Suppose that 20% of the adult women in the

United States dye or highlight their hair. We
would like to know the probability that a SRS of
size 200 would come within plus or minus 3
percentage points of this true value. In other
words, find probability that pˆ takes a value
between 0.17 and 0.23.

qhat

n jb

To find the probability that pˆ (the sample proportion) takes a value between 0.17 and 0.23, we can use the normal approximation to the binomial distribution since the sample size is large (n = 200).

Step 1: Calculate the mean and standard deviation of the sampling distribution.
- The mean of the sampling distribution is equal to the true population proportion, which is 20%, or 0.20.
- The standard deviation of the sampling distribution can be calculated using the formula: standard deviation = sqrt((p * (1-p))/n), where p is the true population proportion and n is the sample size.
- Plugging in the values, the standard deviation is sqrt((0.20 * (1-0.20))/200) = sqrt(0.0004) = 0.02.

Step 2: Calculate the standardized z-scores for the lower and upper bounds.
- The lower bound is 0.17, so the z-score for the lower bound is (0.17 - 0.20) / 0.02 = -0.15 / 0.02 = -7.5.
- The upper bound is 0.23, so the z-score for the upper bound is (0.23 - 0.20) / 0.02 = 0.03 / 0.02 = 1.5.

Step 3: Calculate the area under the standard normal curve between the z-scores.
- We want to find the probability between -7.5 and 1.5, which represents the area under the standard normal curve.
- Using a normal distribution table or calculator, we find that the area to the left of -7.5 is essentially 0, and the area to the left of 1.5 is approximately 0.9332.
- Therefore, the area between -7.5 and 1.5 is approximately 0.9332.

Step 4: Calculate the probability.
- Since the probability is the area under the curve between the two z-scores, the probability is approximately 0.9332.

Therefore, the probability that pˆ takes a value between 0.17 and 0.23 is approximately 0.9332, or 93.32%.

To find the probability that the sample proportion (p̂) takes a value between 0.17 and 0.23 with a sample size of 200, we can use the Central Limit Theorem (CLT) and assume that the distribution of the sample proportion is approximately normal.

The Central Limit Theorem states that, for a large enough sample size (n > 30), the distribution of the sample proportion follows a normal distribution with mean equal to the population proportion (p) and standard deviation equal to the square root of (p * (1 - p)) divided by the square root of the sample size.

In this case, the population proportion (p) is given as 20% or 0.20, and the sample size (n) is 200.

First, calculate the standard deviation (σ) using the formula:
σ = sqrt(p * (1 - p) / n)
σ = sqrt(0.20 * (1 - 0.20) / 200)
σ = sqrt(0.16 / 200)
σ ≈ 0.0283

Next, we need to find the z-scores corresponding to the lower and upper bounds of the desired range of p̂. The z-score formula is:
z = (x - μ) / σ

For the lower bound, the z-score can be calculated as:
z_lower = (0.17 - 0.20) / 0.0283 = -1.059

For the upper bound, the z-score can be calculated as:
z_upper = (0.23 - 0.20) / 0.0283 = 1.059

Now we can use a standard normal distribution table or calculator to find the probabilities associated with these z-scores.

The probability that p̂ takes a value between 0.17 and 0.23 is equal to the area under the normal curve between the corresponding z-scores.

Using a standard normal distribution table or calculator, we find the area to the left of z = -1.059 is approximately 0.1443, and the area to the left of z = 1.059 is approximately 0.8557.

Therefore, the probability that p̂ takes a value between 0.17 and 0.23 is approximately 0.8557 - 0.1443 = 0.7114, or 71.14%.

So, there is approximately a 71.14% probability that a random sample of 200 adult women in the United States would have a sample proportion (p̂) between 0.17 and 0.23.