Question

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.)​

Answer 1

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 $(\sigma _p) = \sqrt{\dfrac{p(1-p)}{n}$

=$\sqrt\dfrac{0.75 (1-0.75)}{300}}$

=0.025

$P(p > 0.71) = P(\dfrac{ p- \mu }{\sigma _p} > \dfrac{0.71-0.75}{0.025})$

P(z > -1.6)

1 - P(z < -1.6)

1 - 0.0548

0.9452 ( from z table )

To know more about Probability

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