“How would you describe your own physical health at this time? Would you say physical health is—excellent, good, only fair, or poor?” The Gallup Poll asked this question of 1005 randomly selected adults, of whom 29% said “excellent.” Suppose that in fact the proportion of the adult population who say their health is excellent is 0.29. What is the probability that the sample proportion of an SRS of size 1000 who say their health is excellent lies between 26% and 32%?

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To find the probability that the sample proportion lies between 26% and 32%, we need to calculate the z-scores for both percentages and then find the area under the normal curve between these z-scores.

First, let's calculate the z-score for 26%:

Sample proportion of 26% corresponds to a z-score of (0.26 - 0.29) / sqrt(0.29 * (1 - 0.29) / 1000) = -1.344.

Next, let's calculate the z-score for 32%:

Sample proportion of 32% corresponds to a z-score of (0.32 - 0.29) / sqrt(0.29 * (1 - 0.29) / 1000) = 1.344.

Now, we need to find the area under the normal curve between these z-scores. This can be done by looking up the z-scores in a standard normal distribution table or by using a statistical software.

Using a z-table or a statistical software, the probability that the sample proportion lies between 26% and 32% is approximately 0.681.

To find the probability that the sample proportion of an SRS of size 1000 lies between 26% and 32% when the actual proportion is 0.29, we can use the normal approximation to the binomial distribution.

First, we need to calculate the standard deviation of the sample proportion. The formula for the standard deviation of a sample proportion is:

√(p(1-p)/n),

where p is the proportion of interest (0.29 in this case), and n is the sample size (1000 in this case).

Standard deviation = √(0.29(1-0.29)/1000) = 0.01549

Next, we standardize the range of 26% to 32% into z-scores using the formula:

z = (x - μ) / σ,

where x is the random variable of interest, μ is the mean (0.29 in this case), and σ is the standard deviation (0.01549 in this case).

For 26%:
z = (0.26 - 0.29) / 0.01549 = -1.939

For 32%:
z = (0.32 - 0.29) / 0.01549 = 1.939

Using a standard normal distribution table or a calculator, we can find the probabilities associated with these z-scores. Since the normal distribution is symmetrical, we can find the probability between -1.939 and 1.939 and subtract it from 1 to get the final probability.

P(-1.939 ≤ Z ≤ 1.939) = P(Z ≤ 1.939) - P(Z ≤ -1.939)

From the standard normal distribution table or calculator, we find:

P(Z ≤ 1.939) = 0.9732
P(Z ≤ -1.939) = 0.0268

Therefore,
P(-1.939 ≤ Z ≤ 1.939) = 0.9732 - 0.0268 = 0.9464

So, the probability that the sample proportion of an SRS of size 1000 who say their health is excellent lies between 26% and 32% when the actual proportion is 0.29 is 0.9464, or approximately 94.64%.