The height of women ages 20–29 is normally distributed, with a mean of 64.5 inches. Assume sigmaσequals=2.6 inches. Are you more likely to randomly select 1 woman with a height less than 66.4 inches, or are you more likely to select a sample of 15 women with a mean height less than 66.4 inches? Explain. What is the probability of randomly selecting 1 woman with a height less than 66.4 inches? (Round to four decimal places as needed.) What is the probability of selecting a sample of 15 women with a mean height less than 66.4 inches? (Round to four decimal places as needed.) Are you more likely to randomly select 1 woman with a height less than 66.4 inches, or are you more likely to select a sample of 15 women with a mean height less than 66.4 inches? Choose the correct answer below. A. It is more likely to select 1 woman with a height less than 66.4 inches, because the probability is lower. B. It is more likely to select a sample of 15 women with a mean height less than 66.4 inches, because the sample of 15 has a higher probability. This is the correct answer. C. It is more likely to select 1 woman with a height less than 66.466.4 inches, because the probability is higher. Your answer is not correct. D. It is more likely to select a sample of 15 women with a mean height less than 66.4 inches, because the sample of 15 has a lower probability.

It is more likely to choose a sample of 15 with mean less than 66.4 inches, as the probability is higher.

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We are given a normal distribution with mean `mu=64.5`inches and standard deviation of `sigma=2.6`.

(a) What is the probability that a randomly chosen individual has height less than 66.4 inches? Since the distribution is (approximately) normal, we can convert the variable to a standard normal variable by `z=(x-mu)/sigma` .

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We are given a normal distribution with mean `mu=64.5`inches and standard deviation of `sigma=2.6`.

Here, `z=(66.4-64.5)/2.6~~0.73`. This essentially says that a height of 66.4 inches is about .73 standard deviations above the mean. We can consult a standard normal table (or technology) to find the probability that a randomly chosen z score is less than 0.73. (Thus, `P(x<66.4) =P(z<0.73)`. )

From a table, we get `P(z<.73)~~0.7673` (my calculator gives .76754).

So, the probability that an individual is 66.4 inches or shorter is about 77%, or .77

(b) What is the probability that a randomly chosen group of 15 has a mean less than 66.4?

Now we use the central limit theorem. The standard error is `sigma/sqrt(n)` .

So, `z=(x-mu)/(sigma/sqrt(n))=(66.4-64.5)/(2.6/sqrt(15))~~2.83`.

Now, `P(bar(x)<66.4)=P(z<2.83)` , which from a table or technology is .9977.

So the probability an individual is under 66.4 inches is about 77%, while the probability that a group of 15 has a mean under 66.4 inches is more than 99%.

The central limit theorem guarantees that as the sample size increases, the distribution of the means gets close to the mean with decreasing error.

This is intuitively obvious—it is far more likely to find one standout individual rather than randomly selecting 15 individuals whose average differs this much from the population mean.

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