OPTIMA Statistics

Obed's rate of $95 an hour is between \(Q_2\text{,}\) the second quartile or median, and \(Q_3\text{,}\) the third quartile. So he charges more than at least 50% of plumbers, but less than at most 75% of plumbers in his region.

The following statements interpret the percentiles used above.

1. Johnny's height was in the 20th percentile, meaning that he was as tall or taller than 20% of newborns and as short or shorter than 80%. So relative to other newborns, he was fairly short.

2. Isabella scored in the 95th percentile, meaning that her score was as good or better than 95% of children who took this achievement test. Stated from the opposite perspective, only 5% of children scored as well or better than Isabella.

The first step is to compute the five number summary. To do that, we must arrange the data in order.

Next we identify the following quartiles:

• Since there are eleven numbers, the second quartile, or median, is right in the middle at position 6. So \(Q_2 = 14\text{.}\)

• We will share this middle number between the upper and lower half of the data, so that there are six values in each half. The first quartile is therefore the average of the 3rd and 4th numbers, and \(Q_1 = \frac{10+12}{2} = 11\text{.}\)

• The third quartile is the median of the top six numbers, so it is \(Q_3 = \frac{16+17}{2} = 16.5\text{.}\)

Next, we compute the interquartile range and calculate where the fences should be placed. We get \(IQR = Q_3 - Q_1 = 16.5 - 11 = 5.5\text{.}\) So the upper and lower fences are at \(Q_1 - 1.5(\text{IQR}) = 11 - 1.5(5.5) = 2.75\) and at \(Q_3 + 1.5(\text{IQR}) = 16.5 + 1.5(5.5) = 24.75\text{.}\)

Note that the maximum value in the data set, 42, is above the upper fence. So it is an outlier and will not be included in the five number summary. The minimum and maximum vales of the remaining data are 9 and 19 respectively. This gives us the following five-number summary.

Using these, we construct the box-plot shown below, remembering to identify the outlier with an asterisk.

There are two ways to approach this problem, depending on which types of measures we want to use.

1. We could use the IQR to determine if a value is an outlier. In this method, we will find the first and third quartiles and then take anything that is more than \(1.5 \times \text{IQR}\) away from these values to be an outlier.

   Arranging the data in order gives:

   \begin{equation*} 4, 9, 10, 10, 12, 13. \end{equation*}

   So the quartiles are \(Q_1 = 9\) and \(Q_3 = 12\text{.}\) Thus, the IQR is \(12-9 = 3\) and our fences are at

   \begin{equation*} 9 - 1.5(3) = 4.5 \qquad \text{and} \qquad 12 + 1.5(3) = 16.5. \end{equation*}

   Therefore, 4 is an outlier on the low side.

2. Using the mean and standard deviation gives us another way to check for outliers. Computing these gives:

   \begin{equation*} \overline{x} = \frac{58}{6} \approx 9.7 \end{equation*}
   \begin{equation*} \scriptstyle s = \sqrt{\frac{(4-9.7)^2 + (10-9.7)^2 + (12-9.7)^2 + (13-9.7)^2 + (9-9.7)^2 + (10-9.7)^2}{5}} \approx 3.1. \end{equation*}

   So outliers will be less than \(\overline{x} - 3s = 9.7 - 3(3.1) = 0.4\) or bigger than \(\overline{x} + 3s = 9.7 + 3(3.1) = 19\text{.}\) Therefore, there are no outliers using this method.

The z-score for 4 is:

This means that 4 lies approximately 1.83 standard deviations below (because the z-score is negative) the mean. Since it is not at least three standard deviations above or below the mean, we do not consider it an outlier.

The first exam score of 70%, or 0.70, has a z-score of

The second exam score of 70%, or 0.70, has a z-score of

Since \(z_1 > z_2\text{,}\) the score on the first exam is “higher” than the score on the second exam relative to the rest of the data.

Even though both scores are exactly the same value, 70%, a 70% on the first exam is actually a better score than a 70% on the second exam. This is because the first exam has a smaller standard deviation, meaning that the scores are more tightly grouped around the mean of 65%.

1. Since \(82.9 = 102.5 - 2(9.8)\) and \(122.1 = 102.5+2(9.8)\text{,}\) the range extends two standard deviations above and below the mean. So the empirical rule says that 95% of students will score in this range.

2. According to the empirical rule, 68% of students scored within one standard deviation of the mean. So this range is from \(102.5 - 1(9.8) = 92.7\) and \(102.5 + 1(9.8) = 112.3\text{.}\)

3. Finally, The top 0.15%, which is half of 0.3%, scored at least 3 standard deviations above the mean. So their score is at least \(102.5 + 3(9.8) = 131.9\text{.}\)

We can no longer use the empirical rule, since we do not know if the distribution is mound shaped. Instead, we use Chebyshev's Inequality.

1. Our first step is to figure out how many standard deviations 82.9 and 122.1 are on either side of the mean. This is the value of \(k\) in Chebyshev's inequality. We compute that by finding the difference between the mean and the lower bound of our range and dividing by \(\sigma\text{.}\)

   \begin{equation*} k = \frac{\mu - \text{lower bound}}{\sigma} = \frac{102.5-82.9}{9.8} = 2 \end{equation*}

   We could have found the same value using the upper bound.

   \begin{equation*} k = \frac{\text{upper bound} - \mu}{\sigma} = \frac{122.1-102.5}{9.8} = 2 \end{equation*}

   Therefore, the percent of values in this range is

   \begin{equation*} P \geq 1-\frac{1}{2^2} = 1 - \frac{1}{4} = \frac{3}{4} = 0.75 \text{ or } 75\% \end{equation*}

   Note that this is less than the 95% the empirical rule gave us. That is because we can not be as certain how the data is distributed if we don't know the shape of the distribution.

2. For the second part, we again start by finding the value of \(k\text{.}\)

   \begin{equation*} k = \frac{102.5 - 87.8}{9.8} = 1.5 \end{equation*}

   So our range is 1.5 standard deviations on either side of the mean. Plugging this in we get

   \begin{equation*} P \geq 1-\frac{1}{(1.5)^2} = 1-\frac{1}{2.25} \approx 0.5556. \end{equation*}

   So at least 55.6% of the data is within 1.5 standard deviations of the mean.

   Note that this problem would not be possible with the empirical rule, even if we knew the distribution was mound shaped. 1.5 standard deviations is not one of the three options given by the empirical rule.

In this problem we are given the value of \(P\text{,}\) the proportion that lies within \(k\) standard deviations of the mean. So we will use it to find \(k\text{.}\)

So the range of monthy sales is at most two standard deviations on either side of the mean, or from a minimum of \(\$92,000-2(12,500) = \$67,000\) to a maximum of \(\$92,000+2(12,500) = \$117,000\)

As in Example 1.4.25, we are given a range of values. In that example, we used it to find \(k\)k, but this required that we know the standard deviation \(\sigma\text{.}\) But we are also given a percent of the population in the range, as in Example 1.4.26. So we can start by finding \(k\) with this information.

Thus, we know that the maximum range will be from 10 standard deviations below the mean to 10 standard deviations above the mean. Or, using the minimum and expressing this symbolically,

Thus, the standard deviation is approximately 0.17 hours, or 10.2 minutes.