Mean and Standard deviation
Problems with Solutions

Mean and standard deviation problems are presented. Problems related to data sets as well as grouped data are discussed. The solutions to these problems are at the bottom of the page.

  1. Consider the following three data sets A, B and C.

    A = {9,10,11,7,13}

    B = {10,10,10,10,10} Find

    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 set has the largest standard deviation?

    d) Is it possible to answer question c) without calculations of the standard deviation?



  2. A given data set has a mean μ and a standard deviation σ.

    a) What are the new values of the mean and the standard deviation if the same constant k is added to each data value in the given set?Explain.

    b) What are the new values of the mean and the standard deviation if each data value of the set is multiplied by the same constant k?Explain.



  3. If the standard deviation of a given data set is equal to zero, what can we say about the data values included in the given data set?



  4. The frequency table of the monthly slaries of 20 people is shown below.

    salary(in $) frequency
    3500 5
    4000 8
    4200 5
    4300 2


    a) Calculate the mean of the salaries of the 20 people.

    b) Calculate the standard deviation of the salaries of the 20 people.



  5. The following table shows the grouped data, in classes, for the heights of 50 people.

    height (in cm) - classes frequency
    120 <- 130 2
    130 <- 140 5
    140 <- 150 25
    150 <- 160 10
    160 <- 170 8


    a) Calculate the mean of the salaries of the 20 people.

    b) Calculate the standard deviation of the salaries of the 20 people.



Solutions to the Above Problems
    1. mean of Data set A = (9+10+11+7+13)/5 = 10

      mean of Data set B = (10+10+10+10+10)/5 = 10

      mean of Data set C = (1+1+10+19+19)/5 = 10


    2. Standard Deviation Data set A

      = √[ ( (9-10)2+(10-10)2+(11-10)2+(7-10)2+(13-10)2 )/5 ] = 2

      Standard Deviation Data set B

      = √[ ( (10-10)2+(10-10)2+(10-10)2+(10-10)2+(10-10)2 )/5 ] = 0

      Standard Deviation Data set C

      = √[ ( (1-10)2+(1-10)2+(10-10)2+(19-10)2+(19-10)2 )/5 ] = 8.05

    3. Data set C has the largest standard deviation.

    4. Yes, since data Set C has data values that are further away from the mean compared to sets A and B.
    1. We limit the discusion to a data set with 3 values for simplicity, but the conclusions are true for any data set with quantitative data.

      Let x, y and z be the data values making a data set.

      The mean μ = (x + y + z) / 3

      The standard deviation σ = √[ ((x - μ)2 + (y - μ)2 + (z - μ)2)/3 ]

      We now add a constant k to each data value and calculate the new mean μ'.

      μ' = ((x + k) + (y + k) + (z + k)) / 3 = (x + y + z) / 3 + 3k/3 = μ + k

      We now calculate the new mean standard deviation σ'.

      σ' = √[ ((x + k - μ')2 +(y + k - μ')2+(z + k - μ')2)/3 ]

      Note that x + k - μ' = x + k - μ - k = x - μ

      also y + k - μ' = y + k - μ - k = y - μ and z + k - μ' = z + k - μ - k = z - μ

      Therefore σ' = √[ ((x - μ)2 +(y - μ)2+(z - μ)2)/3 ] = σ

      If we add the same constant k to all data values included in a data set, we obtain a new data set whose mean is the mean of the original data set PLUS k. The standard deviation does not change.

    2. We now multiply all data values by a constant k and calculate the new mean μ' and the new standard deviation σ'.

      μ' = (kx + ky + kz) / 3 = kμ

      σ' = √[ ((kx - kμ)2 +(ky - kμ)2+(kz - kμ)2)/3 ] = |k| σ

      If we multiply all data values included in a data set by a constant k, we obtain a new data set whose mean is the mean of the original data set TIMES k and standard deviation is the standard deviation of the original data set TIMES the absolute value of k.
    1. Again, we limit the discussion to a data set with 4 values for simplicity, but the conclusions are true for any data set with quantitative data.

      Let x, y, z and w be the data values making a data set with mean μ.

      The standard deviation σ = √[ ((x - μ)2 + (y - μ)2 + (z - μ)2 + (w - μ)2)/3 ]

      Let σ = 0, hence

      √[ ((x - μ)2 + (y - μ)2 + (z - μ)2 + (w - μ)2)/3 ] = 0

      Which gives

      (x - μ)2 + (y - μ)2 + (z - μ)2 + (w - μ)2 = 0

      All terms in the equation are positive and therefore, the above equation is equivalent to

      (x - μ)2 = 0, (y - μ)2 = 0, (z - μ)2 = 0 and (w - μ)2 = 0.

      Which gives

      x = y = z = w = μ : all data values in the set with σ = 0 are equal.


    1. Let xi be the i th salary and fi be the corresponding frequency.

      mean of grouped data = μ = (Σxi*fi) / Σfi

      = (3500*5 + 4000*8 + 4200*5 + 4300*2) /(5 + 8 + 5 + 2)

      = $3955

      b) standard deviation of grouped data = √[ (Σ(xi-μ)2*fi) / Σfi ]

      = √[ (5*(3500-3955)2+8*(4000-3955)2+5*(4200-3955)2+2*(4300-3955)2) /(20) ]

      = 282 (rounded to the nearest unit)
    1. We first find the midpoints of the given classes.

      height (in cm) - classes midpoint frequency
      120 <- 130 125 2
      130 <- 140 135 5
      140 <- 150 145 25
      150 <- 160 155 10
      160 <- 170 165 8


      Let mi be the midpoint of the i th clss and fi be the corresponding frequency.

      mean of grouped data = μ = (Σmi*fi) / Σfi

      = (125*2 + 135*5 + 145*25 + 155*10 + 165*8) /(2+5+25+10+8)

      = 148.4

      b) standard deviation of grouped data = √[ (Σ(mi-μ)2*fi) / Σfi ]

      = √[ (2*(125-148.4)2+5*(135-148.4)2+25*(145-148.4)2+10*(155-148.4)2+8*(165-148.4)2) /(50) ]

      = 9.9



More references on
elementary statistics and probabilities.