Question

Guinea pig survival times Here are the survival times (in days) of 72 guinea pigs after they were injected with infectious bacteria in a medical experiment. 49 Survival times, whether of machines under stress or cancer patients after treatment, usually have distributions that are skewed to the right.

Part (a). Make a histogram of the data. Does it show the expected right skew?

Part (b). Now make a boxplot of the data.

Part (c). Compare the histogram from part (a) with the boxplot from part (b). Identify an aspect of the distribution that one graph reveals but the other does not.

Short answer

Part (a) Yes

Part (b)

The boxplot whiskers are at the minimum and maximum value (not including the outliers). The box begins at the first quartile and ends at the third, with a vertical line at the median.

Part (c)The boxplot reveals six outliers in the data set, whereas the histogram reveals no outliers.

Step-by-step solution

Part (a) Step 1. Given information

Survival times, whether of machines under stress or cancer patients after treatment, usually have distributions that are skewed to the right.

43, 45, 53, 56, 56, 57, 58, 66, 67, 73, 74, 79, 80, 80, 81, 81, 81, 82, 83, 83, 84, 88, 89, 91, 91, 92, 92, 97, 99, 99, 100, 100, 101, 102, 102, 102, 103, 104, 107, 108, 109, 113, 114, 118, 121, 123, 126, 128, 137, 138, 139, 144, 145, 147, 156, 162, 174, 178, 179, 184, 191, 198, 211, 214, 243, 249, 329, 380, 403, 511, 522, 598

Part (b) Step 2. Create a histogram from the data. Is it displaying the expected right skew

Table of frequencies

First, we'll calculate the frequency of each interval, which is the number of data values that fall within each interval. The first interval is 0-<100 because it begins at 0 and has a width of 100. The next intervals will be 100-<200, 200-<300, and so on until all data values are assigned to exactly one interval.

Interval	Frequency
0<100	30
100<200	32
200<300	4
300<400	2
400<500	1
500<600	3

Histogram of Frequency

Each bar must have the same width and be bounded by the interval boundaries, while the height must be equal to the frequency.

We can see that the distribution has the expected right skew because the highest bars in the histogram are to the left, with a tail of smaller bars to the right.

As a result:

Yes

Part (b) Step 1.  Make a boxplot of the data now.

It's worth noting that the data values are already sorted from smallest to largest.

The bare minimum is 43.

The median is the sorted data set's middle value. Because the data set contains 72 data values, the median is the average of the sorted data set's 36th and 37th data values.

$$\begin{gathered} M=Q_2=\frac{102+103}{2} \\ \frac{102+103}{2}=\frac{205}{2} \\ \frac{205}{2}=102.5 \end{gathered}$$

The first quartile is the median of the data values that are less than the median (or at 25 percent of the data). The first quartile is the average of the 18th and 19th data values because there are 36 data values below the median.

$$\begin{gathered} Q_1=\frac{82+83}{2}=\frac{165}{2} \\ \frac{82+83}{2}=\frac{165}{2} \\ \frac{165}{2}=82.5 \end{gathered}$$

The median of the data values above the median is the third quartile (or at 75 percent of the data). The third quartile is the average of the 54th and 55th data values because there are 36 data values above the median.

$$\begin{gathered} Q_3=\frac{147+156}{2} \\ \frac{147+156}{2}=\frac{303}{2} \\ \frac{303}{2}=151.5 \end{gathered}$$

The maximum number is 598.

The IQR is the difference between the third and first quartiles:

$$\begin{gathered} IQR=Q_3-Q_1 \\ 151.5-82.5=69 \end{gathered}$$

Outliers are observations that are more than 1.5 times the IQR higher than $Q_3$or lower than $Q_1$

$$\begin{gathered} Q_3+1.5IQR \\ 151.5+1.5\left(69\right)=255 \\ {Q}_1-1.5IQR \\ 82.5-1.5\left(69\right)=-21 \end{gathered}$$

Boxplot

The boxplot whiskers are at the minimum and maximum value (not including the outliers). The box begins at the first quartile and ends at the third, with a vertical line at the median.

The first quartile represents 25% of the sorted data list, the median 50%, and the third quartile 75%.

"X" denotes an outlier

Part (c) Step 1. Contrast the histogram in part (a) with the boxplot in part (b) (b). Identify a distribution feature that one graph reveals but the other does not.

The boxplot reveals six outliers in the data set (represented by the X's in the boxplot), whereas the histogram reveals no os (as there are no gaps in the histogram).

Both graphs show a right-skewed distribution because the highest bars in the histogram are to the left and the box of the boxplot is to the left between the whiskers.

As a result:

The boxplot reveals six outliers in the data set, whereas the histogram reveals no outliers.