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  1. Prepare
  2. Tutorials
  3. 10 Days of Statistics
  4. Day 1: Standard Deviation
  5. Discussions

Day 1: Standard Deviation

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426 Discussions

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  • shiva_keshavan
     + 0 comments

    Considering that finite sequences of inputs typically are samples from a population, the correct answer should be the implementation of the sample standard deviation,

    i.e. s = Sqrt{ Sum [((x_i-mu)^2)/(N-1)]}.

  • akulbachandra67
     + 0 comments

    import statistics as st def stdDev(arr): m = st.mean(arr) s = [ (i-m)**2 for i in arr ] return print(f"{(sum(s)/len(arr))**0.5:.1f}") `

  • illogicalsleepi1
     + 0 comments
    def stdDev(arr):
    N=len(arr)
    mean=sum(arr)/N
    s=0
    for i in arr:
        s+=(mean-i)**2
    print(round((s/N)**0.5,1))

  • [deleted]     + 0 comments
    def stdDev(arr):
    # Print your answers to 1 decimal place within this function
    s = sum(arr)
    l = len(arr)
    mean = s/l
    
    a=[]
    
    for i in range(l):
        ans = abs(arr[i]-mean)
        new_ans = ans*ans
        a.append(new_ans)
    
    new_sum = sum(a)
    standard = new_sum / l
    answer = math.sqrt(standard)
    print(f"{answer:.1f}")

  • FedeHide
     + 0 comments

    Javascript

    function stdDev(arr) {
    const sum = arr.reduce((acc, cur) => acc + cur, 0);
    const mean = sum / arr.length;
    console.log(Math.sqrt([...arr].reduce((acc, curr) => acc + Math.pow(curr - mean, 2), 0) / arr.length).toFixed(1))
}