Question

Dist. Pdf CDF M ?² uniform \frac{1}{B-A} A?x?B \frac{x-A}{B-A} M = \frac{A+B}{2} ?² = \frac{(B-A)²}{12} Normal \frac{1}{?\sqrt{2?}}e^{-\frac{(x-?)²}{2?²}} ????? M = np ?² = npq Gamma \frac{1}{???(?)}x^{?-1}e^{-\frac{x}{?}} x>0 ?? ??² Exp \frac{1}{?}e^{-\frac{x}{?}} 1-e^{-\frac{x}{?}} ? ?² Chi \frac{1}{2^{\frac{?}{2}}?(\frac{?}{2})}x^{\frac{?}{2}-1}e^{-\frac{x}{2}} x>0 ????? ? 2? New New

          Dist. Pdf CDF M ?²
uniform \frac{1}{B-A} A?x?B \frac{x-A}{B-A} M = \frac{A+B}{2} ?² = \frac{(B-A)²}{12}
Normal \frac{1}{?\sqrt{2?}}e^{-\frac{(x-?)²}{2?²}} ????? M = np ?² = npq
Gamma \frac{1}{???(?)}x^{?-1}e^{-\frac{x}{?}} x>0 ?? ??²
Exp \frac{1}{?}e^{-\frac{x}{?}} 1-e^{-\frac{x}{?}} ? ?²
Chi \frac{1}{2^{\frac{?}{2}}?(\frac{?}{2})}x^{\frac{?}{2}-1}e^{-\frac{x}{2}} x>0 ????? ? 2?
New New

Dist. Pdf CDF M ?²
uniform (1)/(B-A) A?x?B (x-A)/(B-A) M = (A+B)/(2) ?² = ((B-A)²)/(12)
Normal (1)/(?√(2?))e^-((x-?)²)/(2?²) ????? M = np ?² = npq
Gamma (1)/(???(?))x^?-1e^-(x)/(?) x>0 ?? ??²
Exp (1)/(?)e^-(x)/(?) 1-e^-(x)/(?) ? ?²
Chi (1)/(2^(?)/(2)?((?)/(2)))x^(?)/(2)-1e^-(x)/(2) x>0 ????? ? 2?
New New

Added by Benjamin C.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

Instant Answer

Solved by Expert Aishwarya Krishnakumar

06/11/2023

Step 1

Step 1: Normal Distribution (pdf) The probability density function (pdf) of a normal distribution is given by: f(x) = (1/(σ√(2π))) * exp(-(x-μ)^2/(2σ^2)) where μ is the mean and σ is the standard deviation. Show more…

Show all steps

Thanks for your feedback!

For the normal-uniform-gamma-exp distributions, I want the (pdf-cdf-mean-variance). Dis e. Pdf JaJ uniform X - A B-A MA+B 2 6B-A 12 B- A A<X<B Normal M-np np - Gamma xe Ba(x) X>0 2 ExP B 2V New Chi 2T() q<x New