Question

3 from unit 1, lesson 14
the mean vitamin c level for 20 dogs was 7.6 milligrams per liter, with a standard deviation of 2.1 milligrams per liter.
one dogs vitamin c level was not in the normal range. it was 0.9 milligrams per liter, which is a very low level of vitamin c.
a. if the value 0.9 is eliminated from the data set, does the mean increase or decrease?
b. if the value 0.9 is eliminated from the data set, does the standard deviation increase or decrease?
4 from unit 1, lesson 14
the data set represents the number of hours that 15 students walked during a two - week period.
6 6 7 8 8 8 9 10 10 12 13 14 15 16 30
the median is 10 hours, q1 is 8, q3 is 14, and the iqr is 6 hours. are there any outliers in the data?
explain or show your reasoning.

Explanation:

Step1: Analyze effect on mean

The original mean is 7.6. The value 0.9 is less than the mean. Removing a value less than the mean from the data - set will increase the sum of the remaining values relative to the number of remaining values. So the mean will increase.

Step2: Analyze effect on standard - deviation

The standard deviation measures the spread of the data. The value 0.9 is an outlier (far from the mean). Removing an outlier that is far from the mean will make the data more clustered around the mean, thus decreasing the standard deviation.

Step3: Check for outliers in second data - set

The formula to check for outliers is: Lower fence = Q1−1.5×IQR and Upper fence = Q3 + 1.5×IQR.
First, calculate the lower fence: Q1 = 8, IQR = 6, so Lower fence=8−1.5×6=8 - 9=-1.
Then, calculate the upper fence: Q3 = 14, so Upper fence=14 + 1.5×6=14 + 9 = 23.
The value 30 is greater than 23, so 30 is an outlier.

Answer:

a. The mean increases.
b. The standard deviation decreases.

1. Yes, 30 is an outlier. The lower fence is - 1 and the upper fence is 23, and 30 is greater than 23.