Question

unit 1: one variable data l6 practice - measures of center and spread

1. a random sample of fruit juices in the store shows the following calories per serving: 150, 110, 100, 35, 60, 130, 40, 140, 120, 160, 110.

a) find and interpret the mean of the data.
b) find and interpret the median of the data.
c) which is a better measure of center? why?

Explanation:

Step1: Calculate the mean for part a

The formula for the mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$. For the data set $130,40,140,120,160,110$, $n = 6$, and $\sum_{i=1}^{6}x_{i}=130 + 40+140+120+160+110=700$. So, $\bar{x}=\frac{700}{6}\approx116.67$. This value represents the average number of calories per - serving in the sample of fruit juices.

Step2: Calculate the median for part a

First, order the data set: $40,110,120,130,140,160$. Since $n = 6$ (an even number), the median $M=\frac{120 + 130}{2}=125$. The median is the middle - value of the ordered data set, splitting the data into two equal - sized halves.

Step3: Calculate the mean for part b

For the data set $150,110,100,35,60$, $n = 5$, and $\sum_{i = 1}^{5}x_{i}=150+110+100+35+60=455$. So, $\bar{x}=\frac{455}{5}=91$.

Step4: Calculate the median for part b

Order the data set: $35,60,100,110,150$. Since $n = 5$ (an odd number), the median $M = 100$.

Step5: Determine the better measure of center

The presence of the outlier $40$ in the first data set and $35$ in the second data set affects the mean. The median is less affected by outliers. So, the median is a better measure of center when there are outliers in the data set.

Answer:

a) Mean: $\approx116.67$, represents the average calories per serving. Median: $125$, is the middle - value of the ordered data.
b) Mean: $91$, represents the average calories per serving. Median: $100$, is the middle - value of the ordered data.
c) The median is a better measure of center because it is less affected by outliers.