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MAT 208 (Intro Statistics)

Solution. Notice that this is a binomial distribution, since:

1. There are a fixed number of trials (\(n=12\)).

2. The trials are independent. (Whether one customer orders a beverage doesn't affect the probability of whether another customer orders a beverage.)

3. Each trial results in a success (ordering a beverage) or a failure (not ordering a beverage).

4. The probability of success in each trial is fixed (\(p=.41\), the decimal form of 41%).

We will use the "binomcdf" command in the TI-84 Plus calculator to compute the probability of at most 6 successes. (This will not give us our final answer.)

To navigate, we use [2ND] [VARS/DISTR] [binomcdf]:

Next we enter \(n=12\) (trials), \(p=.41\), and \(x=6\) (since we are computing the probability of "at most 6 sucesses"):

Then we [Paste] the command into the new window and press [ENTER], which gives us a result of .8235436756 (the probability of "at most 6 successes"). To convert to the probability of "at least 7 successes", we subtract that result from 1 (this is called the complement rule):

Note that rounding to four decimal places is typical. Hence, the probability of at least 7 of the first 12 customers ordering a beverage is:


click for video explanation

Solution. Since the durations are normally-distributed, we can sketch a diagram of the situation. The probability we seek is represented by the shaded area:

Notice that the lower bound is \(-\infty\) and that the upper bound is \(60\). In the TI-84 Plus, we can navigate to the \(\verb|DISTR|\) menu and use the command \(\verb|normalcdf|\).

Since the calculator can't handle negative infinity, for the lower bound we use the largest negative number it can handle: \(\verb|-10^99|\)

The mean \((\mu=65.6)\) and standard deviation \((\sigma=3.7)\) are given in the problem statement. When we enter the values into the calculator, it looks like this:

Now we can compute the probability:

Rounding to four decimal places, our final answer is:


click for video explanation

Solution. Note that .075 is the decimal form of 7.5%. Since the data is normally-distributed and the top 7.5% (or .075) gives us a right-tailed area, that means we are looking for the following cutoff value:

However, in order to use the TI-84 Plus, we need to determine the corresponding left-tailed area, which is

\(\begin{array}{ll}\text{left-tailed area}&=1-.075 \\&=.925\end{array}\)

When given an area and asked for a cutoff value, we must use the "inverse" normal function. In the TI-84 Plus, this is found by going to [2ND] [VARS/DISTR] [invNorm]:

Next, we enter \(.925\) for the area, and also \(\mu=3510\) and \(\sigma=385\). (These are the mean and standard deviation given in the problem statement.)

Now we can [Paste] this into the main window and press [Enter] to find the cutoff value:

Rounding to the nearest integer, we find that the top 7.5% of all birth weights among these infants are:

\(4064\text{ grams and above.}\)

click for video explanation

Solution. Here we must use the "sampling distribution of the mean" (symbolized \(\bar{x}\)), which has formulas

\(\mu_{\bar{x}}=\mu\;\;\;\text{ and }\;\;\;\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}},\)

where \(\mu=12\) and \(\sigma=3.6\) are the population mean and standard deviation, and \(n=40\) is the sample size. Hence,

\(\mu_{\bar{x}}=12\;\;\;\text{ and }\;\;\;\sigma_{\bar{x}}=\frac{3.6}{\sqrt{40}},\)

Since the sample size is at least 30, that means the sampling distribution is approximately normal. Hence we are looking for the following area:

Note that the lower bound of the area is \(12.5\), and the upper bound is \(\infty\). To find this area, we use the "normalcdf" command in the TI-84 Plus calculator, which we find by navigating to [2ND] [VARS/DISTR] [normalcdf]:

The lower bound is, as mentioned above, \(12.5\). For the upper bound since the TI-84 Plus cannot handle \(\infty\) directly, we use an approximation of \(\verb|10^99|\), which is one of the largest numbers that the calculator can handle. We also use \(\mu_{\bar{x}}=12\) and \(\sigma_{\bar{x}}=3.6/\sqrt{40}\), as discussed above.

Having entered these values, we [Paste] the command into the main window and press [ENTER] to obtain our area:

It is typical to round this result to four decimal places. Hence, the probability that a random sample of 40 deer has a mean lifespan of at least 12.5 years is:


click for video explanation