Mean and Standard Deviation
Practice Problems with Solutions

This page presents practice problems on mean and standard deviation, with complete solutions provided at the bottom of the page. Problems include raw data sets, frequency tables, and grouped data.

Problems

1. Consider the following three data sets:
   
   \( A = \{9, 10, 11, 7, 13\} \)
   \( B = \{10, 10, 10, 10, 10\} \)
   \( C = \{1, 1, 10, 19, 19\} \)
   
   a) Calculate the mean of each data set.
   b) Calculate the standard deviation of each data set.
   c) Which data set has the largest standard deviation?
   d) Can question (c) be answered without calculations?
2. A data set has mean \( \mu \) and standard deviation \( \sigma \).
   
   a) What happens to the mean and standard deviation if a constant \( k \) is added to every data value? Explain.
   b) What happens if every data value is multiplied by \( k \)? Explain.
3. If the standard deviation of a data set is zero, what can be concluded about the data values?
4. The frequency table below shows the monthly salaries of 20 people.
   
   

   Salary ($)	Frequency
   3500	5
   4000	8
   4200	5
   4300	2

   
   a) Calculate the mean salary.
   b) Calculate the standard deviation.
5. The following table shows grouped height data for 50 people.
   
   

   Height (cm)	Frequency
   \((120,130]\)	2
   \((130,140]\)	5
   \((140,150]\)	25
   \((150,160]\)	10
   \((160,170]\)	8

   
   a) Calculate the mean height.
   b) Calculate the standard deviation.

Solutions

1. a) Means
   \[ \bar{x}_A = \frac{9+10+11+7+13}{5} = 10 \] \[ \bar{x}_B = \frac{10+10+10+10+10}{5} = 10 \] \[ \bar{x}_C = \frac{1+1+10+19+19}{5} = 10 \] b) Standard deviations
   \[ \sigma_A = \sqrt{\frac{(9-10)^2+(10-10)^2+(11-10)^2+(7-10)^2+(13-10)^2}{5}} = 2 \] \[ \sigma_B = 0 \] \[ \sigma_C = \sqrt{\frac{(1-10)^2+(1-10)^2+(10-10)^2+(19-10)^2+(19-10)^2}{5}} \approx 8.05 \] c) Data set \( C \) has the largest standard deviation.
   d) Yes. The values in set \( C \) are farther from the mean.
2. Let the data values be \( x,y,z \). \[ \mu = \frac{x+y+z}{3}, \quad \sigma = \sqrt{\frac{(x-\mu)^2+(y-\mu)^2+(z-\mu)^2}{3}} \] a) Adding \( k \): \[ \mu' = \mu + k, \quad \sigma' = \sigma \] b) Multiplying by \( k \): \[ \mu' = k\mu, \quad \sigma' = |k|\sigma \]
3. If \( \sigma = 0 \), then \[ (x-\mu)^2+(y-\mu)^2+(z-\mu)^2+(w-\mu)^2 = 0 \] which implies \[ x=y=z=w=\mu \] All data values are equal.
4. \[ \mu = \frac{\sum x_i f_i}{\sum f_i} = \frac{3500(5)+4000(8)+4200(5)+4300(2)}{20} = 3955 \] \[ \sigma \approx 282 \]
5. Using class midpoints \( m_i \): \[ \mu = \frac{\sum m_i f_i}{\sum f_i} = 148.4 \] \[ \sigma \approx 9.9 \]

More References