=√(p*(1-p)/n)= | 0.0645 |
probability that the proportion of teenagers in the sample who play videos games on their phone is less than 0.486 :
probability =P(X<0.486)=(Z<(0.486-0.48)/0.064)=P(Z<0.093)=0.5359 |
(please try 0.5371 if this comes wrong)
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