7. Math 210- Module 6 - Show all Work - Ask me for help if needed. The sampling distribution of proportions is normal under certain conditions with a mean of p(1-p) H = p and o n Assume the conditions are met. Suppose a weighted coin has a 44.0% chance of getting heads. Use the normal distribution to find the probability that when flipping the coin 110 times the proportion of heads is more than 50.0%.

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**Math 210 - Module 6: Sampling Distribution of Proportions**

This module covers the sampling distribution of proportions, which is normal under certain conditions.

**Mean and Standard Deviation:**

- The mean (\( \mu_{\hat{p}} \)) of the sampling distribution is equal to the population proportion (\( p \)).
- The standard deviation (\( \sigma_{\hat{p}} \)) is given by:
  \[
  \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}
  \]

**Example Problem:**

Assume the conditions are met. Suppose a weighted coin has a 44.0% chance of getting heads. Use the normal distribution to find the probability that when flipping the coin 110 times, the proportion of heads is more than 50.0%.
Transcribed Image Text:**Math 210 - Module 6: Sampling Distribution of Proportions** This module covers the sampling distribution of proportions, which is normal under certain conditions. **Mean and Standard Deviation:** - The mean (\( \mu_{\hat{p}} \)) of the sampling distribution is equal to the population proportion (\( p \)). - The standard deviation (\( \sigma_{\hat{p}} \)) is given by: \[ \sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}} \] **Example Problem:** Assume the conditions are met. Suppose a weighted coin has a 44.0% chance of getting heads. Use the normal distribution to find the probability that when flipping the coin 110 times, the proportion of heads is more than 50.0%.
Expert Solution
Step 1: Write the given information.

Given information:

mu subscript p with hat on top end subscript equals p
sigma subscript p with hat on top end subscript equals square root of fraction numerator p open parentheses 1 minus p close parentheses over denominator n end fraction end root

p=The chance of getting heads= 44.0%=0.44

n=number of coins flipped = 110

Therefore,

mu subscript p with hat on top end subscript equals p equals 0.44
sigma subscript p with hat on top end subscript equals square root of fraction numerator p open parentheses 1 minus p close parentheses over denominator n end fraction end root equals square root of fraction numerator 0.44 open parentheses 1 minus 0.44 close parentheses over denominator 110 end fraction end root equals 0.047329


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