Question

decide if each statement is true or false. explain your reasoning.

1. the standard deviation of christchurchs temperatures is zero because the data is symmetric.
2. the standard deviation of st. louiss temperatures is equal to the standard deviation of chicagos temperatures.
3. the standard deviation of chicagos temperatures is less than the standard deviation of londons temperatures.

Explanation:

Step1: Recall standard - deviation concept

Standard deviation measures the amount of variation or dispersion in a data - set. Just because a data - set is symmetric does not mean the standard deviation is zero. There is still variation in the data.

Step2: Analyze Christchurch data

The Christchurch data has non - zero variation as the data points are spread out around the mean, so the statement "The standard deviation of Christchurch's temperatures is zero because the data is symmetric" is false.

Step3: Understand the effect of a constant shift

If one data - set is a constant shift of another data - set, the standard deviation remains the same. Chicago's temperatures are the same as St. Louis's but 3 degrees cooler each day. This is a constant shift, and the two data sets have the same variability in temperature, so their standard deviations are equal.

Step4: Compare Chicago and London data

London has the same temperatures as Chicago except on the last day, where London's temperature is 3 degrees warmer than Chicago's. This means London's data has more variability than Chicago's data, so the standard deviation of Chicago's temperatures is less than the standard deviation of London's temperatures.

Answer:

1. False. Standard deviation measures variability. There is variability in the Christchurch data set even though it is symmetric.
2. True. Chicago's data is a constant shift of St. Louis's data, so they have the same variability and thus the same standard deviation.
3. True. London has more variability than Chicago due to the difference on the last day, so Chicago's standard deviation is less than London's.