OPTIMA Statistics

We start by drawing a graph of the density curve for \(X\text{.}\) Because the distribution is uniform, this means the curve will be a horizontal line above the range of possible values of \(X\text{,}\) which is from 5 to 10. Since the area under the curve must equal the probability and the probability that \(X\) is between 5 and 10 is 1 (this is the sample space because all possible values of \(X\) are between 5 and 10), we must make the height \(h\) the number for which \(h \times (10-5) = 1\text{.}\) That means, \(h \times 5 = 1\text{,}\) so \(h = \frac{1}{5}\text{.}\) This yields the first graph.

Now let's look at how this graph can help us to find the probability that \(X\) has a value in the ranges listed above. For each of the ranges of values of \(X\) given, we shade in the area under the graph and then find that area using \(\text{width } \times \text{ height}\) (since this is a rectangular region).

1. The probability of \(X\) between 7 and 8 is found by taking the width of the range \((8 - 7) = 1\) and multiplying by the height, which is \(\frac{1}{5}\text{.}\) This gives a probability of

   \begin{equation*} P(7 \leq X \leq 8) = 1 \times \frac{1}{5} = \frac{1}{5}\text{.} \end{equation*}

   

2. The probability that \(X\) is more than 8.5 is the area in the shaded region shown. That has a width of \((10 - 8.5) = 1.5 = \frac{3}{2}\) and a height of \(\frac{1}{5}\text{.}\) Therefore,

   \begin{equation*} P(X \geq 8.5) = \frac{3}{2} \times \frac{1}{5} = \frac{3}{10}\text{.} \end{equation*}

   

Based on the distributions above, there are two observations that can be made:

• It is clear that the average woman is shorter than the average man. Note that he peak for women is at \(\mu_X = 63.6\) inches while for men it is at \(\mu_Y = 69.0\) inches. Graphically, this is represented by the fact that the peak of the blue curve is to the left of the peak of the red curve.

• The variability in womens' heights is less than that in mens' heights. This can be seen either by observing that for women \(\sigma_X = 2.5\) is less than \(\sigma_Y = 2.8\) for men. Graphically, notice that the blue graph is taller and does not spread as wide as the red graph.

Using the normal distribution from the last example, we have shaded the appropriate area. This area is the probability we are looking for.

The tables give us the probability of the shaded area in the picture. The left-most column gives the first two digits of the z-score, and the column headers give the second decimal place.

1. \(P(Z \lt 1.10)\).

   This is the probability that \(Z\) is less than 1.10. To find this we need to look up 1.10 in the table. We go to the 1.10 in the left-most column and then go over to the 0.00 column. The blue lines show these entries. The entry in the blue box is then the probability that \(Z \lt 1.10\text{.}\) So, our answer is \(0.8643\text{.}\)

2. \(P(Z \gt 1.24)\).

   This is the probability that \(Z\) is greater than 1.24. We must find this in two steps. First, we find \(P(Z \lt 1.24)\) using the table. The red lines show how this is done, by going to the 1.20 entry in the left-most column, and then over to the 0.04 entry in the column headings. The boxed entry, \(0.8925\text{,}\) is \(P(Z \lt 1.24)\text{.}\) But we want \(P(Z \gt 1.24)\text{.}\) This is the complement, so using the complement rule,

   \begin{equation*} P(Z \gt 1.24) = 1 - P(Z \lt 1.24) = 1 - 0.8925 = 0.1075\text{.} \end{equation*}

1. \(P(Z \lt -1.82)\).

   The probability that \(Z\) is less than -1.82 can be found by looking up -1.8 row, and then going over to the 0.02 column. The red lines in the table above show this computation. The boxed probability gives \(P(Z \lt -1.82) = 0.0344\text{.}\)

2. \(P(Z \gt -1.68)\).

   The blue lines in the table above show how the probability 0.0465 is found. Note, however, that this is the probability that \(Z \lt -1.68\text{.}\) To get the probability that \(Z\) is greater than -1.68, we again use the complement rule to get:

   \begin{equation*} P(Z \gt -1.68) = 1 - P(Z \lt - 1.68) = 1 - 0.0465 = 0.9535\text{.} \end{equation*}

We begin by sketching the standard normal distribution and the area corresponding to the probability we want.

Now, according to the positive standard normal distribution table, \(P(Z \lt 1.24) = 0.8925\text{.}\) That means that the area under the curve from \(z=1.24\) all the way to the left is \(0.8925\text{.}\) From the negative standard normal distribution table, \(P(Z \lt -1.82) = 0.0344\text{.}\) So the area from \(z=-1.82\) all the way to the left is \(0.0344\text{.}\) Subtracting these leaves us with the area of the shaded region as shown below.

