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 is between 0.489 and
0.580?
Write only a number as your answer. Round to 4 decimal places (for
example 0.3748). Do not write as a percentage.
Solution :
Given that ,
p = 0.50
1 - p = 0.50
n = 36
= p = 0.50
= (p*(1-p))/n = (0.50*0.50)/ 36= 0.08333
P(0.489 < <0.580 ) = P((0.489-0.50)/0.08333 ) < ( - ) / < (0.580-0.50) /0.08333 ) )
= P(-1.09 < z < 0.96 )
= P(z < 0.96) - P(z < -1.09) using z - table,
= 0.8315 - 0.1379
= 0.6936
The proportion of students who take advantage of free tutoring services in the sample is between 0.489 and 0.580 is 0.6936
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