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  1. Prepare
  2. Tutorials
  3. 10 Days of Statistics
  4. Day 6: The Central Limit Theorem I
  5. Discussions

Day 6: The Central Limit Theorem I

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71 Discussions

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  • bemof25579
     + 0 comments

    R: cat(round(pnorm(9800, mean=49*205, sd=sqrt(49)*15),4),fill=TRUE)

  • blackspring
     + 0 comments
    #python
import math
def cdf(m, std, x):
    return 0.5 * (1 + math.erf((x-m)/(std*2**.5)))
n, m0, std0, x=49, 205, 15, 9800
m=n*m0
std=math.sqrt(n)*std0
print(round(cdf(m, std, x), 4))

  • lucasopoka
     + 1 comment
    import math as m
def cdf(mn, std, x):
    return 1/2 * (1 + m.erf((x-mn)/(std*m.sqrt(2))))
print(f'{cdf(205*49, 15*m.sqrt(49), 9800):.4f}')

  • asilichenko
     + 0 comments
    double maxLimit = 9800;
int n = 49;

double mean = 205;
double stdDev = 15;

double meanX = n * mean;
double stdDevX = Math.sqrt(n) * stdDev;

double p = probability(meanX, stdDevX, maxLimit);

    Explanation

    If you need to calculate whether the sum of random values fit the limit then you need:

    • multiply the mean in number of random values;
    • multiply standard deviation in the square root of number of random values;
    • the probability is calculated in the same way as in the previous problem.

  • shashinp1996
     + 0 comments

    JS

       function erf(x) {
        var a1 = 0.254829592;
        var a2 = -0.284496736;
        var a3 = 1.421413741;
        var a4 = -1.453152027;
        var a5 = 1.061405429;
        var p = 0.3275911;
        var sign = 1;
        if (x < 0) {
            sign = -1;
        }
        x = Math.abs(x);
        var t = 1.0 / (1.0 + p * x);
        var y = 1.0 - (((((a5 * t + a4) * t) + a3) * t + a2) * t + a1) * t * Math.exp(-x * x);
        return sign * y;
    }
    let arr = input.split(/[\n]/g);
    let h = parseFloat(arr[0]);
    let b = parseFloat(arr[1]);
    let c = parseFloat(arr[2]);
    let std = Math.sqrt(b) * parseFloat(arr[3]);
    let res = 0.5 * (1+ erf((h - (b * c)) / (std * Math.sqrt(2))));
    console.log(res.toFixed(4));