Answer to Question #149406 in Statistics and Probability for jee

Question #149406
a. The probability of success on any trial of a binomial
experiment is 25%. Find the probability
that the proportion of successes in a sample of
500 is less than 22%.
b. Repeat part (a) with n = 800.
c. Repeat part (a) with n = 1,000.
1
Expert's answer
2020-12-15T08:51:21-0500

a. Find the probability of success that the proportion of success in a sample of 500 is less than 22% .

The probability of success for any trail of a binomial experiment is, p=0.25.

The sample size, n = 500

The probability of success that the proportion of success in a sample of 500 is less than 22% is,

"P(\\bar{P}<0.22)=P( \\frac{\\bar{P}-p}{\\sqrt{\\frac{p(1-p)}{n}}}<\\frac{0.22-0.25}{\\sqrt{\\frac{0.25(1-0.25)}{500}}} ) \\\\\n\n=P(Z<\\frac{0.22-0.25}{0.0194} \\\\\n\n=p(Z<-1.55) \\\\\n\n= 0.0606"

Therefore, the probability of success that the proportion of success in a sample of 500 is less than 22% is 0.0606.

b. Find the probability of success that the proportion of success in a sample of 800 is less than 22% .

The probability of success for any trail of a binomial experiment is, p=0.25.

The sample size, n=800

The probability of success that the proportion of success in a sample of 800 is less than 22% is,

"P(\\bar{P}<0.22)=P( \\frac{\\bar{P}-p}{\\sqrt{\\frac{p(1-p)}{n}}}<\\frac{0.22-0.25}{\\sqrt{\\frac{0.25(1-0.25)}{800}}} ) \\\\\n\n=P(Z<\\frac{0.22-0.25}{0.0153} \\\\\n\n=p(Z<-1.96) \\\\\n\n= 0.0250"

Therefore, the probability of success that the proportion of success in a sample of 800 is less than 22% is 0.0250.

c. Find the probability of success that the proportion of success in a sample of 1000 is less than 22%.

The probability of success for any trail of a binomial experiment is, p=0.25.

The sample size, n=1000

The probability of success that the proportion of success in a sample of 1000 is less than 22% is,

"P(\\bar{P}<0.22)=P( \\frac{\\bar{P}-p}{\\sqrt{\\frac{p(1-p)}{n}}}<\\frac{0.22-0.25}{\\sqrt{\\frac{0.25(1-0.25)}{1000}}} ) \\\\\n\n=P(Z<\\frac{0.22-0.25}{0.0137} \\\\\n\n=p(Z<-2.19) \\\\\n\n= 0.0143"

Therefore, the probability of success that the proportion of success in a sample of 1000 is less than 22% is 0.0143.


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