Question

In an election, suppose that 55% of voters support a school levy increase. If we poll 195 of these voters at random, the probability distribution for the proportion of the polled voters that support a school levy increase can be modeled by the normal distibution pictured below. Complete the boxes accurate to two decimal places (give your answers in decimal form, not percentage form). The boxes are at -2 standard deviations from the mean, the mean, and +2 standard

Answer 1

The probability distribution for the proportion of voters supporting a school levy increase can be approximated by a normal distribution. The mean and standard deviation can be determined using the given information. The -2 standard deviations, mean, and +2 standard deviations can then be calculated.

Step-by-step explanation:

The probability distribution for the proportion of the polled voters that support a school levy increase can be approximated by a normal distribution. To determine the mean and standard deviation of this distribution, we use the given information that 55% of voters support the school levy increase. This means that the mean (μ) is 0.55 and the standard deviation (σ) can be calculated as follows:

σ = √(p(1-p)/n)

Where p is the proportion of voters supporting the school levy increase (0.55) and n is the number of voters polled (195). Using this formula, we can find that the standard deviation is approximately 0.0427. Now, we can answer the questions based on this normal distribution.

1. The -2 standard deviations from the mean is μ - 2σ = 0.55 - 2(0.0427) = 0.4646
2. The mean is μ = 0.55
3. The +2 standard deviations from the mean is μ + 2σ = 0.55 + 2(0.0427) = 0.6374