Question

ap 13. a carton of eggs is usually classified by the size of the eggs: medium, large, or extra large. bill raises chickens and sells the eggs at a farmstand. he weighs each egg in a carton of 12 so he can determine the size classification. the results are shown in the dot - plot. a. find the five - number summary for the distribution of egg weights. b. determine if there are any outliers using the 1.5×iqr rule. show your work. c. construct a boxplot of the distribution. d. without doing any calculations, how will the mean compare to the median? explain.

Explanation:

Step1: Arrange data in ascending order

From the dot - plot, the data set (in grams) is: 56, 56, 57, 57, 58, 59, 60, 61, 64, 65, 67, 68

Step2: Find the minimum

The minimum value of the data set is 56 grams.

Step3: Find the first quartile ($Q_1$)

Since $n = 12$, the position of $Q_1$ is $\frac{n + 1}{4}=\frac{12+1}{4}=3.25$. So, $Q_1$ is the value between the 3rd and 4th ordered data points. $Q_1=\frac{57 + 57}{2}=57$ grams.

Step4: Find the median

Since $n = 12$, the median is the average of the 6th and 7th ordered data points. Median$=\frac{59+60}{2}=59.5$ grams.

Step5: Find the third quartile ($Q_3$)

The position of $Q_3$ is $\frac{3(n + 1)}{4}=\frac{3\times(12 + 1)}{4}=9.75$. So, $Q_3$ is the value between the 9th and 10th ordered data points. $Q_3=\frac{64+65}{2}=64.5$ grams.

Step6: Find the maximum

The maximum value of the data set is 68 grams.

Step7: Calculate the inter - quartile range (IQR)

$IQR=Q_3 - Q_1=64.5 - 57 = 7.5$ grams.

Step8: Determine the lower and upper fences for outliers

Lower fence$=Q_1-1.5\times IQR=57-1.5\times7.5=57 - 11.25 = 45.75$ grams.
Upper fence$=Q_3 + 1.5\times IQR=64.5+1.5\times7.5=64.5 + 11.25 = 75.75$ grams. Since all data points are within 45.75 and 75.75, there are no outliers.

Step9: Analyze the relationship between mean and median

The distribution is skewed to the right (because the tail on the right - hand side is longer as there are larger values like 67 and 68). In a right - skewed distribution, the mean is greater than the median because the larger values in the right - tail pull the mean in that direction.

Answer:

a. Minimum: 56 grams, $Q_1$: 57 grams, Median: 59.5 grams, $Q_3$: 64.5 grams, Maximum: 68 grams
b. There are no outliers.
c. To construct a boxplot: Draw a number line from 56 to 68. Draw a box from 57 to 64.5 with a vertical line at 59.5 inside the box. Draw whiskers from the box to 56 and 68.
d. The mean is greater than the median because the distribution is right - skewed. The larger values in the right - tail pull the mean in the positive direction.