1. \(P(X \lt 40.1)\).

   We use the z-score formula to convert our range involving \(X\) into one involving \(Z\) and then use the standard normal distribution table to look up that probability.

   \begin{equation*} P(X \lt 40.1) = P\left(\frac{X-\mu}{\sigma} \lt \frac{40.1-37.5}{4.32}\right) = P(Z \lt 0.60) = 0.7257\text{.} \end{equation*}

2. \(P(X \gt 29.7)\).

   Since this range is in the right tail of the distribution, we make use of the complelemtn rule before we convert to a standard normal distribution.

   \begin{align*} P(X \gt 29.7) \amp = 1 - P(X \lt 29.7)\\ \amp = 1 - P\left(\frac{X-\mu}{\sigma} \lt \frac{29.7 - 37.5}{4.32}\right)\\ \amp = 1 - P(Z \lt -1.81)\\ \amp = 1 - 0.0351\\ \amp = 0.9649\text{.} \end{align*}

3. \(P(12.5 \lt X \lt 43.2)\).

   For this problem we need z-scores for both the bottom and top of the range of values.

   \begin{align*} P(12.5 \lt X \lt 43.2) \amp = P\left(\frac{12.5 - 37.5}{4.32} \lt \frac{X-\mu}{\sigma} \lt \frac{43.2 - 37.5}{4.32}\right)\\ \amp = P(-5.79 \lt Z \lt 1.32)\\ \amp = 0.9066 - 0.0001\\ \amp = 0.9065\text{.} \end{align*}

   We didn't find the z-score -5.79 in the table. But the note at the bottom says that for z-scores less than -3.49, we should use a probability of 0.0001.

If the manufacturer offers the 2-year warranty, then only computers that go bad within the first 24 months will be eligible for service. Therefore, we want to know what the probability is that \(X \lt 25\) (within the first 24 months). This has a probability of

So 22.96%, or approximately 23% of their computers will require service.

On the other hand, if they offer a 3-year warranty, then computers that go bad within the first 36 months will receive warranty service. The probability of this is

In this case, 74.86%, or almost 75% of the computers will require warranty service.

The company should probably stick with the 2-year plan.

Since we are using the table in reverse, we will look in the body of the table for the given probability, and find the z-scores in the row and column headings.

1. The bottom 80% of data in a normal distribution would have probability 0.8000, as shown in the picture below.

   

   We look in the body of the table to find the probability closest to 0.8000. This is 0.7995 going with the z-score of 0.84.

2. The second problem may look familiar from the empirical rule. Since 68% of the data is in this middle range, that leaves \(1 - 0.68 = 0.32\text{,}\) or 32% of the data in the two tails. Since the distribution is symmetric, the left and right tails will contain the same proportion, which is \(0.32 \div 2 = 0.16\text{.}\) This is shown in the following picture.

   

   • Looking up 0.1600 in the table we find 0.1611 goes with z = -0.99 and 0.1587 goes with z=-1.00. So the z-score is very close to -1.00 for the bottom of this range of values.

   • If there is 0.16 in the right tail, then we need to look up 1 - 0.16 = 0.84 in the table. the closest probability in the table is 0.8413. This goes with the z-score of 1.00.

   Therefore, the middle 68% lies within the range of 1 standard deviation above and 1 standard deviation below the mean.

We will first determine the z-scores that separate the A's and B's from the rest of the grades, and then turn those z-scores into exam scores. The image below will serve as our guide.

• The top 10% of the class will get A's, meaning the bottom 90% will not. Therefore, we are looking for the z-score that separates the bottom 90% from the top 10%. Looking for 0.9000 in the table, the closest we find is 0.8997, which goes with a z-score of \(z_A=1.28\text{.}\)

• The top 25% will be separated by the z-score which has probability 0.7500 in the table. The closest probability is 0.7486. The associated z-score is \(z_B=0.67\text{.}\)

Now we need to turn these z-scores back into exam scores. That means we need to use the z-score formula in reverse. This yields, for A's, the following.

It appears that those with a score of 80.14 or better will get A's. In practical terms, this probably means you need to score 80.5 or 81 to get an A, depending on if the professor gives half points.

For B's, we repeat the above using the z-score 0.67.

So a student needs to score between 73.072 and 80.14 to get a B.