Questions: In an election, suppose that 60% of voters support creating a new fire district. If we poll 139 of these voters at random, the probability distribution for the proportion of the polled voters that support creating a new fire district can be modeled by the normal distribution pictured below. Complete the boxes accurate to two decimal places.

In an election, suppose that 60% of voters support creating a new fire district. If we poll 139 of these voters at random, the probability distribution for the proportion of the polled voters that support creating a new fire district can be modeled by the normal distribution pictured below. Complete the boxes accurate to two decimal places.
Transcript text: In an election, suppose that $60 \%$ of voters support creating a new fire district. If we poll 139 of these voters at random, the probability distribution for the proportion of the polled voters that support creating a new fire district can be modeled by the normal distibution pictured below. Complete the boxes accurate to two decimal places.
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Solution

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Solution Steps

Step 1: Identify the given information
  • Proportion of voters supporting the new fire district (p) = 0.60
  • Sample size (n) = 139
Step 2: Calculate the mean of the sampling distribution

The mean (μ) of the sampling distribution of the sample proportion is equal to the population proportion (p). μ=p=0.60 \mu = p = 0.60

Step 3: Calculate the standard deviation of the sampling distribution

The standard deviation (σ) of the sampling distribution of the sample proportion is calculated using the formula: σ=p(1p)n \sigma = \sqrt{\frac{p(1-p)}{n}} σ=0.60×0.40139 \sigma = \sqrt{\frac{0.60 \times 0.40}{139}} σ=0.24139 \sigma = \sqrt{\frac{0.24}{139}} σ=0.001726 \sigma = \sqrt{0.001726} σ0.04154 \sigma \approx 0.04154

Step 4: Determine the values for the normal distribution

The normal distribution is centered at the mean (0.60) and extends three standard deviations on either side. The values for the boxes are:

  • Mean (center): 0.60
  • One standard deviation below the mean: 0.600.041540.56 0.60 - 0.04154 \approx 0.56
  • One standard deviation above the mean: 0.60+0.041540.64 0.60 + 0.04154 \approx 0.64
  • Two standard deviations below the mean: 0.602×0.041540.52 0.60 - 2 \times 0.04154 \approx 0.52
  • Two standard deviations above the mean: 0.60+2×0.041540.68 0.60 + 2 \times 0.04154 \approx 0.68

Final Answer

  • The values for the boxes, accurate to two decimal places, are:
    • 0.52 0.52
    • 0.56 0.56
    • 0.64 0.64
    • 0.68 0.68
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