From Łukasz Graczykowski
(Difference between revisions)
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| Output: | | Output: |
- | P(5<X<10) z gestosci: 0.245723 | + | P(5<X<10) from the density: 0.245723 |
- | P(5<X<10) z dystrybuanty: 0.245723 | + | P(5<X<10) from the cumulative distribution: 0.245723 |
- | Srednia z funkcji TF1: 5.98798 | + | Mean from the function TF1: 5.98798 |
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- | Srednia analitycznie: 6 | + | Mean analytically: 6 |
- | Wariancja z funkcji TF1: 35.3988 | + | Variance from TF1: 35.3988 |
- | Wariancja analitycznie: 36 | + | Variance analytically: 36 |
- | Moda: 0 | + | Mode: 0 |
| x0.25: 1.74 | | x0.25: 1.74 |
| x0.5: 4.16 | | x0.5: 4.16 |
| x0.75: 8.32 | | x0.75: 8.32 |
Latest revision as of 12:14, 7 March 2022
Zadanie
Let X be a random variable that has a probability distribution function f(x):
Please do:
- Draw the probability distribution function f(x) (0.5 pkt)
- Draw the cumulative distribution F(x). Cumulative distribution has to be drawn numerically and analytically. (1 pkt)
- Calculate the probability P(5 <= X <= 10) (0.5 pkt)
- Calculate the expected value E(X) and compare with the analytic result (1 pkt)
- Calculate the variance V(X) and compare with the analytic result(1 pkt)
- Calculate mode (0.5 pkt)
- Calculate quantiles: (0.5 pkt)
- lower quartile
- median
- upper quartile
Results
Plots:
Output:
P(5<X<10) from the density: 0.245723
P(5<X<10) from the cumulative distribution: 0.245723
Mean from the function TF1: 5.98798
Mean analytically: 6
Variance from TF1: 35.3988
Variance analytically: 36
Mode: 0
x0.25: 1.74
x0.5: 4.16
x0.75: 8.32