Question

# You may need to use the appropriate appendix table or technology to answer this question. A population proportion is 0.40. A sample of size 100 will be taken and the sample proportion p will be used to estimate the population proportion. $Round your answers to four decimal places.$ $a$ What is the probability that the sample proportion will be within ±0.03 of the population proportion? .7291 Incorrect: Your answer is incorrect. $b$ What is the probability that the sample proportion will be within ±0.05 of the population proportion? .8485 Incorrect: Your answer is incorrect.

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## Answer to a math question You may need to use the appropriate appendix table or technology to answer this question. A population proportion is 0.40. A sample of size 100 will be taken and the sample proportion p will be used to estimate the population proportion. $Round your answers to four decimal places.$ $a$ What is the probability that the sample proportion will be within ±0.03 of the population proportion? .7291 Incorrect: Your answer is incorrect. $b$ What is the probability that the sample proportion will be within ±0.05 of the population proportion? .8485 Incorrect: Your answer is incorrect.

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To find the probability that the sample proportion will be within ±0.03 and ±0.05 of the population proportion, we can use the normal distribution approximation for the sample proportion.

The standard error of the sample proportion is given by:
SE = \sqrt{\frac{p$1-p$}{n}}

Given that the population proportion, p = 0.40 and sample size n = 100, we can calculate the standard error:
SE = \sqrt{\frac{0.40$1-0.40$}{100}} = \sqrt{\frac{0.24}{100}} = 0.0490

$a$ For ±0.03 of the population proportion:
To find the probability that the sample proportion will be within ±0.03 of the population proportion, we need to find the z-scores for ±0.03:
z = \frac{0.03}{SE} = \frac{0.03}{0.0490} = 0.6122

Using the z-table, the probability that the sample proportion will be within ±0.03 of the population proportion is:
P$-0.6122 < Z < 0.6122$ = \text{approximately 0.7291}

$b$ For ±0.05 of the population proportion:
Similarly, for ±0.05 of the population proportion:
z = \frac{0.05}{SE} = \frac{0.05}{0.0490} = 1.0204

Using the z-table, the probability that the sample proportion will be within ±0.05 of the population proportion is:
P$-1.0204 < Z < 1.0204$ = \text{approximately 0.8485}

$a$ The probability that the sample proportion will be within ±0.03 of the population proportion is approximately 0.7291.
$b$ The probability that the sample proportion will be within ±0.05 of the population proportion is approximately 0.8485.
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