Assume that women's heights are normally distributed with a mean of 63.6 and a standard deviation of 2.5 a) If 90 woman is randomly selected, find the probability that they have a mean height between 62.9 and 64.0 inches.
To find the probability that the mean height of a randomly selected group of 90 women is between 62.9 and 64.0 inches, we need to use the Central Limit Theorem and the properties of the normal distribution.
First, let's determine the mean (μ) and the standard deviation (σ) of the sampling distribution of the sample means. The mean of the sample means will be the same as the population mean, which is 63.6 inches. The standard deviation of the sample means can be calculated using the formula σ/√n, where σ is the population standard deviation and n is the sample size.
Given that the population standard deviation (σ) is 2.5 and the sample size (n) is 90, we can calculate the standard deviation of the sample means:
σ_sample_means = σ / √n = 2.5 / √90 ≈ 0.263
Now we can use the Central Limit Theorem to approximate the sampling distribution of the sample means as a normal distribution. The mean of the sampling distribution is still the population mean, which is 63.6 inches, and the standard deviation is the standard deviation of the sample means we calculated above, approximately 0.263 inches.
To find the probability that the mean height of the 90 randomly selected women falls between 62.9 and 64.0 inches, we need to calculate the z-scores for these values and then use the z-table or a statistical calculator to find the corresponding probabilities.
To calculate the z-scores, we use the formula z = (x - μ) / σ, where x is the given value, μ is the mean, and σ is the standard deviation.
For x = 62.9:
z1 = (62.9 - 63.6) / 0.263 ≈ -2.66
For x = 64.0:
z2 = (64.0 - 63.6) / 0.263 ≈ 1.52
Now we can use the z-table or a statistical calculator to find the probabilities associated with these z-scores:
P(-2.66 < z < 1.52)
Using a z-table or a statistical calculator, we find that P(z < -2.66) ≈ 0.0035 and P(z < 1.52) ≈ 0.9362.
To find the probability between these z-scores, we subtract the smaller probability from the larger:
P(-2.66 < z < 1.52) ≈ 0.9362 - 0.0035 ≈ 0.9327
Therefore, the probability that the mean height of a randomly selected group of 90 women is between 62.9 and 64.0 inches is approximately 0.9327.
you can play around with Z table stuff at
http://davidmlane.com/hyperstat/z_table.html