Suppose that women's heights in the United States have a distribution that is normal with mean 65.2 inches and standard deviation 2.4 inches. If we randomly select 60 women from the US population, what is the probability that the sample mean, x-bar, will be greater than 66 inches?

Give your answer rounded to four decimal places.

Question

Suppose that women's heights in the United States have a distribution that is normal with mean 65.2 inches and standard deviation 2.4 inches. If we randomly select 60 women from the US population, what is the probability that the sample mean, x-bar, will be greater than 66 inches?

Give your answer rounded to four decimal places.

BlackTom AI · Accepted Answer

We start by identifying the given population parameters and the sampling distribution for the sample mean.

- Population mean (mu): 65.2 inches
- Population standard deviation (sigma): 2.4 inches
- Sample size (n): 60

For the sampling distribution of the sample mean x-bar, the mean is mu_xbar = mu = 65.2, and the standard deviation is sigma_xbar = sigma / sqrt(n) = 2.4 / sqrt(60).

Compute sqrt(60): sqrt(60) ≈ 7.745966.
Then sigma_xbar ≈ 2.4 / 7.745966 ≈ 0.3098 (rounded to four decimals, it is about 0.3098).

We want P(x-bar > 66). Convert to a Z-score using the standard normal distribution:
Z = (x-bar - mu_xbar) / sigma_xbar = (66 - 65.2) / 0.3098.
Compute the numerator: 66 - 65.2 = 0.8.
Thus Z ≈ 0.8 / 0.3098 ≈ 2.58 (more precisely about 2.582).

Now consult the standard normal distribution: P(Z > 2.58) is the upper tail probability beyond 2.58. The cumulative probability at 2.58 is approximately 0.995, so the tail probability is about 1 - 0.995 ≈ 0.005. Using a more precise value for Z ≈ 2.582 gives P(Z > 2.582) ≈ 0.0049.

Therefore, the probability that the sample mean of 60 women exceeds 66 inches is about 0.0049. Note: this result assumes the Central Limit Theorem applies nicely here due to the reasonably large sample size and the underlying population distribution being normal, so the sampling distribution of x-bar is Normal with the computed mean and standard deviation.

(If you were to show steps explicitly, you would include the calculation of sigma_xbar, the Z-score, and then the standard normal tail probability as shown above.)

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