Respuesta :
The mean is 0.75 , standard deviation is 0.025 and the probability that more than 71% is 0.9452 or 94.52%
The complete question is
In a memory test, the test subjects are given a large number and are asked to memorize it. Historical records show that
75% of test subjects pass the test. To pass the test, a subject must exactly repeat all the digits in the number after two hours.
A random sample of 300 people to take the memory test is going to be chosen. Let p be the proportion of people in the sample who pass the test. Answer the following. (If necessary, consult a list of formulas.)
(a)Find the mean of p
.(b)Find the standard deviation of p
.(c)Compute an approximation for P(P>0.71) , which is the probability that more than 71%
of the people in the sample pass the test. Round your answer to four decimal places.
What is Probability ?
Probability is the study of likelihood of an event to happen.
It is given in the question that
n = 300
p = 0.75
The mean is therefore = p = 0.75
Standard Deviation is [tex](\sigma _p) = \sqrt{\dfrac{p(1-p)}{n}[/tex]
=[tex]\sqrt\dfrac{0.75 (1-0.75)}{300}}[/tex]
=0.025
[tex]P(p > 0.71) = P(\dfrac{ p- \mu }{\sigma _p} > \dfrac{0.71-0.75}{0.025})[/tex]
P(z > -1.6)
1 - P(z < -1.6)
1 - 0.0548
0.9452 ( from z table )
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