]> Brase Sections 7.4 and 8.2

Special: Sections 7.4 and 8.2

Section 7.4: The Normal Approximation to the Binomial Distribution

When we dealt with the Binomial Distribution in Chapter 6, we had a couple of tools that made it much more palatable—binompdf( and binomcdf(. However, there is another way to deal with the binomial—the normal!

If np > 5 and nq > 5, then the binomial distribution (with n and p) can be approximated by a normal distribution with $μ = n \cdot p$ and $σ = \sqrt{n \cdot p \cdot q}$ .

There is one catch—the binomial distribution is discrete; you can ask about having 4 successes, or 5 successes, but not 4.5 successes! The normal distribution is continuous—a value of 4.5 won't cause any problems at all. To fix this, we need to adjust values of r (the number of successes in the binomial) when we convert them to x (a value in a normal distribution). In particular, you'll need to subtract 0.5 from a value of r on the left of a desired area, and you'll need to add 0.5 to an r on the right of a desired area.

You need to watch the wording! In particular, you want/need to write all inequalities using the "or equal to" option.

Examples:

[1.] At one time, Southwest Airlines reported that 16% of ticketed passengers do not show up for flights. If this is true, then what is the probability that at least 5 of the next 42 ticketed passengers do not show for their flights?

n = 42, p = 0.16, and r ≥ 5. 5 is the lowest possible value of r for the question—that means that this is on the left side of the desired area. The continuity correction will make that r ≥ 4.5.

$μ = 42 \cdot 0.16 = 6.72$ and $σ = \sqrt{42 \cdot 0.16 \cdot 0.84} = 2.38$ . $z = \frac{4.5 - 6.72}{2.68} = - 0.93$ . r ≥ 4.5 is the same as z ≥ -0.93. The table entry for this z is 0.1762—but that's area to the left! The answer is 82.38%.

[2.] Back in 2001, USA Today reported that Anchorage was the city with the highest cell-phone ownership rate (at a whopping 56%!). Assuming that this is still true, then what is the probability that fewer than 220 out of 400 residents will have cell-phones?

n = 400, p = 0.56, and r ≤ 219. Since 219 is the largest possible value for r, it is on the right-hand side of the desired area. The continuity correction will make that r ≤ 219.5.

$μ = 400 \cdot 0.56 = 224$ and $σ = \sqrt{400 \cdot 0.56 \cdot 0.44} = 9.93$ . $z = \frac{219.5 - 224}{9.93} = - 0.45$ . r ≤ 219.5 is the same as z ≤ -0.45. The table entry for that z is 0.3264. The answer is 32.64%.

Section 8.2: The Sampling Distribution of the Sample Mean

Back in Chapter 7, we talked about the normal distribution—but only in the context of single measurements. Think about it—the probability that one IQ is higher than 115; the probability that one pregnancy lasts less than 218 days…in reality, we (almost) never take just one measurement; we never measure just one individual. We (almost) always take measurements from a large group of individuals. Afterwards, we summarize those measurements with numbers—the mean, the median, the standard deviation, etc.

So the distribution of a single measurement (x) isn't really interesting…what we really need is the distribution of a summary number, like $\overline{x}$ !

Fortunately, we have The Central Limit Theorem: as the sample size (number of measurements) increases, the distribution of $\overline{x}$ approaches a normal distribution, with mean $μ_{\bar{x}} = μ_{x}$ and standard deviation $σ_{\bar{x}} = \frac{σ_{x}}{\sqrt{n}}$ . Thus, for large sample sizes (say, at least 30), the distribution of $\overline{x}$ is just another normal distribution. When the sample size is small (say, less than 30), then you need to know that the distribution of the original variable (x) is normal, in order to say that the distribution of $\overline{x}$ is approximately normal.

Fortunately, most of this is at the very edge of what we need to know—we won't dwell on the details, or the distinction. Let it suffice to say that when a question asks about the probability of obtaining a certain mean value, then you need these new formulas.

Let's look at a couple of examples…

Examples:

[3.] IQ scores are approximately normally distributed, with mean 100 and standard deviation 15. What is the probability that (a) a single IQ score is at least 120, and (b) the mean IQ score of 4 people is at least 120?

Note the difference in the two parts of the question!

The probability that a single IQ score is above 120 means stuff from Chapter 7… $z = \frac{120 - 100}{15} = 1.33$ , and P(x > 120) = P(z > 1.33) = 0.0912.

The probability that a means score is above 120…that's the new stuff; the question is asking about $\overline{x}$ ! Note that the mean for this distribution is the same as the original—100. The standard deviation, however, is different— $σ_{\bar{x}} = \frac{σ_{x}}{\sqrt{n}} = \frac{15}{\sqrt{4}} = 7.5$ . Thus, the z calculation will come out differently: $z = \frac{120 - 100}{7.5} = 2.67$ . $P (\bar{x} > 120) = P (z > 2.67) = 0.0038$ .

[4.] The number of people living in a U.S. household has mean 2.685 and standard deviation 1.47. What is the probability that a sample of 30 households will have a mean number of people of less than 2.5?

This is asking about the mean—so the mean stays the same (2.685), but the standard deviation becomes $\frac{1.47}{\sqrt{30}} = 0.2684$ . $z = \frac{2.5 - 2.685}{0.2684} = - 0.6893$ . $P (\bar{x} < 2.5) = P (z < - 0.6893) = 0.2453$ .

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