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` .

Here,...

## See

This Answer NowStart your **48-hour free trial** to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Already a member? Log in here.

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` .

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.

**Further Reading**