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Notes

Practice Problems

\(\textbf{1)}\) At a local store, on average, 6.2 customers buy carrots per hour. What is \(\lambda\)?

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\(\lambda = 6.2\)

\(\textbf{2)}\) At a local store, on average, 6.2 customers buy carrots per hour. What is the probability exactly 0 customers buy carrots during any hour?

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\(P(x=0)\approx 2.02\%\)

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\(\,\,\,\,\,P(x)=\displaystyle\frac{\lambda^{x}e^{-\lambda}}{x!}\)

\(\,\,\,\,\,P(x=0)=\displaystyle\frac{(6.2)^{0}e^{-(6.2)}}{0!}\approx 0.00202=\)

\(\,\,\,\,\,P(x=0)\approx 2.02\%\)

\(\textbf{3)}\) At a local store, on average, 6.2 customers buy carrots per hour. What is the probability exactly 1 customers buy carrots during any hour?

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\(P(x=1)\approx 1.26\%\)

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\(\,\,\,\,\,P(x)=\displaystyle\frac{\lambda^{x}e^{-\lambda}}{x!}\)

\(\,\,\,\,\,P(x=1)=\displaystyle\frac{(6.2)^{1}e^{-(6.2)}}{1!}\approx 0.01258\)

\(\,\,\,\,\,P(x=1)\approx 1.26\%\)

\(\textbf{4)}\) At a local store, on average, 6.2 customers buy carrots per hour. What is the probability exactly 4 customers buy carrots during any hour?

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\(P(x=4)\approx 12.49\%\)

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\(\,\,\,\,\,P(x)=\displaystyle\frac{\lambda^{x}e^{-\lambda}}{x!}\)

\(\,\,\,\,\,P(x=4)=\displaystyle\frac{(6.2)^{4}e^{-(6.2)}}{4!}\approx 0.12494\)

\(\,\,\,\,\,P(x=4)\approx 12.49\%\)

\(\textbf{5)}\) At a local store, on average, 6.2 customers buy carrots per hour. What is the probability of 5 or fewer people buying carrots during any hour?

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\(P(x\le5)\approx 41.41\%\)

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\(\,\,\,\,\,P(x)=\displaystyle\frac{\lambda^{x}e^{-\lambda}}{x!}\)

\(\,\,\,\,\,P(x=0)=\displaystyle\frac{(6.2)^{0}e^{-(6.2)}}{0!}\approx 0.00202=\)

\(\,\,\,\,\,P(x=1)=\displaystyle\frac{(6.2)^{1}e^{-(6.2)}}{1!}\approx 0.01258\)

\(\,\,\,\,\,P(x=2)=\displaystyle\frac{(6.2)^{2}e^{-(6.2)}}{2!}\approx 0.03900\)

\(\,\,\,\,\,P(x=3)=\displaystyle\frac{(6.2)^{3}e^{-(6.2)}}{3!}\approx 0.08061\)

\(\,\,\,\,\,P(x=4)=\displaystyle\frac{(6.2)^{4}e^{-(6.2)}}{4!}\approx 0.12494\)

\(\,\,\,\,\,P(x=5)=\displaystyle\frac{(6.2)^{5}e^{-(6.2)}}{5!}\approx 0.15493\)

\(\,\,\,\,\,P(x\le5)=P(x=0)+P(x=1)+P(x=2)+P(x=3)+P(x=4)+P(x=5)\)

\(\,\,\,\,\,P(x\le5)=0.00202+0.01258+0.03900+0.08061+0.12494+0.15493=0.41408\)

\(\,\,\,\,\,P(x\le5)\approx 41.41\%\)

\(\textbf{6)}\) At a local store, on average, 6.2 customers buy carrots per hour. What is the probability of at least 1 person buying carrots during any hour?

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\(P(x\ge1)=99.80\%\)

Show Work

\(\,\,\,\,\,P(x)=\displaystyle\frac{\lambda^{x}e^{-\lambda}}{x!}\)

\(\,\,\,\,\,P(x=0)=\displaystyle\frac{(6.2)^{0}e^{-(6.2)}}{0!}\approx 0.00202=\)

\(\,\,\,\,\,P(x\ge1)=1-P(x=0)\)

\(\,\,\,\,\,P(x\ge1)=1-0.00202=0.99798\)

\(\,\,\,\,\,P(x\ge1)=99.80\%\)

See Related Pages\(\)

\(\bullet\text{ Poisson Distribution Calculator }\)
\(\,\,\,\,\,\,\,\,\text{(Stattrek.com)}\)

\(\bullet\text{ Statistics Homepage}\)
\(\,\,\,\,\,\,\,\,\text{All the Best Topics…}\)

\(\bullet\text{ Uniform Distribution}\)
\(\,\,\,\,\,\,\,\,p(x)=\frac{1}{b-a}…\)

\(\bullet\text{ Binomial Distribution}\)
\(\,\,\,\,\,\,\,\,p(r)={}_{n}C_{r}(p)^r(1-p)^{n-r}…\)

\(\bullet\text{ Geometric Distribution}\)
\(\,\,\,\,\,\,\,\,P(X=n)=p(1-p)^{n-1}…\)

\(\bullet\text{ Continuity Correction}\)
\(\,\,\,\,\,\,\,\,c-.5\lt x\lt c+.5…\)

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