Question

name
cows milk. here are a histogram and summary statistics of data on the percentage of butterfat in milk from a random sample of 100 three - year - old ayshire cows.

1. determine whether the has low and/or high outliers using:

a. the 1.5 * iqr rule
b. the 2 * sd rule
c. which measures of center and variability should we choose to summarize this distribution? explain your answer.
directions: for each data set with q1 and q3 given, determine whether x is an outlier using the iqr rule.

1. $q_1 = 20, q_3 = 29, x = 35$

Explanation:

Step1: Calculate IQR for the first - part (a)

Given $Q_1 = 3.97$, $Q_3=4.37$. The inter - quartile range $IQR = Q_3 - Q_1=4.37 - 3.97 = 0.4$.

Step2: Find lower and upper bounds for outliers using 1.5 * IQR rule

Lower bound for outliers: $Q_1-1.5\times IQR=3.97 - 1.5\times0.4=3.97 - 0.6 = 3.37$.
Upper bound for outliers: $Q_3 + 1.5\times IQR=4.37+1.5\times0.4=4.37 + 0.6 = 4.97$. Since the minimum value is $3.52$ and the maximum value is $4.91$, there are no low or high outliers according to the 1.5 * IQR rule.

Step3: Use 2 * SD rule (part b)

Given $SD = 0.291$. The range within 2 * SD of the mean:
Lower limit: $\bar{x}-2\times SD$, upper limit: $\bar{x}+2\times SD$.
Since $\bar{x}=4.173$, lower limit: $4.173-2\times0.291=4.173 - 0.582 = 3.591$, upper limit: $4.173+2\times0.291=4.173 + 0.582 = 4.755$. Since the minimum value is $3.52$ and the maximum value is $4.91$, there are high outliers according to the 2 * SD rule.

Step4: Choose measures of center and variability (part c)

Since the distribution may have outliers (as shown by the 2 * SD rule), the median is a better measure of center than the mean because the mean is affected by outliers. The inter - quartile range (IQR) is a better measure of variability than the standard deviation (SD) as the SD is also affected by outliers.

Step5: Solve part 2

Given $Q_1 = 20$, $Q_3 = 29$, then $IQR=Q_3 - Q_1=29 - 20 = 9$.
Lower bound for outliers: $Q_1-1.5\times IQR=20-1.5\times9=20 - 13.5 = 6.5$.
Upper bound for outliers: $Q_3 + 1.5\times IQR=29+1.5\times9=29 + 13.5 = 42.5$. Since $x = 35$ and $6.5<35<42.5$, $x$ is not an outlier.

Answer:

1. a. No low or high outliers according to the 1.5 * IQR rule.

b. There are high outliers according to the 2 * SD rule.
c. Median for center and IQR for variability as the distribution may have outliers.

1. $x = 35$ is not an outlier.