In section 8.2 we saw that the **Normal Approximation** works for calculating probabilities about sample proportions or sample means that arise from a large number of independent trials. How large is large? A few rules of thumb are useful:

For sample proportions: the normal approximation works pretty well as long as \((n)(p)\) and \((n)(1-p)\) are both at least five.

For sample means: the normal approximation works pretty well if \(n > 15\) as long as the distribution isn’t too strongly skewed and there are no outliers. It will typically work well for \( n > 25\) or \(30\) if the population distribution is moderately skewed or at least 40 if it is more strongly skewed.

**: The normal approximation works better when the original population distribution is closer to a normal curve. The key point**

##
Example 8.4
Section* *

The median household income in Franklin County Ohio is about \$43,000 although the average household income is closer to \$54,000. The standard deviation of household incomes is about \$30,000. You pick a random sample of 50 households. What is the chance the average household income in your sample is over \$60,000?

#### Solution

Because the median and mean are not equal, this suggests that the population distribution of household incomes is quite skewed. However, even though the distribution is not bell-shaped, this is a random sample and the sample size of 50 households is large enough for us to use the normal approximation.The sampling distribution of the sample mean is approximately normal with mean \$54,000 and...

standard deviation =\(\dfrac{\$30,000}{\sqrt{50}}\) = \$4,243.

The standard score of \$60,000 is*...*

\(z = \dfrac{(60,000 – 54,000)}{ 4243} = +1.41\)

From Table 8.1, the probability of being larger that +1.4 is about 1 – 0.92 = 0.08 or about 8%.

##
Example 8.5
Section* *

You pick a random sample of 4 households. What is the chance the average household income in your sample is over \$60,000?

#### Solution* *

Because the distribution is so right-skewed (mean much bigger than median), the normal approximation may not be appropriate for such a small sample size.##
Example 8.6
Section* *

Seventy-five percent (75%) of the flights from Philadelphia International airport leave on time. Would it be appropriate to use the Normal Approximation to calculate the chance that more than 80 of the next 100 flights leave on time?

#### Solution

No; whether different flights leave on time will not be independent since the reasons for delayed flights often affect many flights (e.g. the weather is bad in Philadelphia or at a major connection).##
Example 8.7
Section* *

The GPAs of the students in a large statistics class average 3.0 with a standard deviation of 0.5 and roughly follow the normal curve. What’s the chance that 4 randomly selected students to have an average GPA between 2.85 and 3.15?

#### Solution

Even though n is only 4, it is okay to use the normal approximation since the population distribution follows the normal distribution already. Here, the standard deviation of the sample mean is \(\frac{0.50}{\sqrt{4}} = 0.25\). Thus the standard scores are

\(z = \dfrac{(2.85-3) }{ 0.25} = -0.6\) and \(z = \dfrac{(3.15-3)}{ 0.25} = 0.6\)

From Table 8.1 the answer is 72.5% - 27.5% = 45%.