At a certain college, 50% of all students take advantage of free tutoring services. A sample of 36 students is selected. what is the probability that the proportion of students who take advantage of free tutoring services in the sample between .485 and .580? Show the functions using the TI-84 calculator
for normal distribution z score =(p̂-p)/σ_{p} | |
here population proportion= p= | 0.500 |
sample size =n= | 36 |
std error of proportion=σ_{p}=√(p*(1-p)/n)= | 0.0833 |
probability =P(0.485<X<0.58)=P((0.485-0.5)/0.083)<Z<(0.58-0.5)/0.083)=P(-0.18<Z<0.96)=0.8315-0.4286=0.4029 |
NOte:
if using ti-84 press 2nd - vars -use command :normalcdf(0.485,0.58,0.5,0.0833) |
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