Question

Can someone please help me solve this? In a set of data that show a normal distribution the mean is 7.8 and the standard deviation is 1.7. What is the proportion of numbers that are greater than 9? Show all work. (Hint: Find the z-score and use the z-table.)

Answer 1

I don't understand how to find the data set or how to find which ones are over 9....

Answer 2

You don't need to know any more about the data set than its mean and standard deviation.

What you're trying to do is essentially find the probability that any number you pick from the data set is larger than \(9\). So if \(X\) is a random variable denoting any number in the data set, then you are looking for the value of \(\mathbb P(X>9)\).

To do this, you need to transform the current distribution to the standard normal distribution. Both distributions are normal, but the current one is trickier to work with than the standard one which has mean \(0\) and standard deviation \(1\).

We use the following rule: \(Z=\dfrac{X-\mu}{\sigma}\), where \(\mu\) is the mean and \(\sigma\) is the standard deviation of the distribution for \(X\).

Now,
\[\mathbb P(X>9)=\mathbb P\left(\frac{X-7.8}{1.7}>\frac{9-7.8}{1.7}\right)\approx\mathbb P(Z>0.7059)\]You can consult a right-tailed \(z\) table to determine the answer, or you can use the calculator here:
http://www.wolframalpha.com/input/?i=p(z%3E.7059)