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6.5: Distribución de Bernoulli

  • Page ID
    151557
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    Ahora exploraremos variables aleatorias específicas que se utilizan frecuentemente en la práctica. Estas variables aleatorias serán generalizadas por parámetros. Comenzaremos con la más simple de todas las variables aleatorias, la Distribución de Bernoulli, también conocida como la variable indicadora. Esta variable aleatoria,\(X\), está diseñada para una pregunta de sí/no o de éxito/fracaso. Si la respuesta es Sí/Éxito, entonces\(X = 1\). Si la respuesta es No/Fracaso, entonces\(X = 0\). La probabilidad de éxito es\(p\), y la probabilidad de fracaso es\(q = 1‐p\).

    \(X\) \(P(X)\)
    \ (X\)” class="lt-estados-20891">0 \ (P (X)\)” class="lt-stats-20891">\(q = 1‐p\)
    \ (X\)” class="lt-estados-20891">1 \ (P (X)\)” class="lt-stats-20891">\(p\)

    Ejemplo: Tiro libre

    clipboard_e46bdb5f4ab72b564be9ce94cb2119dbb.png

    Draymond Green 66, jugador de basquetbol de la NBA para los Golden State Warriors, es un tirador de tiros libres del 70%. Esto significa que cuando dispara un tiro libre, hay un 70% de probabilidad de que haga el tiro. La variable aleatoria\(X\) = el número de éxitos cuando Draymond Green toma un tiro libre sigue a una Distribución de Bernoulli con\(p =0.7\) (éxito) y\(q = 0.3\) (fracaso). Determinar el pdf, media y varianza de la variable aleatoria.

    \(x\) \(P(x)\) \(x \cdot P(x)\) \(x-\mu\) \((x-\mu)^{2}\) \((x-\mu)^{2} \cdot P(x)\)
    \ (x\) ">0 \ (P (x)\) ">0.30 \ (x\ cdot P (x)\) ">0.00 \ (x-\ mu\) ">‐0.70 \ ((x-\ mu) ^ {2}\) ">0.49 \ ((x-\ mu) ^ {2}\ cdot P (x)\) ">0.147
    \ (x\) ">1 \ (P (x)\) ">0.70 \ (x\ cdot P (x)\) ">0.70 \ (x-\ mu\) ">0.30 \ ((x-\ mu) ^ {2}\) ">0.09 \ ((x-\ mu) ^ {2}\ cdot P (x)\) ">0.063
    \ (x\) "> Total \ (P (x)\) "> 1 \ (x\ cdot P (x)\) "> \(0.7=\mu\) \ (x-\ mu\) "> \ ((x-\ mu) ^ {2}\) "> \ ((x-\ mu) ^ {2}\ cdot P (x)\) "> \(0.21=\sigma^{2}\)

    Solución

    La media y varianza se pueden calcular directamente para la Variable Aleatoria de Bernoulli.

    \(x\) \(P(x)\) \(x \cdot P(x)\) \(x-\mu\) \((x-\mu)^{2}\) \((x-\mu)^{2} \cdot P(x)\)
    \ (x\) ">0 \ (P (x)\) ">\(1‐p\) \ (x\ cdot P (x)\) ">0 \ (x-\ mu\) ">\(‐p\) \ ((x-\ mu) ^ {2}\) ">\(p^{2}\) \ ((x-\ mu) ^ {2}\ cdot P (x)\) ">\((1-p) p^{2}\)
    \ (x\) ">1 \ (P (x)\) ">\(p\) \ (x\ cdot P (x)\) ">\(p\) \ (x-\ mu\) ">\(1‐p\) \ ((x-\ mu) ^ {2}\) ">\((1-p)^{2}\) \ ((x-\ mu) ^ {2}\ cdot P (x)\) ">\(P(1-p)^{2}\)
    \ (x\) "> Total \ (P (x)\) "> 1 \ (x\ cdot P (x)\) "> \(\mu = p\) \ (x-\ mu\) "> \ ((x-\ mu) ^ {2}\) "> \ ((x-\ mu) ^ {2}\ cdot P (x)\) "> \(\sigma^{2}=p(1-p)=p q\)

    Para el ejemplo de Draymond Green,\(\mu=p=0.7\) y\(\sigma^{2}=p q=(0.7)(0.3)=.21\), que coincide con la respuesta cuando se calcula manualmente

    Distribución de probabilidad de Bernoulli (parámetro =\(p\))

    Un ensayo, dos posibles resultados (Éxito/Fracaso) o (Sí/No)

    \(\mathbf{P} = P\)(sí/éxito)

    \(q=1-p=P\)(no/falla)

    X = Número de Sí/Éxitos {0, 1}

    \(\mu=p\)

    \(\sigma^{2}=p(1-p)=p q\)


    This page titled 6.5: Distribución de Bernoulli is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Maurice A. Geraghty via source content that was edited to the style and standards of the LibreTexts platform.